Q2: For the reaction: 2 ~A+B → A_2 ~B the rate =k[~A][B]^2 with k=2.0 × 10^-6 ~mol^-2 ~L^2 ~s^-1. Calculate the initial rate of the reaction when [A]=0.1…

Part 1: Calculation of the initial rate Given: Rate law: Rate =k[A][B]^2 Rate constant, k = 2.0 × 10^-6 ~mol^-2 ~L^2 ~s^-1 Initial concentration of A, [A] = 0.1 ~mol ~L^-1 Initial concentration of B, [B] = 0.2 ~mol ~L^-1 Formula: Rate = k[A][B]^2 Calculation: Substitute the given values into the…

Let the reactant be 'A' and the reaction be second order with respect to A. The rate law for this reaction can be written as: Rate = k[A]^2 where k is the rate constant and [A] is the concentration of the reactant A. Let the initial rate be R_1 when the concentration is [A]. R_1 = k[A]^2 (i) If the…

Q6 (ii): reduced to half?

Let the reactant be 'A' and the reaction be second order with respect to A. The rate law for this reaction can be written as: Rate = k[A]^2 where k is the rate constant and [A] is the concentration of the reactant A. Let the initial rate be R_1 when the concentration is [A]. R_1 = k[A]^2 (i) If the…

Chemical Kinetics Class 12 Chemistry NCERT Solutions

Q1Exercises

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i)

$3 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ Rate $=k[\mathrm{NO}]^{2}$

(ii)

$\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{H}^{+} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{I}_{3}^{-}$ Rate $=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[\mathrm{I}^{-}\right]$

(iii)

$\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$ Rate $=k\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2}$

(iv)

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})$ Rate $=k\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right]$

Solution

The order of a reaction is the sum of the powers of the concentration terms in the rate law expression. The dimensions (units) of the rate constant

k

for a reaction of order

n

are given by

(\text{concentration})^{1-n} \text{time}^{-1}

. Assuming concentration is in

\text{mol L}^{-1}

and time is in

\text{s}

.

(i) $3 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ , Rate $=k[\mathrm{NO}]^{2}$

Order of reaction: The power of the concentration term $[\mathrm{NO}]$ is 2. So, the order of the reaction is 2.
Dimensions of rate constant: For a second-order reaction ( $n=2$ ), the units of $k$ are $(\text{mol L}^{-1})^{1-2} \text{s}^{-1} = \text{mol}^{-1} \text{L s}^{-1}$ .

(ii) $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{H}^{+} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{I}_{3}^{-}$ , Rate $=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[\mathrm{I}^{-}\right]$

Order of reaction: The power of $[\mathrm{H}_{2} \mathrm{O}_{2}]$ is 1 and the power of $[\mathrm{I}^{-}]$ is 1. The overall order is $1 + 1 = 2$ .
Dimensions of rate constant: For a second-order reaction ( $n=2$ ), the units of $k$ are $(\text{mol L}^{-1})^{1-2} \text{s}^{-1} = \text{mol}^{-1} \text{L s}^{-1}$ .

(iii) $\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$ , Rate $=k\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2}$

Order of reaction: The power of the concentration term $[\mathrm{CH}_{3} \mathrm{CHO}]$ is $3/2$ . So, the order of the reaction is $3/2$ .
Dimensions of rate constant: For a reaction of order $n=3/2$ , the units of $k$ are $(\text{mol L}^{-1})^{1-3/2} \text{s}^{-1} = (\text{mol L}^{-1})^{-1/2} \text{s}^{-1} = \text{mol}^{-1/2} \text{L}^{1/2} \text{s}^{-1}$ .

(iv) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})$ , Rate $=k\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right]$

Order of reaction: The power of the concentration term $[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}]$ is 1. So, the order of the reaction is 1.
Dimensions of rate constant: For a first-order reaction ( $n=1$ ), the units of $k$ are $(\text{mol L}^{-1})^{1-1} \text{s}^{-1} = \text{s}^{-1}$ .

Q2Exercises

For the reaction: $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{A}_{2} \mathrm{~B}$ the rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$ with $\mathrm{k}=2.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~s}^{-1}$ . Calculate the initial rate of the reaction when $[\mathrm{A}]=0.1 \mathrm{~mol} \mathrm{~L}^{-1},[\mathrm{B}]=0.2 \mathrm{~mol} \mathrm{~L}^{-1}$ . Calculate the rate of reaction after $[\mathrm{A}]$ is reduced to $0.06 \mathrm{~mol} \mathrm{~L}^{-1}$ .

Solution

Part 1: Calculation of the initial rate

Given:

Rate law: Rate $=k[\mathrm{A}][\mathrm{B}]^{2}$
Rate constant, $k = 2.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~s}^{-1}$
Initial concentration of A, $[\mathrm{A}] = 0.1 \mathrm{~mol} \mathrm{~L}^{-1}$
Initial concentration of B, $[\mathrm{B}] = 0.2 \mathrm{~mol} \mathrm{~L}^{-1}$

Formula:

\text{Rate} = k[\mathrm{A}][\mathrm{B}]^{2}

Calculation: Substitute the given values into the rate law equation:

\text{Rate} = (2.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~s}^{-1}) \times (0.1 \mathrm{~mol} \mathrm{~L}^{-1}) \times (0.2 \mathrm{~mol} \mathrm{~L}^{-1})^{2}

\text{Rate} = (2.0 \times 10^{-6}) \times (0.1) \times (0.04) \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

\text{Rate} = 8.0 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

Final Answer (Initial Rate): The initial rate of the reaction is

8.0 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

.

Part 2: Calculation of the rate after [A] is reduced

Given:

The new concentration of A is $0.06 \mathrm{~mol} \mathrm{~L}^{-1}$ .

Calculation of new concentrations:

Calculate the amount of A that has reacted: Change in $[\mathrm{A}] = \text{Initial } [\mathrm{A}] - \text{Final } [\mathrm{A}] = 0.1 - 0.06 = 0.04 \mathrm{~mol} \mathrm{~L}^{-1}$ .
Calculate the amount of B that has reacted: From the stoichiometry of the reaction $2 \mathrm{A}+\mathrm{B} \rightarrow \mathrm{A}_{2} \mathrm{~B}$ , 2 moles of A react with 1 mole of B. So, the concentration of B reacted is half the concentration of A reacted. Concentration of B reacted $= \frac{1}{2} \times (\text{Concentration of A reacted}) = \frac{1}{2} \times 0.04 = 0.02 \mathrm{~mol} \mathrm{~L}^{-1}$ .
Calculate the new concentration of B: New $[\mathrm{B}] = \text{Initial } [\mathrm{B}] - (\text{Concentration of B reacted}) = 0.2 - 0.02 = 0.18 \mathrm{~mol} \mathrm{~L}^{-1}$ .

Calculation of the new rate: Now use the new concentrations in the rate law:

\text{New Rate} = k[\mathrm{A}]_{\text{new}}[\mathrm{B}]_{\text{new}}^{2}

\text{New Rate} = (2.0 \times 10^{-6} \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{~s}^{-1}) \times (0.06 \mathrm{~mol} \mathrm{~L}^{-1}) \times (0.18 \mathrm{~mol} \mathrm{~L}^{-1})^{2}

\text{New Rate} = (2.0 \times 10^{-6}) \times (0.06) \times (0.0324) \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

\text{New Rate} = 3.888 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

Final Answer (New Rate): The rate of reaction after

[\mathrm{A}]

is reduced to

0.06 \mathrm{~mol} \mathrm{~L}^{-1}

is

3.89 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

.

Q3Exercises

The decomposition of $\mathrm{NH}_{3}$ on platinum surface is zero order reaction. What are the rates of production of $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$ if $k=2.5 \times 10^{-4} \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~s}^{-1}$ ?

Solution

First, it is important to note that the units for the rate constant

k

for a zero-order reaction should be

\mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}

. The units given in the question (

"mol^{-1} L s^{-1}"

) correspond to a second-order reaction. We will assume the correct units for a zero-order reaction, i.e.,

k=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

.

Given:

Reaction: Decomposition of $\mathrm{NH}_{3}$ on a platinum surface.
Order of reaction: Zero order.
Rate constant, $k = 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ .

Balanced Chemical Equation:

2 \mathrm{NH}_{3}(\mathrm{g}) \xrightarrow{\text{Pt}} \mathrm{N}_{2}(\mathrm{g}) + 3 \mathrm{H}_{2}(\mathrm{g})

Rate Expression: For a zero-order reaction, the rate is independent of the concentration of the reactant.

\text{Rate} = k[\mathrm{NH}_{3}]^{0} = k

So, the rate of reaction is equal to the rate constant.

\text{Rate} = 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

The rate of reaction can also be expressed in terms of the rate of disappearance of reactants and the rate of appearance of products:

\text{Rate} = -\frac{1}{2} \frac{d[\mathrm{NH}_{3}]}{dt} = +\frac{d[\mathrm{N}_{2}]}{dt} = +\frac{1}{3} \frac{d[\mathrm{H}_{2}]}{dt}

where

\frac{d[\mathrm{N}_{2}]}{dt}

is the rate of production of

\mathrm{N}_{2}

and

\frac{d[\mathrm{H}_{2}]}{dt}

is the rate of production of

\mathrm{H}_{2}

.

Calculation of Rate of Production of $\mathrm{N}_{2}$ : From the rate expression:

\text{Rate of production of } \mathrm{N}_{2} = \frac{d[\mathrm{N}_{2}]}{dt} = \text{Rate of reaction}

\frac{d[\mathrm{N}_{2}]}{dt} = 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

Calculation of Rate of Production of $\mathrm{H}_{2}$ : From the rate expression:

\text{Rate of reaction} = \frac{1}{3} \frac{d[\mathrm{H}_{2}]}{dt}

\text{Rate of production of } \mathrm{H}_{2} = \frac{d[\mathrm{H}_{2}]}{dt} = 3 \times (\text{Rate of reaction})

\frac{d[\mathrm{H}_{2}]}{dt} = 3 \times (2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1})

\frac{d[\mathrm{H}_{2}]}{dt} = 7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

Final Answer:

The rate of production of $\mathrm{N}_{2}$ is $2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ .
The rate of production of $\mathrm{H}_{2}$ is $7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ .

Q4Exercises

The decomposition of dimethyl ether leads to the formation of $\mathrm{CH}_{4}, \mathrm{H}_{2}$ and $\mathrm{CO}$ and the reaction rate is given by $\text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{OCH}_{3}\right]^{3 / 2}$ The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., $\text { Rate }=k\left(p_{\mathrm{CH}_{3} \mathrm{OCH}_{3}}\right)^{3 / 2}$ If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

Solution

1. Units of Rate of Reaction

The rate of reaction is defined as the change in concentration (or pressure for gases) of a reactant or product per unit time.

In this case, the rate is expressed in terms of the change in pressure over time.

Pressure is measured in bar.
Time is measured in minutes.

Therefore, the units of the rate of reaction are bar min⁻¹.

2. Units of Rate Constant (k)

The rate law is given as:

\text{Rate} = k (p_{\mathrm{CH}_{3}OCH_{3}})^{3/2}

To find the units of the rate constant,

k

, we can rearrange the equation:

k = \frac{\text{Rate}}{(p_{\mathrm{CH}_{3}OCH_{3}})^{3/2}}

Now, substitute the units for Rate and Pressure into the rearranged equation:

Units of Rate = $\text{bar min}^{-1}$
Units of Pressure ( $p$ ) = $\text{bar}$

\text{Units of } k = \frac{\text{bar min}^{-1}}{(\text{bar})^{3/2}}

Using the laws of exponents (

a^m / a^n = a^{m-n}

):

\text{Units of } k = \text{bar}^{(1 - 3/2)} \text{min}^{-1}

\text{Units of } k = \text{bar}^{(-1/2)} \text{min}^{-1}

So, the units of the rate constant are bar⁻¹/² min⁻¹.

Final Answer:

The units of the rate of reaction are bar min⁻¹.
The units of the rate constant are bar⁻¹/² min⁻¹.

Q5Exercises

Mention the factors that affect the rate of a chemical reaction.

Solution

The rate of a chemical reaction is influenced by several factors. The primary factors are:

Concentration of Reactants: Generally, the rate of reaction increases with an increase in the concentration of the reactants. According to the collision theory, a higher concentration leads to more frequent collisions between reactant molecules per unit time, which increases the probability of effective collisions.
Temperature: The rate of reaction increases significantly with an increase in temperature. A rise in temperature increases the kinetic energy of the reactant molecules. This leads to two effects: molecules collide more frequently, and more importantly, a larger fraction of molecules possess energy equal to or greater than the activation energy, resulting in a higher number of effective collisions.
Presence of a Catalyst: A catalyst is a substance that increases the rate of a reaction without being consumed in the process. It provides an alternative reaction pathway with a lower activation energy. A lower activation energy means that a larger fraction of reactant molecules can overcome the energy barrier, thus increasing the reaction rate.
Nature of Reactants: The physical state, chemical bonding, and complexity of the reactant molecules affect the reaction rate. For example, reactions involving ionic species in aqueous solutions are usually very fast, while reactions involving the breaking of strong covalent bonds in large molecules are often slower.
Surface Area of Reactants: For reactions involving reactants in different phases (heterogeneous reactions), such as a solid reacting with a liquid or gas, the rate of reaction increases with an increase in the surface area of the solid reactant. A larger surface area provides more sites for the reaction to occur.
Presence of Light (Radiation): Some reactions, known as photochemical reactions, are initiated by the absorption of light of a suitable wavelength. The rate of such reactions depends on the intensity of the radiation. For example, the reaction between hydrogen and chlorine is very slow in the dark but occurs explosively in sunlight.

Q6Exercises

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

(i)

doubled

(ii)

reduced to half ?

Solution

Let the reactant be 'A' and the reaction be second order with respect to A. The rate law for this reaction can be written as:

\text{Rate} = k[\mathrm{A}]^2

where

k

is the rate constant and

[\mathrm{A}]

is the concentration of the reactant A.

Let the initial rate be

R_1

when the concentration is

[\mathrm{A}]

.

R_1 = k[\mathrm{A}]^2

(i) If the concentration of the reactant is doubled

The new concentration,

[\mathrm{A}']

, will be

2[\mathrm{A}]

. The new rate,

R_2

, will be:

R_2 = k[\mathrm{A}']^2 = k(2[\mathrm{A}])^2

R_2 = k(4[\mathrm{A}]^2) = 4 \times (k[\mathrm{A}]^2)

Since

R_1 = k[\mathrm{A}]^2

, we have:

R_2 = 4 \times R_1

Answer: When the concentration of the reactant is doubled, the rate of reaction increases by a factor of 4 (it becomes four times the original rate).

(ii) If the concentration of the reactant is reduced to half

The new concentration,

[\mathrm{A}'']

, will be

\frac{1}{2}[\mathrm{A}]

. The new rate,

R_3

, will be:

R_3 = k[\mathrm{A}'']^2 = k\left(\frac{1}{2}[\mathrm{A}]\right)^2

R_3 = k\left(\frac{1}{4}[\mathrm{A}]^2\right) = \frac{1}{4} \times (k[\mathrm{A}]^2)

Since

R_1 = k[\mathrm{A}]^2

, we have:

R_3 = \frac{1}{4} \times R_1

Answer: When the concentration of the reactant is reduced to half, the rate of reaction becomes one-fourth of the original rate.

Q7Exercises

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Solution

Effect of Temperature on the Rate Constant:

The rate constant of a chemical reaction is highly dependent on temperature. For the vast majority of chemical reactions, the value of the rate constant, and therefore the rate of reaction, increases as the temperature increases. A common rule of thumb for many reactions is that the rate constant approximately doubles for every 10°C (or 10 K) rise in temperature. This increase is due to the fact that at higher temperatures, reactant molecules have higher kinetic energies. This leads to more frequent collisions and, more importantly, a significantly larger fraction of molecules possessing the minimum energy required for a reaction to occur, known as the activation energy (

E_a

).

Quantitative Representation:

The effect of temperature on the rate constant can be represented quantitatively by the Arrhenius equation, proposed by Svante Arrhenius. The equation is:

k = A e^{-E_a / RT}

Where:

$k$ is the rate constant.
$A$ is the Arrhenius factor or the pre-exponential factor. It is a constant for a given reaction and is related to the frequency of collisions between reactant molecules.
$E_a$ is the activation energy of the reaction, which is the minimum energy required for the reaction to occur (in J mol⁻¹).
$R$ is the universal gas constant ( $8.314 \text{ J K}^{-1} \text{mol}^{-1}$ ).
$T$ is the absolute temperature in Kelvin (K).

The Arrhenius equation can also be expressed in a logarithmic form by taking the natural logarithm of both sides:

\ln k = \ln A - \frac{E_a}{RT}

This equation is in the form of a straight line (

y = c + mx

), where a plot of

\ln k

(y-axis) versus

1/T

(x-axis) gives a straight line with a slope of

-E_a/R

and a y-intercept of

\ln A

. This graphical method allows for the experimental determination of activation energy and the pre-exponential factor.

For comparing rate constants (

k_1

and

k_2

) at two different temperatures (

T_1

and

T_2

), the following form of the Arrhenius equation is used:

\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]

Q8Exercises

In a pseudo first order reaction in water, the following results were obtained:
$\mathrm{t} / \mathrm{s}$ 0 30 60 90
$[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}$ 0.55 0.31 0.17 0.085
Calculate the average rate of reaction between the time interval 30 to 60 seconds.

$\mathrm{t} / \mathrm{s}$	0	30	60	90
$[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}$	0.55	0.31	0.17	0.085

Solution

Given:

At time $t_1 = 30 \text{ s}$ , the concentration of reactant A, $[\mathrm{A}]_1 = 0.31 \mathrm{~mol} \mathrm{~L}^{-1}$ .
At time $t_2 = 60 \text{ s}$ , the concentration of reactant A, $[\mathrm{A}]_2 = 0.17 \mathrm{~mol} \mathrm{~L}^{-1}$ .

To Find: The average rate of reaction between the time interval 30 s to 60 s.

Formula: The average rate of a reaction is defined as the change in concentration of a reactant or product divided by the time interval over which the change occurs. For a reactant A, the average rate (

r_{av}

) is given by:

r_{av} = -\frac{\Delta[\mathrm{A}]}{\Delta t} = -\frac{[\mathrm{A}]_2 - [\mathrm{A}]_1}{t_2 - t_1}

The negative sign is used because the concentration of a reactant decreases with time, and the rate of reaction is always a positive quantity.

Calculation: Substitute the given values into the formula:

r_{av} = -\frac{(0.17 \mathrm{~mol} \mathrm{~L}^{-1}) - (0.31 \mathrm{~mol} \mathrm{~L}^{-1})}{60 \mathrm{~s} - 30 \mathrm{~s}}

r_{av} = -\frac{-0.14 \mathrm{~mol} \mathrm{~L}^{-1}}{30 \mathrm{~s}}

r_{av} = \frac{0.14}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

r_{av} = 0.004666... \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

r_{av} = 4.67 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

Final Answer: The average rate of reaction between the time interval 30 to 60 seconds is

4.67 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

.

Q9Exercises

A reaction is first order in A and second order in B.

(i)

Write the differential rate equation.

(ii)

How is the rate affected on increasing the concentration of B three times?

(iii)

How is the rate affected when the concentrations of both A and B are doubled?

Solution

(i) Write the differential rate equation.

Given that the reaction is first order in reactant A and second order in reactant B, the rate of the reaction is proportional to the first power of the concentration of A and the second power of the concentration of B.

The differential rate equation (or rate law) is:

\text{Rate} = k[\mathrm{A}]^1 [\mathrm{B}]^2

or simply

\text{Rate} = k[\mathrm{A}][\mathrm{B}]^2

where

k

is the rate constant.

(ii) How is the rate affected on increasing the concentration of B three times?

Let the initial rate be

R_1

with concentrations

[\mathrm{A}]

and

[\mathrm{B}]

:

R_1 = k[\mathrm{A}][\mathrm{B}]^2

Now, the concentration of B is increased three times, so the new concentration is

[\mathrm{B}'] = 3[\mathrm{B}]

. The concentration of A remains unchanged. The new rate,

R_2

, will be:

R_2 = k[\mathrm{A}][\mathrm{B}']^2 = k[\mathrm{A}](3[\mathrm{B}])^2

R_2 = k[\mathrm{A}](9[\mathrm{B}]^2) = 9 \times (k[\mathrm{A}][\mathrm{B}]^2)

Since

R_1 = k[\mathrm{A}][\mathrm{B}]^2

,

R_2 = 9 \times R_1

Answer: The rate of the reaction will increase by a factor of 9 (it will become 9 times the original rate).

(iii) How is the rate affected when the concentrations of both A and B are doubled?

Let the initial rate be

R_1

with concentrations

[\mathrm{A}]

and

[\mathrm{B}]

:

R_1 = k[\mathrm{A}][\mathrm{B}]^2

Now, the concentrations of both A and B are doubled. The new concentrations are

[\mathrm{A}'] = 2[\mathrm{A}]

and

[\mathrm{B}'] = 2[\mathrm{B}]

. The new rate,

R_3

, will be:

R_3 = k[\mathrm{A}'][\mathrm{B}']^2 = k(2[\mathrm{A}])(2[\mathrm{B}])^2

R_3 = k(2[\mathrm{A}])(4[\mathrm{B}]^2) = (2 \times 4) \times (k[\mathrm{A}][\mathrm{B}]^2)

R_3 = 8 \times (k[\mathrm{A}][\mathrm{B}]^2)

Since

R_1 = k[\mathrm{A}][\mathrm{B}]^2

,

R_3 = 8 \times R_1

Answer: The rate of the reaction will increase by a factor of 8 (it will become 8 times the original rate).

Q10Exercises

In a reaction between A and B, the initial rate of reaction ( $r_0$ ) was measured for different initial concentrations of A and B as given below:
$\mathrm{A} / \mathrm{mol} \mathrm{L}^{-1}$ 0.20 0.20 0.40
$\mathrm{B} / \mathrm{mol} \mathrm{L}^{-1}$ 0.30 0.10 0.05
$\mathrm{r}_{0} / \mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}$ $5.07 \times 10^{-5}$ $5.07 \times 10^{-5}$ $1.43 \times 10^{-4}$
What is the order of the reaction with respect to A and B?

$\mathrm{A} / \mathrm{mol} \mathrm{L}^{-1}$	0.20	0.20	0.40
$\mathrm{B} / \mathrm{mol} \mathrm{L}^{-1}$	0.30	0.10	0.05
$\mathrm{r}_{0} / \mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}$	$5.07 \times 10^{-5}$	$5.07 \times 10^{-5}$	$1.43 \times 10^{-4}$

Solution

Let the rate law for the reaction be:

\text{Rate} = k[\mathrm{A}]^x [\mathrm{B}]^y

where

x

is the order with respect to A and

y

is the order with respect to B.

We will use the data from the experiments to determine the values of

x

and

y

.

Step 1: Determine the order with respect to B (y)

Compare Experiment 1 and Experiment 2. In these two experiments, the concentration of A is kept constant (

[\mathrm{A}] = 0.20 \mathrm{~mol} \mathrm{~L}^{-1}

), while the concentration of B is changed.

Rate in Exp 1 ( $r_1$ ) = $k(0.20)^x (0.30)^y = 5.07 \times 10^{-5}$
Rate in Exp 2 ( $r_2$ ) = $k(0.20)^x (0.10)^y = 5.07 \times 10^{-5}$

Divide the rate equation of Exp 1 by that of Exp 2:

\frac{r_1}{r_2} = \frac{k(0.20)^x (0.30)^y}{k(0.20)^x (0.10)^y} = \frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}

\left(\frac{0.30}{0.10}\right)^y = 1

3^y = 1

Any number raised to the power of 0 is 1. Therefore,

y=0

.

The reaction is zero order with respect to B.

Step 2: Determine the order with respect to A (x)

Now we know the rate law is Rate

= k[\mathrm{A}]^x

. We can compare Experiment 1 and Experiment 3 to find

x

. (Note: We must include the B term even if its order is zero, for completeness).

Rate in Exp 1 ( $r_1$ ) = $k(0.20)^x (0.30)^0 = 5.07 \times 10^{-5}$
Rate in Exp 3 ( $r_3$ ) = $k(0.40)^x (0.05)^0 = 1.43 \times 10^{-4}$

Divide the rate equation of Exp 3 by that of Exp 1:

\frac{r_3}{r_1} = \frac{k(0.40)^x (0.05)^0}{k(0.20)^x (0.30)^0} = \frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}

\left(\frac{0.40}{0.20}\right)^x = 2.82

2^x = 2.82

To solve for

x

, we can take the logarithm of both sides:

x \log(2) = \log(2.82)

x = \frac{\log(2.82)}{\log(2)} = \frac{0.450}{0.301} \approx 1.5

The reaction is of order 1.5 with respect to A.

Final Answer:

The order of the reaction with respect to A is 1.5.
The order of the reaction with respect to B is 0.

Q11Exercises

The following results have been obtained during the kinetic studies of the reaction: $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}$
Experiment $[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}$ $[\mathrm{B}] / \mathrm{mol} \mathrm{L}^{-1}$ Initial rate of formation of $\mathrm{D} / \mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}$
I 0.1 0.1 $6.0 \times 10^{-3}$
II 0.3 0.2 $7.2 \times 10^{-2}$
III 0.3 0.4 $2.88 \times 10^{-1}$
IV 0.4 0.1 $2.40 \times 10^{-2}$
Determine the rate law and the rate constant for the reaction.

Experiment	$[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}$	$[\mathrm{B}] / \mathrm{mol} \mathrm{L}^{-1}$	Initial rate of formation of $\mathrm{D} / \mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}$
I	0.1	0.1	$6.0 \times 10^{-3}$
II	0.3	0.2	$7.2 \times 10^{-2}$
III	0.3	0.4	$2.88 \times 10^{-1}$
IV	0.4	0.1	$2.40 \times 10^{-2}$

Solution

Let the rate law for the reaction be:

\text{Rate} = k[\mathrm{A}]^x [\mathrm{B}]^y

where

x

is the order with respect to A and

y

is the order with respect to B.

Step 1: Determine the order with respect to B (y)

Compare Experiment II and Experiment III. In these experiments, the concentration of A is kept constant (

[\mathrm{A}] = 0.3 \mathrm{~mol} \mathrm{~L}^{-1}

), while the concentration of B is doubled (from 0.2 to 0.4

\mathrm{~mol} \mathrm{~L}^{-1}

).

Rate in Exp II ( $r_{II}$ ) = $k(0.3)^x (0.2)^y = 7.2 \times 10^{-2}$
Rate in Exp III ( $r_{III}$ ) = $k(0.3)^x (0.4)^y = 2.88 \times 10^{-1}$

Divide the rate equation of Exp III by that of Exp II:

\frac{r_{III}}{r_{II}} = \frac{k(0.3)^x (0.4)^y}{k(0.3)^x (0.2)^y} = \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}}

\left(\frac{0.4}{0.2}\right)^y = 4

2^y = 4

Therefore,

y=2

. The reaction is second order with respect to B.

Step 2: Determine the order with respect to A (x)

Compare Experiment I and Experiment IV. In these experiments, the concentration of B is kept constant (

[\mathrm{B}] = 0.1 \mathrm{~mol} \mathrm{~L}^{-1}

), while the concentration of A is quadrupled (from 0.1 to 0.4

\mathrm{~mol} \mathrm{~L}^{-1}

).

Rate in Exp I ( $r_I$ ) = $k(0.1)^x (0.1)^y = 6.0 \times 10^{-3}$
Rate in Exp IV ( $r_{IV}$ ) = $k(0.4)^x (0.1)^y = 2.40 \times 10^{-2}$

Divide the rate equation of Exp IV by that of Exp I:

\frac{r_{IV}}{r_I} = \frac{k(0.4)^x (0.1)^y}{k(0.1)^x (0.1)^y} = \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}}

\left(\frac{0.4}{0.1}\right)^x = 4

4^x = 4

Therefore,

x=1

. The reaction is first order with respect to A.

Step 3: Write the Rate Law

Based on the values of

x

and

y

found, the rate law is:

\text{Rate} = k[\mathrm{A}]^1 [\mathrm{B}]^2 \quad \text{or} \quad \text{Rate} = k[\mathrm{A}][\mathrm{B}]^2

Step 4: Calculate the Rate Constant (k)

We can use the data from any of the experiments to calculate

k

. Let's use Experiment I.

\text{Rate} = k[\mathrm{A}][\mathrm{B}]^2

6.0 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{min}^{-1} = k (0.1 \mathrm{~mol} \mathrm{~L}^{-1}) (0.1 \mathrm{~mol} \mathrm{~L}^{-1})^2

6.0 \times 10^{-3} = k (0.1) (0.01) = k (0.001)

k = \frac{6.0 \times 10^{-3}}{0.001} = 6.0

The units of

k

for a third-order reaction (

n=1+2=3

) are

(\text{mol L}^{-1})^{1-3} \text{min}^{-1} = \text{mol}^{-2} \text{L}^2 \text{min}^{-1}

.

Final Answer:

The rate law for the reaction is Rate $= k[\mathrm{A}][\mathrm{B}]^2$ .
The rate constant for the reaction is $k = 6.0 \mathrm{~mol}^{-2} \mathrm{~L}^{2} \mathrm{min}^{-1}$ .

Q12Exercises

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
Experiment $[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}$ $[\mathrm{B}] / \mathrm{mol} \mathrm{L}^{-1}$ Initial rate/ $\mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}$
I 0.1 0.1 $2.0 \times 10^{-2}$
II - 0.2 $4.0 \times 10^{-2}$
III 0.4 0.4 -
IV - 0.2 $2.0 \times 10^{-2}$

Solution

Step 1: Determine the rate law and calculate the rate constant (k)

Given that the reaction is first order with respect to A and zero order with respect to B, the rate law is:

\text{Rate} = k[\mathrm{A}]^1 [\mathrm{B}]^0 = k[\mathrm{A}]

We can use the data from Experiment I to find the value of

k

.

2.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{min}^{-1} = k (0.1 \mathrm{~mol} \mathrm{~L}^{-1})

k = \frac{2.0 \times 10^{-2}}{0.1} \mathrm{min}^{-1} = 0.2 \mathrm{min}^{-1}

Step 2: Fill in the blank for Experiment II

We need to find

[\mathrm{A}]

for Experiment II. Using the rate law: Rate

= k[\mathrm{A}]

4.0 \times 10^{-2} = (0.2) \times [\mathrm{A}]

[\mathrm{A}] = \frac{4.0 \times 10^{-2}}{0.2} = 0.2 \mathrm{~mol} \mathrm{~L}^{-1}

Step 3: Fill in the blank for Experiment III

We need to find the initial rate for Experiment III. Using the rate law: Rate

= k[\mathrm{A}]

\text{Rate} = (0.2) \times (0.4)

\text{Rate} = 0.08 = 8.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{min}^{-1}

Step 4: Fill in the blank for Experiment IV

We need to find

[\mathrm{A}]

for Experiment IV. Using the rate law: Rate

= k[\mathrm{A}]

2.0 \times 10^{-2} = (0.2) \times [\mathrm{A}]

[\mathrm{A}] = \frac{2.0 \times 10^{-2}}{0.2} = 0.1 \mathrm{~mol} \mathrm{~L}^{-1}

Completed Table:

Experiment	$[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}$	$[\mathrm{B}] / \mathrm{mol} \mathrm{L}^{-1}$	Initial rate/ $\mathrm{mol} \mathrm{L}^{-1} \mathrm{min}^{-1}$
I	0.1	0.1	$2.0 \times 10^{-2}$
II	0.2	0.2	$4.0 \times 10^{-2}$
III	0.4	0.4	$8.0 \times 10^{-2}$
IV	0.1	0.2	$2.0 \times 10^{-2}$

Q13Exercises

Calculate the half-life of a first order reaction from their rate constants given below:

(i)

$200 \mathrm{~s}^{-1}$

(ii)

$2 \mathrm{~min}^{-1}$

(iii)

4 years $^{-1}$

Solution

The half-life (

t_{1/2}

) of a first-order reaction is related to its rate constant (

k

) by the following formula:

t_{1/2} = \frac{0.693}{k}

This formula shows that for a first-order reaction, the half-life is independent of the initial concentration of the reactant.

(i) $k = 200 \mathrm{~s}^{-1}$

Calculation:

t_{1/2} = \frac{0.693}{200 \mathrm{~s}^{-1}}

t_{1/2} = 0.003465 \mathrm{~s}

t_{1/2} = 3.465 \times 10^{-3} \mathrm{~s}

Final Answer: The half-life is

3.465 \times 10^{-3} \mathrm{~s}

.

(ii) $k = 2 \mathrm{~min}^{-1}$

Calculation:

t_{1/2} = \frac{0.693}{2 \mathrm{~min}^{-1}}

t_{1/2} = 0.3465 \mathrm{~min}

Final Answer: The half-life is

0.3465 \mathrm{~min}

.

(iii) $k = 4 \mathrm{~years}^{-1}$

Calculation:

t_{1/2} = \frac{0.693}{4 \mathrm{~years}^{-1}}

t_{1/2} = 0.17325 \mathrm{~years}

Final Answer: The half-life is

0.17325 \mathrm{~years}

.

Q14Exercises

The half-life for radioactive decay of $^{14} \mathrm{C}$ is 5730 years. An archaeological artifact containing wood had only $80 \%$ of the $^{14} \mathrm{C}$ found in a living tree. Estimate the age of the sample.

Solution

Radioactive decay follows first-order kinetics.

Given:

Half-life of $^{14} \mathrm{C}$ , $t_{1/2} = 5730$ years.
The amount of $^{14} \mathrm{C}$ in the artifact is 80% of the amount in a living tree. Let the initial amount of $^{14} \mathrm{C}$ (in a living tree) be $[R]_0$ . Then the amount of $^{14} \mathrm{C}$ remaining in the artifact, $[R]$ , is $0.80 [R]_0$ .

Step 1: Calculate the rate constant (k)

For a first-order reaction, the rate constant is related to the half-life by:

k = \frac{0.693}{t_{1/2}}

Calculation:

k = \frac{0.693}{5730 \text{ years}} = 1.209 \times 10^{-4} \text{ years}^{-1}

Step 2: Calculate the age of the sample (t)

The integrated rate law for a first-order reaction is:

t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}

Calculation: Substitute the values of

k

,

[R]_0

, and

[R]

into the equation:

t = \frac{2.303}{1.209 \times 10^{-4} \text{ years}^{-1}} \log \frac{[R]_0}{0.80 [R]_0}

t = \frac{2.303}{1.209 \times 10^{-4}} \log \frac{1}{0.80}

t = \frac{2.303}{1.209 \times 10^{-4}} \log(1.25)

Using the value

\log(1.25) = 0.0969

:

t = (19049 \text{ years}) \times 0.0969

t = 1845.8 \text{ years}

Final Answer: The estimated age of the sample is approximately 1846 years.

Q15Exercises

The experimental data for decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}$ $\left[2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\right]$ in gas phase at 318 K are given below:
$t / \mathrm{s}$ 0 400 800 1200 1600 2000 2400 2800 3200
$10^{2} \times\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] / \mathrm{mol} \mathrm{L}^{-1}$ 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35

$t / \mathrm{s}$	0	400	800	1200	1600	2000	2400	2800	3200
$10^{2} \times\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] / \mathrm{mol} \mathrm{L}^{-1}$	1.63	1.36	1.14	0.93	0.78	0.64	0.53	0.43	0.35

(i)

Plot $\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$ against $t$ .

(ii)

Find the half-life period for the reaction.

(iii)

Draw a graph between $\log \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$ and $t$ .

(iv)

What is the rate law ?

(v)

Calculate the rate constant.

(vi)

Calculate the half-life period from $k$ and compare it with (ii).

Solution

(i) Plot of $[\mathrm{N}_{2} \mathrm{O}_{5}]$ against $t$

A plot of

[\mathrm{N}_{2} \mathrm{O}_{5}]

(y-axis) versus time

t

(x-axis) would show a curve where the concentration decreases exponentially with time. It would not be a straight line.

(ii) Find the half-life period for the reaction

The initial concentration

[\mathrm{N}_{2} \mathrm{O}_{5}]_0

at

t=0

is

1.63 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

. The half-life (

t_{1/2}

) is the time taken for the concentration to reduce to half of its initial value, which is:

\frac{[\mathrm{N}_{2} \mathrm{O}_{5}]_0}{2} = \frac{1.63 \times 10^{-2}}{2} = 0.815 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

Looking at the data table, this concentration value lies between

t = 1200 \text{ s}

(

[\mathrm{N}_{2} \mathrm{O}_{5}] = 0.93 \times 10^{-2}

) and

t = 1600 \text{ s}

(

[\mathrm{N}_{2} \mathrm{O}_{5}] = 0.78 \times 10^{-2}

). By interpolation, the half-life is approximately 1440-1450 s.

(iii) Draw a graph between $\log [\mathrm{N}_{2} \mathrm{O}_{5}]$ and $t$

First, we calculate the values of

\log [\mathrm{N}_{2} \mathrm{O}_{5}]

:

$t / \mathrm{s}$	$[\mathrm{N}_{2} \mathrm{O}_{5}] / \mathrm{mol} \mathrm{L}^{-1}$	$\log [\mathrm{N}_{2} \mathrm{O}_{5}]$
0	$1.63 \times 10^{-2}$	-1.788
400	$1.36 \times 10^{-2}$	-1.866
800	$1.14 \times 10^{-2}$	-1.943
1200	$0.93 \times 10^{-2}$	-2.031
1600	$0.78 \times 10^{-2}$	-2.108
2000	$0.64 \times 10^{-2}$	-2.194
2400	$0.53 \times 10^{-2}$	-2.276
2800	$0.43 \times 10^{-2}$	-2.367
3200	$0.35 \times 10^{-2}$	-2.456

A graph of

\log [\mathrm{N}_{2} \mathrm{O}_{5}]

(y-axis) versus

t

(x-axis) will yield a straight line with a negative slope. This indicates that the reaction is a first-order reaction.

(iv) What is the rate law?

Since the plot of

\log [\mathrm{N}_{2} \mathrm{O}_{5}]

vs

t

is a straight line, the reaction is first order with respect to

\mathrm{N}_{2} \mathrm{O}_{5}

. Therefore, the rate law is:

\text{Rate} = k[\mathrm{N}_{2} \mathrm{O}_{5}]

(v) Calculate the rate constant

For a first-order reaction, the slope of the

\log [\mathrm{N}_{2} \mathrm{O}_{5}]

vs

t

graph is equal to

-k/2.303

. Let's calculate the slope using the first and last data points:

\text{Slope} = \frac{\Delta(\log [\mathrm{N}_{2} \mathrm{O}_{5}])}{\Delta t} = \frac{(-2.456) - (-1.788)}{3200 - 0} = \frac{-0.668}{3200} = -2.0875 \times 10^{-4} \mathrm{~s}^{-1}

Now, we find

k

:

\text{Slope} = -\frac{k}{2.303} \implies k = -2.303 \times \text{Slope}

k = -2.303 \times (-2.0875 \times 10^{-4} \mathrm{~s}^{-1}) = 4.807 \times 10^{-4} \mathrm{~s}^{-1}

(vi) Calculate the half-life period from $k$ and compare it with (ii)

Using the formula for the half-life of a first-order reaction:

t_{1/2} = \frac{0.693}{k}

t_{1/2} = \frac{0.693}{4.807 \times 10^{-4} \mathrm{~s}^{-1}} = 1441.6 \mathrm{~s}

This calculated value of

t_{1/2} \approx 1442 \text{ s}

is in excellent agreement with the value estimated graphically in part (ii).

Q16Exercises

The rate constant for a first order reaction is $60 \mathrm{~s}^{-1}$ . How much time will it take to reduce the initial concentration of the reactant to its $1 / 16^{\text {th }}$ value?

Solution

Method 1: Using the Integrated Rate Law

Given:

The reaction is first order.
Rate constant, $k = 60 \mathrm{~s}^{-1}$ .
The final concentration, $[R]$ , is $1/16$ of the initial concentration, $[R]_0$ . So, $[R] = \frac{1}{16}[R]_0$ .

Formula: The integrated rate law for a first-order reaction is:

t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}

Calculation: Substitute the given values into the equation:

t = \frac{2.303}{60 \mathrm{~s}^{-1}} \log \frac{[R]_0}{(1/16)[R]_0}

t = \frac{2.303}{60} \log(16)

We know that

\log(16) = \log(2^4) = 4 \log(2) = 4 \times 0.3010 = 1.204

.

t = \frac{2.303}{60} \times 1.204 \mathrm{~s}

t = 0.03838 \times 1.204 \mathrm{~s}

t = 0.0462 \mathrm{~s}

Method 2: Using the Concept of Half-Lives

The reduction of the concentration to

1/16

of its initial value can be seen as a series of half-life periods.

\frac{1}{16} = \frac{1}{2^4} = \left(\frac{1}{2}\right)^4

This means that the time required for the concentration to fall to

1/16

of its initial value is equal to 4 half-lives (

4 \times t_{1/2}

).

Step 1: Calculate the half-life ( $t_{1/2}$ )

t_{1/2} = \frac{0.693}{k} = \frac{0.693}{60 \mathrm{~s}^{-1}} = 0.01155 \mathrm{~s}

Step 2: Calculate the total time (t)

t = 4 \times t_{1/2} = 4 \times 0.01155 \mathrm{~s} = 0.0462 \mathrm{~s}

Both methods give the same result.

Final Answer: It will take 0.0462 s to reduce the initial concentration of the reactant to its

1/16^{\text{th}}

value.

Q17Exercises

During nuclear explosion, one of the products is $^{90} \mathrm{Sr}$ with half-life of 28.1 years. If $1 \mu \mathrm{g}$ of $^{90} \mathrm{Sr}$ was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.

Solution

Radioactive decay, such as that of

^{90} \mathrm{Sr}

, follows first-order kinetics.

Given:

Half-life of $^{90} \mathrm{Sr}$ , $t_{1/2} = 28.1$ years.
Initial amount of $^{90} \mathrm{Sr}$ , $N_0 = 1 \mu \mathrm{g}$ .

Step 1: Calculate the decay constant (k)

For a first-order process, the decay constant is related to the half-life by:

k = \frac{0.693}{t_{1/2}}

Calculation:

k = \frac{0.693}{28.1 \text{ years}} = 0.02466 \text{ years}^{-1}

Step 2: Use the integrated rate law to find the remaining amount (N) after a certain time (t)

The integrated rate law for first-order decay is:

t = \frac{2.303}{k} \log \frac{N_0}{N} \quad \text{or} \quad \log \frac{N_0}{N} = \frac{kt}{2.303}

Case 1: Amount remaining after 10 years ( $t = 10$ years)

\log \frac{1 \mu \mathrm{g}}{N} = \frac{(0.02466 \text{ years}^{-1}) \times (10 \text{ years})}{2.303}

\log \frac{1}{N} = \frac{0.2466}{2.303} = 0.10708

To find

\frac{1}{N}

, we take the antilogarithm:

\frac{1}{N} = 10^{0.10708} = 1.2796

N = \frac{1}{1.2796} \mu \mathrm{g} = 0.7815 \mu \mathrm{g}

Case 2: Amount remaining after 60 years ( $t = 60$ years)

\log \frac{1 \mu \mathrm{g}}{N} = \frac{(0.02466 \text{ years}^{-1}) \times (60 \text{ years})}{2.303}

\log \frac{1}{N} = \frac{1.4796}{2.303} = 0.6424

To find

\frac{1}{N}

, we take the antilogarithm:

\frac{1}{N} = 10^{0.6424} = 4.389

N = \frac{1}{4.389} \mu \mathrm{g} = 0.2278 \mu \mathrm{g}

Final Answer:

After 10 years, $0.782 \mu \mathrm{g}$ of $^{90} \mathrm{Sr}$ will remain.
After 60 years, $0.228 \mu \mathrm{g}$ of $^{90} \mathrm{Sr}$ will remain.

Q18Exercises

For a first order reaction, show that time required for $99 \%$ completion is twice the time required for the completion of $90 \%$ of reaction.

Solution

For a first-order reaction, the integrated rate law is given by:

t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}

where:

$t$ is the time
$k$ is the rate constant
$[R]_0$ is the initial concentration of the reactant
$[R]$ is the concentration of the reactant at time $t$

Step 1: Calculate the time required for 90% completion ( $t_{90\%}$ )

For 90% completion, the amount of reactant that has reacted is 90% of the initial amount. The concentration of reactant remaining,

[R]

, is:

[R] = [R]_0 - 0.90[R]_0 = 0.10[R]_0

So,

\frac{[R]_0}{[R]} = \frac{[R]_0}{0.10[R]_0} = 10

.

Substitute this into the integrated rate law:

t_{90\%} = \frac{2.303}{k} \log(10)

Since

\log(10) = 1

:

t_{90\%} = \frac{2.303}{k} \times 1 = \frac{2.303}{k} \quad \cdots(1)

Step 2: Calculate the time required for 99% completion ( $t_{99\%}$ )

For 99% completion, the amount of reactant that has reacted is 99% of the initial amount. The concentration of reactant remaining,

[R]

, is:

[R] = [R]_0 - 0.99[R]_0 = 0.01[R]_0

So,

\frac{[R]_0}{[R]} = \frac{[R]_0}{0.01[R]_0} = 100

.

Substitute this into the integrated rate law:

t_{99\%} = \frac{2.303}{k} \log(100)

Since

\log(100) = \log(10^2) = 2

:

t_{99\%} = \frac{2.303}{k} \times 2 = 2 \left( \frac{2.303}{k} \right) \quad \cdots(2)

Step 3: Compare $t_{99\%}$ and $t_{90\%}$

Now, divide equation (2) by equation (1):

\frac{t_{99\%}}{t_{90\%}} = \frac{2 \left( \frac{2.303}{k} \right)}{\left( \frac{2.303}{k} \right)} = 2

Therefore:

t_{99\%} = 2 \times t_{90\%}

This shows that for a first-order reaction, the time required for 99% completion is twice the time required for 90% completion.

Q19Exercises

A first order reaction takes 40 min for $30 \%$ decomposition. Calculate $t_{1 / 2}$ .

Solution

Given:

The reaction is first order.
Time, $t = 40$ min.
The reaction is 30% decomposed. This means 30% of the reactant has been consumed.

Step 1: Calculate the rate constant (k)

If the reaction is 30% decomposed, the amount of reactant remaining is

100\% - 30\% = 70\%

of the initial amount. Let the initial concentration be

[R]_0

. The concentration at time

t

,

[R]

, is

0.70 [R]_0

. Therefore, the ratio

\frac{[R]_0}{[R]} = \frac{[R]_0}{0.70 [R]_0} = \frac{1}{0.70}

.

The integrated rate law for a first-order reaction is:

k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}

Calculation of k:

k = \frac{2.303}{40 \text{ min}} \log \left( \frac{1}{0.70} \right)

k = \frac{2.303}{40} \log(1.4286)

Using

\log(1.4286) \approx 0.1549

:

k = \frac{2.303}{40} \times 0.1549 \text{ min}^{-1}

k = 0.057575 \times 0.1549 \text{ min}^{-1}

k = 0.008918 \text{ min}^{-1}

Step 2: Calculate the half-life ( $t_{1/2}$ )

The half-life of a first-order reaction is related to the rate constant by:

t_{1/2} = \frac{0.693}{k}

Calculation of $t_{1/2}$ :

t_{1/2} = \frac{0.693}{0.008918 \text{ min}^{-1}}

t_{1/2} = 77.71 \text{ min}

Final Answer: The half-life (

t_{1/2}

) of the reaction is 77.7 min.

Q20Exercises

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
$t(\mathrm{sec})$ $\mathrm{P}(\mathrm{mm} \text { of } \mathrm{Hg})$
0 35.0
360 54.0
720 63.0
Calculate the rate constant.

$t(\mathrm{sec})$	$\mathrm{P}(\mathrm{mm} \text { of } \mathrm{Hg})$
0	35.0
360	54.0
720	63.0

Solution

The decomposition of azoisopropane is a first-order gas-phase reaction. The chemical equation is:

(\mathrm{CH}_{3})_{2} \mathrm{CHN}=\mathrm{NCH}(\mathrm{CH}_{3})_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g}) + \mathrm{C}_{6} \mathrm{H}_{14}(\mathrm{g})

Let the reactant azoisopropane be A. The reaction is of the type:

\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g})

.

Let

p_i

be the initial pressure of A, and

p_t

be the total pressure at time

t

. The partial pressure of A at time

t

,

p_A

, is related to

p_i

and

p_t

by the expression:

p_A = 2p_i - p_t

.

The integrated rate law for a first-order gas-phase reaction is:

k = \frac{2.303}{t} \log \frac{p_i}{p_A} = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}

Given:

Initial pressure, $p_i = 35.0 \text{ mm of Hg}$ (at $t=0$ ).

Calculation of k at $t = 360$ s:

Total pressure, $p_t = 54.0 \text{ mm of Hg}$ . $k_1 = \frac{2.303}{360} \log \frac{35.0}{2(35.0) - 54.0}$ $k_1 = \frac{2.303}{360} \log \frac{35.0}{70.0 - 54.0} = \frac{2.303}{360} \log \frac{35.0}{16.0}$ $k_1 = \frac{2.303}{360} \log(2.1875) = (0.006397) \times (0.340) = 2.175 \times 10^{-3} \mathrm{s}^{-1}$

Calculation of k at $t = 720$ s:

Total pressure, $p_t = 63.0 \text{ mm of Hg}$ . $k_2 = \frac{2.303}{720} \log \frac{35.0}{2(35.0) - 63.0}$ $k_2 = \frac{2.303}{720} \log \frac{35.0}{70.0 - 63.0} = \frac{2.303}{720} \log \frac{35.0}{7.0}$ $k_2 = \frac{2.303}{720} \log(5) = (0.003198) \times (0.699) = 2.235 \times 10^{-3} \mathrm{s}^{-1}$

The values of the rate constant calculated at different times are very close, which confirms that the reaction is first order.

Average Rate Constant: To get a more accurate value, we can average the calculated rate constants:

k_{avg} = \frac{k_1 + k_2}{2} = \frac{(2.175 + 2.235) \times 10^{-3}}{2} \mathrm{s}^{-1}

k_{avg} = \frac{4.410 \times 10^{-3}}{2} \mathrm{s}^{-1} = 2.205 \times 10^{-3} \mathrm{s}^{-1}

Final Answer: The rate constant for the reaction is approximately

2.21 \times 10^{-3} \mathrm{~s}^{-1}

.

Q21Exercises

The following data were obtained during the first order thermal decomposition of $\mathrm{SO}_{2} \mathrm{Cl}_{2}$ at a constant volume. $\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$
Experiment Time $/ \mathrm{s}^{-1}$ Total pressure/atm
1 0 0.5
2 100 0.6
Calculate the rate of the reaction when total pressure is 0.65 atm. (Note: The unit for time in the table should be 's', not 's⁻¹').

Experiment	Time $/ \mathrm{s}^{-1}$	Total pressure/atm
1	0	0.5
2	100	0.6

Solution

The thermal decomposition of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

is a first-order gas-phase reaction. The reaction is of the type:

\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g})

.

Step 1: Calculate the rate constant (k)

Let

p_i

be the initial pressure of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

, and

p_t

be the total pressure at time

t

. The partial pressure of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

at time

t

,

p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}}

, is related to

p_i

and

p_t

by the expression:

p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}} = 2p_i - p_t

.

The integrated rate law for this type of first-order reaction is:

k = \frac{2.303}{t} \log \frac{p_i}{p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}}} = \frac{2.303}{t} \log \frac{p_i}{2p_i - p_t}

Given:

Initial pressure, $p_i = 0.5$ atm (at $t=0$ ).
At $t = 100$ s, total pressure, $p_t = 0.6$ atm.

Calculation of k:

k = \frac{2.303}{100 \text{ s}} \log \frac{0.5}{2(0.5) - 0.6}

k = \frac{2.303}{100} \log \frac{0.5}{1.0 - 0.6} = \frac{2.303}{100} \log \frac{0.5}{0.4}

k = 0.02303 \times \log(1.25) = 0.02303 \times 0.0969 = 2.23 \times 10^{-3} \mathrm{s}^{-1}

Step 2: Calculate the rate of reaction when total pressure is 0.65 atm

The rate law for this reaction in terms of pressure is:

\text{Rate} = k \times p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}}

First, we need to find the partial pressure of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

when the total pressure is 0.65 atm. Let

x

be the decrease in pressure of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

.

At time t: $p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}} = p_i - x = 0.5 - x$
At time t: $p_{\mathrm{SO}_{2}} = x$ and $p_{\mathrm{Cl}_{2}} = x$

Total pressure

p_t = p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}} + p_{\mathrm{SO}_{2}} + p_{\mathrm{Cl}_{2}} = (0.5 - x) + x + x = 0.5 + x

.

When

p_t = 0.65

atm:

0.65 = 0.5 + x \implies x = 0.65 - 0.5 = 0.15 \text{ atm}

Now, find the partial pressure of

\mathrm{SO}_{2} \mathrm{Cl}_{2}

at this time:

p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}} = p_i - x = 0.5 - 0.15 = 0.35 \text{ atm}

Finally, calculate the rate:

\text{Rate} = k \times p_{\mathrm{SO}_{2} \mathrm{Cl}_{2}} = (2.23 \times 10^{-3} \mathrm{s}^{-1}) \times (0.35 \text{ atm})

\text{Rate} = 7.805 \times 10^{-4} \text{ atm s}^{-1}

Final Answer: The rate of the reaction when the total pressure is 0.65 atm is

7.81 \times 10^{-4} \text{ atm s}^{-1}

.

Q22Exercises

The rate constant for the decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}$ at various temperatures is given below:
$T /{ }^{\circ} \mathrm{C}$ 0 20 40 60 80
$10^{5} \times \mathrm{k} / \mathrm{s}^{-1}$ 0.0787 1.70 25.7 178 2140
Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{\text {a. Predict the rate constant at }} 30^{\circ}$ and $50^{\circ} \mathrm{C}$ .

$T /{ }^{\circ} \mathrm{C}$	0	20	40	60	80
$10^{5} \times \mathrm{k} / \mathrm{s}^{-1}$	0.0787	1.70	25.7	178	2140

Solution

The relationship between the rate constant (

k

), temperature (

T

), activation energy (

E_a

), and the pre-exponential factor (

A

) is given by the Arrhenius equation:

\ln k = \ln A - \frac{E_a}{RT}

This equation is in the form of a straight line (

y = c + mx

), where a plot of

\ln k

(y-axis) versus

1/T

(x-axis) gives a straight line with slope

m = -E_a/R

and y-intercept

c = \ln A

.

Step 1: Process the data We need to convert temperature to Kelvin (

T(\text{K}) = T(^{\circ}\text{C}) + 273

) and calculate

1/T

and

\ln k

.

$T /{ }^{\circ} \mathrm{C}$	$T / \mathrm{K}$	$1/T \times 10^3 (\mathrm{K}^{-1})$	$k (\mathrm{s}^{-1})$	$\ln k$
0	273	3.66	$0.0787 \times 10^{-5}$	-16.35
20	293	3.41	$1.70 \times 10^{-5}$	-10.98
40	313	3.19	$25.7 \times 10^{-5}$	-8.27
60	333	3.00	$178 \times 10^{-5}$	-6.33
80	353	2.83	$2140 \times 10^{-5}$	-3.84

A graph of

\ln k

vs

1/T

would be a straight line.

Step 2: Calculate the Activation Energy ( $E_a$ ) We calculate the slope of the line using two points, for example, the first and the last points.

\text{Slope} = \frac{\Delta(\ln k)}{\Delta(1/T)} = \frac{-3.84 - (-16.35)}{(2.83 - 3.66) \times 10^{-3}} = \frac{12.51}{-0.83 \times 10^{-3}} = -15072 \mathrm{K}

Since Slope

= -E_a/R

:

E_a = -\text{Slope} \times R = -(-15072 \mathrm{K}) \times (8.314 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1})

E_a = 125284 \mathrm{J} \mathrm{mol}^{-1} = 125.3 \mathrm{kJ} \mathrm{mol}^{-1}

Step 3: Calculate the Pre-exponential Factor (A) Using the data for

T=273

K:

\ln k = \ln A - \frac{E_a}{RT}

-16.35 = \ln A - \frac{125284}{8.314 \times 273}

-16.35 = \ln A - 55.14

\ln A = 55.14 - 16.35 = 38.79

A = e^{38.79} = 7.78 \times 10^{16} \mathrm{s}^{-1}

Step 4: Predict the rate constant at $30^{\circ}\mathrm{C}$ and $50^{\circ}\mathrm{C}$ We use the calculated values of

A

and

E_a

in the Arrhenius equation.

At $30^{\circ}\mathrm{C}$ ( $T = 303$ K): $\ln k = 38.79 - \frac{125284}{8.314 \times 303} = 38.79 - 49.69 = -10.90$ $k = e^{-10.90} = 1.84 \times 10^{-5} \mathrm{s}^{-1}$
At $50^{\circ}\mathrm{C}$ ( $T = 323$ K): $\ln k = 38.79 - \frac{125284}{8.314 \times 323} = 38.79 - 46.69 = -7.90$ $k = e^{-7.90} = 3.72 \times 10^{-4} \mathrm{s}^{-1}$

Final Answer:

Activation Energy, $E_a = 125.3 \mathrm{kJ} \mathrm{mol}^{-1}$ .
Pre-exponential Factor, $A = 7.78 \times 10^{16} \mathrm{s}^{-1}$ .
Rate constant at $30^{\circ}\mathrm{C}$ is $1.84 \times 10^{-5} \mathrm{s}^{-1}$ .
Rate constant at $50^{\circ}\mathrm{C}$ is $3.72 \times 10^{-4} \mathrm{s}^{-1}$ .

Q23Exercises

The rate constant for the decomposition of hydrocarbons is $2.418 \times 10^{-5} \mathrm{~s}^{-1}$ at 546 K. If the energy of activation is $179.9 \mathrm{~kJ} / \mathrm{mol}$ , what will be the value of pre-exponential factor.

Solution

Given:

Rate constant, $k = 2.418 \times 10^{-5} \mathrm{~s}^{-1}$ .
Temperature, $T = 546$ K.
Activation energy, $E_a = 179.9 \mathrm{~kJ} / \mathrm{mol} = 179900 \mathrm{~J} / \mathrm{mol}$ .
Gas constant, $R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}$ .

To Find: The value of the pre-exponential factor, A.

Formula: The Arrhenius equation relates the rate constant, temperature, activation energy, and the pre-exponential factor:

k = A e^{-E_a / RT}

To solve for A, we can take the natural logarithm of both sides and rearrange:

\ln k = \ln A - \frac{E_a}{RT} \implies \ln A = \ln k + \frac{E_a}{RT}

Alternatively, using base-10 logarithms:

\log k = \log A - \frac{E_a}{2.303 RT} \implies \log A = \log k + \frac{E_a}{2.303 RT}

We will use the second form.

Calculation: First, calculate the term

\frac{E_a}{2.303 RT}

:

\frac{E_a}{2.303 RT} = \frac{179900 \mathrm{~J} \mathrm{mol}^{-1}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1} \times 546 \mathrm{~K}} = \frac{179900}{10488.5} = 17.152

Now, find the value of

\log k

:

\log k = \log(2.418 \times 10^{-5}) = -4.6165

Substitute these values back into the equation for

\log A

:

\log A = -4.6165 + 17.152 = 12.5355

Finally, find A by taking the antilogarithm:

A = 10^{12.5355} = 3.43 \times 10^{12}

The units of A are the same as the units of the rate constant, which is

\mathrm{s}^{-1}

.

Final Answer: The value of the pre-exponential factor is

3.43 \times 10^{12} \mathrm{~s}^{-1}

.

Q24Exercises

Consider a certain reaction $\mathrm{A} \rightarrow$ Products with $k=2.0 \times 10^{-2} \mathrm{~s}^{-1}$ . Calculate the concentration of A remaining after 100 s if the initial concentration of A is $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$ .

Solution

Given:

Rate constant, $k = 2.0 \times 10^{-2} \mathrm{~s}^{-1}$ .
Time, $t = 100$ s.
Initial concentration of A, $[\mathrm{A}]_0 = 1.0 \mathrm{~mol} \mathrm{~L}^{-1}$ .

Step 1: Identify the order of the reaction

The units of the rate constant,

k

, are

\mathrm{s}^{-1}

. This is characteristic of a first-order reaction.

Step 2: Use the integrated rate law for a first-order reaction

The formula relating concentration, time, and rate constant for a first-order reaction is:

t = \frac{2.303}{k} \log \frac{[\mathrm{A}]_0}{[\mathrm{A}]}

where

[\mathrm{A}]

is the concentration of A remaining at time

t

.

We can rearrange this formula to solve for

[\mathrm{A}]

:

\log \frac{[\mathrm{A}]_0}{[\mathrm{A}]} = \frac{kt}{2.303}

Calculation: Substitute the given values into the rearranged equation:

\log \frac{1.0}{[\mathrm{A}]} = \frac{(2.0 \times 10^{-2} \mathrm{~s}^{-1}) \times (100 \mathrm{~s})}{2.303}

\log \frac{1.0}{[\mathrm{A}]} = \frac{2.0}{2.303} = 0.8684

To find the value of

\frac{1.0}{[\mathrm{A}]}

, we take the antilogarithm of both sides:

\frac{1.0}{[\mathrm{A}]} = 10^{0.8684} = 7.386

Now, solve for

[\mathrm{A}]

:

[\mathrm{A}] = \frac{1.0}{7.386} \mathrm{~mol} \mathrm{~L}^{-1} = 0.1354 \mathrm{~mol} \mathrm{~L}^{-1}

Final Answer: The concentration of A remaining after 100 s is $0.135 \mathrm{~mol} \mathrm{~L}^{-1}$ .

Q25Exercises

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t_{1/2}=3.00$ hours. What fraction of sample of sucrose remains after 8 hours ?

Solution

Given:

The reaction follows a first-order rate law.
Half-life, $t_{1/2} = 3.00$ hours.
Time, $t = 8$ hours.

To Find: The fraction of the sample remaining after 8 hours, which is the ratio

\frac{[A]}{[A]_0}

, where

[A]

is the concentration at time

t

and

[A]_0

is the initial concentration.

Step 1: Calculate the rate constant (k)

For a first-order reaction, the rate constant is related to the half-life by:

k = \frac{0.693}{t_{1/2}}

Calculation of k:

k = \frac{0.693}{3.00 \text{ hours}} = 0.231 \text{ hours}^{-1}

Step 2: Use the integrated rate law to find the fraction remaining

The integrated rate law for a first-order reaction is:

t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}

We can rearrange this equation to find the ratio

\frac{[A]_0}{[A]}

:

\log \frac{[A]_0}{[A]} = \frac{kt}{2.303}

Calculation of the ratio:

\log \frac{[A]_0}{[A]} = \frac{(0.231 \text{ hours}^{-1}) \times (8 \text{ hours})}{2.303}

\log \frac{[A]_0}{[A]} = \frac{1.848}{2.303} = 0.8024

Now, take the antilogarithm to find the value of the ratio:

\frac{[A]_0}{[A]} = 10^{0.8024} = 6.345

Step 3: Calculate the fraction remaining

The fraction remaining is the reciprocal of the ratio we just calculated:

\text{Fraction remaining} = \frac{[A]}{[A]_0} = \frac{1}{6.345} = 0.1576

Final Answer: The fraction of the sample of sucrose remaining after 8 hours is 0.158 (or approximately 15.8%).

Q26Exercises

The decomposition of hydrocarbon follows the equation $k=\left(4.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{-28000 K / T}$ Calculate $E_{\mathrm{a}}$ .

Solution

Given: The equation for the rate constant is:

k = (4.5 \times 10^{11} \mathrm{~s}^{-1}) e^{-28000 K / T}

To Find: The activation energy,

E_a

.

Formula: The Arrhenius equation provides the general relationship between the rate constant (

k

), activation energy (

E_a

), and temperature (

T

):

k = A e^{-E_a / RT}

where:

$A$ is the pre-exponential factor.
$R$ is the universal gas constant ( $8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}$ ).

Comparison and Calculation: We can find the value of

E_a

by comparing the given equation with the standard Arrhenius equation.

Given equation:

k = (4.5 \times 10^{11}) e^{-28000/T}

Arrhenius equation:

k = A e^{-E_a/RT}

By comparing the exponential terms of both equations, we can see that:

\frac{E_a}{RT} = \frac{28000 \text{ K}}{T}

We can cancel

T

from both sides of the equation:

\frac{E_a}{R} = 28000 \text{ K}

Now, we can solve for

E_a

by multiplying by

R

:

E_a = 28000 \text{ K} \times R

Substitute the value of the gas constant,

R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}

:

E_a = 28000 \text{ K} \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}

E_a = 232792 \mathrm{~J} \mathrm{mol}^{-1}

To express the activation energy in kilojoules per mole (kJ/mol), we divide by 1000:

E_a = \frac{232792}{1000} \mathrm{~kJ} \mathrm{mol}^{-1} = 232.792 \mathrm{~kJ} \mathrm{mol}^{-1}

Final Answer: The activation energy,

E_a

, for the reaction is $232.79 \mathrm{~kJ} \mathrm{mol}^{-1}$ .

Q27Exercises

The rate constant for the first order decomposition of $\mathrm{H}_{2} \mathrm{O}_{2}$ is given by the following equation: $\log k=14.34-1.25 \times 10^{4} K / T$ Calculate $E_{\mathrm{a}}$ for this reaction and at what temperature will its half-period be 256 minutes?

Solution

Part 1: Calculate Activation Energy ( $E_a$ )

Given equation:

\log k = 14.34 - \frac{1.25 \times 10^4 \text{ K}}{T}

The Arrhenius equation in logarithmic form is:

\log k = \log A - \frac{E_a}{2.303 RT}

By comparing the given equation with the standard Arrhenius equation, we can equate the terms containing

T

:

\frac{E_a}{2.303 R} = 1.25 \times 10^4 \text{ K}

Now, we solve for

E_a

:

E_a = 1.25 \times 10^4 \text{ K} \times 2.303 \times R

Substitute the value of the gas constant,

R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}

:

E_a = 1.25 \times 10^4 \times 2.303 \times 8.314 \mathrm{~J} \mathrm{mol}^{-1}

E_a = 239339 \mathrm{~J} \mathrm{mol}^{-1}

E_a = 239.34 \mathrm{~kJ} \mathrm{mol}^{-1}

Part 2: Calculate the temperature for a half-period of 256 minutes

Given:

Half-life, $t_{1/2} = 256$ minutes.

First, we need to find the rate constant,

k

, that corresponds to this half-life. We must convert the half-life to seconds to be consistent with the units of A (implied by the Arrhenius equation).

t_{1/2} = 256 \text{ min} \times 60 \frac{\text{s}}{\text{min}} = 15360 \text{ s}

For a first-order reaction:

k = \frac{0.693}{t_{1/2}} = \frac{0.693}{15360 \text{ s}} = 4.51 \times 10^{-5} \mathrm{s}^{-1}

Now, substitute this value of

k

into the given equation to find the temperature

T

:

\log(4.51 \times 10^{-5}) = 14.34 - \frac{1.25 \times 10^4}{T}

-4.346 = 14.34 - \frac{1.25 \times 10^4}{T}

Rearrange the equation to solve for

T

:

\frac{1.25 \times 10^4}{T} = 14.34 + 4.346 = 18.686

T = \frac{1.25 \times 10^4}{18.686} \text{ K} = 668.9 \text{ K}

Final Answer:

The activation energy, $E_a$ , is $239.34 \mathrm{~kJ} \mathrm{mol}^{-1}$ .
The temperature at which the half-period will be 256 minutes is 668.9 K.

Q28Exercises

The decomposition of A into product has value of $k$ as $4.5 \times 10^{3} \mathrm{~s}^{-1}$ at $10^{\circ} \mathrm{C}$ and energy of activation $60 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . At what temperature would $k$ be $1.5 \times 10^{4} \mathrm{~s}^{-1}$ ?

Solution

Given:

At temperature $T_1 = 10^{\circ} \mathrm{C} = 10 + 273 = 283$ K, the rate constant $k_1 = 4.5 \times 10^{3} \mathrm{~s}^{-1}$ .
At temperature $T_2$ , the rate constant $k_2 = 1.5 \times 10^{4} \mathrm{~s}^{-1}$ .
Activation energy, $E_a = 60 \mathrm{~kJ} \mathrm{mol}^{-1} = 60000 \mathrm{~J} \mathrm{mol}^{-1}$ .
Gas constant, $R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}$ .

To Find: The temperature

T_2

.

Formula: We use the two-point form of the Arrhenius equation, which relates rate constants at two different temperatures:

\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]

Calculation: First, substitute the known values into the equation:

\log \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}} = \frac{60000 \mathrm{~J} \mathrm{mol}^{-1}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}} \left[ \frac{T_2 - 283}{283 T_2} \right]

Simplify the terms:

\log \left( \frac{15}{4.5} \right) = \frac{60000}{19.147} \left[ \frac{T_2 - 283}{283 T_2} \right]

\log(3.333) = 3133.65 \left[ \frac{T_2 - 283}{283 T_2} \right]

0.5228 = 3133.65 \left[ \frac{T_2 - 283}{283 T_2} \right]

Now, rearrange to solve for the term containing

T_2

:

\frac{0.5228}{3133.65} = \frac{T_2 - 283}{283 T_2}

1.668 \times 10^{-4} = \frac{T_2}{283 T_2} - \frac{283}{283 T_2}

1.668 \times 10^{-4} = \frac{1}{283} - \frac{1}{T_2}

0.0001668 = 0.0035335 - \frac{1}{T_2}

Isolate

\frac{1}{T_2}

:

\frac{1}{T_2} = 0.0035335 - 0.0001668 = 0.0033667

Finally, calculate

T_2

:

T_2 = \frac{1}{0.0033667} = 297.02 \text{ K}

To express the temperature in degrees Celsius:

T_2(^{\circ} \mathrm{C}) = 297.02 - 273 = 24.02^{\circ} \mathrm{C}

Final Answer: The temperature would be 297 K (or

24^{\circ} \mathrm{C}

) for the rate constant to be

1.5 \times 10^{4} \mathrm{~s}^{-1}

.

Q29Exercises

The time required for $10 \%$ completion of a first order reaction at 298 K is equal to that required for its $25 \%$ completion at 308 K. If the value of $A$ is $4 \times 10^{10} \mathrm{~s}^{-1}$ . Calculate $k$ at 318 K and $E_{\mathrm{a}}$ .

Solution

Let

k_1

be the rate constant at

T_1 = 298

K and

k_2

be the rate constant at

T_2 = 308

K.

Step 1: Relate $k_1$ and $k_2$

The time required for a certain percentage completion of a first-order reaction is given by

t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}

.

Time for 10% completion at 298 K ( $t_{10\%}$ ): $[R] = 0.90[R]_0$ . $t_{10\%} = \frac{2.303}{k_1} \log \frac{[R]_0}{0.90[R]_0} = \frac{2.303}{k_1} \log(1.111)$
Time for 25% completion at 308 K ( $t_{25\%}$ ): $[R] = 0.75[R]_0$ . $t_{25\%} = \frac{2.303}{k_2} \log \frac{[R]_0}{0.75[R]_0} = \frac{2.303}{k_2} \log(1.333)$

Given that

t_{10\%} = t_{25\%}

:

\frac{2.303}{k_1} \log(1.111) = \frac{2.303}{k_2} \log(1.333)

\frac{\log(1.111)}{k_1} = \frac{\log(1.333)}{k_2} \implies \frac{k_2}{k_1} = \frac{\log(1.333)}{\log(1.111)} = \frac{0.1249}{0.0457} = 2.733

Step 2: Calculate Activation Energy ( $E_a$ )

Using the Arrhenius equation:

\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)

\log(2.733) = \frac{E_a}{2.303 \times 8.314} \left( \frac{308 - 298}{298 \times 308} \right)

0.4366 = \frac{E_a}{19.147} \left( \frac{10}{91784} \right)

E_a = \frac{0.4366 \times 19.147 \times 91784}{10} = 76650 \mathrm{~J} \mathrm{mol}^{-1} = 76.65 \mathrm{~kJ} \mathrm{mol}^{-1}

Step 3: Calculate k at 318 K

Let

k_3

be the rate constant at

T_3 = 318

K. We can relate

k_3

to

k_1

(at

T_1 = 298

K).

\log \frac{k_3}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_3 - T_1}{T_1 T_3} \right)

First, we need to find

k_1

using the given value of A (

4 \times 10^{10} \mathrm{~s}^{-1}

) and the calculated

E_a

.

\log k_1 = \log A - \frac{E_a}{2.303 R T_1}

\log k_1 = \log(4 \times 10^{10}) - \frac{76650}{2.303 \times 8.314 \times 298}

\log k_1 = 10.602 - \frac{76650}{5705.8} = 10.602 - 13.433 = -2.831

k_1 = 10^{-2.831} = 1.475 \times 10^{-3} \mathrm{~s}^{-1}

Now, we can find

k_3

:

\log \frac{k_3}{1.475 \times 10^{-3}} = \frac{76650}{2.303 \times 8.314} \left( \frac{318 - 298}{298 \times 318} \right)

\log \frac{k_3}{1.475 \times 10^{-3}} = 4002.2 \left( \frac{20}{94764} \right) = 0.8446

\frac{k_3}{1.475 \times 10^{-3}} = 10^{0.8446} = 6.992

k_3 = 6.992 \times 1.475 \times 10^{-3} = 1.031 \times 10^{-2} \mathrm{~s}^{-1}

Final Answer:

Activation Energy, $E_a = 76.65 \mathrm{~kJ} \mathrm{mol}^{-1}$ .
Rate constant at 318 K, $k = 1.03 \times 10^{-2} \mathrm{~s}^{-1}$ .

Q30Exercises

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Solution

Given:

Initial temperature, $T_1 = 293$ K.
Final temperature, $T_2 = 313$ K.
The rate of reaction quadruples, which means the rate constant also quadruples. So, if $k_1$ is the rate constant at $T_1$ and $k_2$ is the rate constant at $T_2$ , then $k_2 = 4k_1$ , or $\frac{k_2}{k_1} = 4$ .
Gas constant, $R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}$ .

To Find: The energy of activation,

E_a

.

Formula: We use the two-point form of the Arrhenius equation:

\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]

Calculation: Substitute the given values into the formula:

\log(4) = \frac{E_a}{2.303 \times 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol}^{-1}} \left[ \frac{313 \mathrm{K} - 293 \mathrm{K}}{(293 \mathrm{K}) \times (313 \mathrm{K})} \right]

Simplify the terms:

0.6021 = \frac{E_a}{19.147} \left[ \frac{20}{91709} \right]

Now, rearrange the equation to solve for

E_a

:

E_a = \frac{0.6021 \times 19.147 \times 91709}{20} \mathrm{~J} \mathrm{mol}^{-1}

E_a = \frac{1058097.3}{20} \mathrm{~J} \mathrm{mol}^{-1}

E_a = 52904.8 \mathrm{~J} \mathrm{mol}^{-1}

To express the activation energy in kilojoules per mole (kJ/mol), we divide by 1000:

E_a = \frac{52904.8}{1000} \mathrm{~kJ} \mathrm{mol}^{-1} = 52.90 \mathrm{~kJ} \mathrm{mol}^{-1}

Final Answer: The energy of activation of the reaction is $52.9 \mathrm{~kJ} \mathrm{mol}^{-1}$ .

Chemical KineticsClass 12 Chemistry NCERT Solutions

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

3NO(g)→N2O(g)3 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})3NO(g)→N2​O(g) Rate =k[NO]2=k[\mathrm{NO}]^{2}=k[NO]2

CH3CHO(g)→CH4(g)+CO(g)\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})CH3​CHO(g)→CH4​(g)+CO(g) Rate =k[CH3CHO]3/2=k\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2}=k[CH3​CHO]3/2

C2H5Cl(g)→C2H4(g)+HCl(g)\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})C2​H5​Cl(g)→C2​H4​(g)+HCl(g) Rate =k[C2H5Cl]=k\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right]=k[C2​H5​Cl]

Solution

Solution

Solution

Solution

Mention the factors that affect the rate of a chemical reaction.

Solution

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

doubled

reduced to half ?

Solution

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Solution

In a pseudo first order reaction in water, the following results were obtained: t/s\mathrm{t} / \mathrm{s}t/s0306090[A]/molL−1[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}[A]/molL−10.550.310.170.085 Calculate the average rate of reaction between the time interval 30 to 60 seconds.

Solution

A reaction is first order in A and second order in B.

Write the differential rate equation.

How is the rate affected on increasing the concentration of B three times?

How is the rate affected when the concentrations of both A and B are doubled?

Solution

Solution

Solution

Solution

Calculate the half-life of a first order reaction from their rate constants given below:

200 s−1200 \mathrm{~s}^{-1}200 s−1

2 min−12 \mathrm{~min}^{-1}2 min−1

4 years−1^{-1}−1

Solution

The half-life for radioactive decay of 14C^{14} \mathrm{C}14C is 5730 years. An archaeological artifact containing wood had only 80%80 \%80% of the 14C^{14} \mathrm{C}14C found in a living tree. Estimate the age of the sample.

Solution

Plot [N2O5]\left[\mathrm{N}_{2} \mathrm{O}_{5}\right][N2​O5​] against ttt.

Find the half-life period for the reaction.

Draw a graph between log⁡[N2O5]\log \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]log[N2​O5​] and ttt.

What is the rate law ?

Calculate the rate constant.

Calculate the half-life period from kkk and compare it with (ii).

Solution

The rate constant for a first order reaction is 60 s−160 \mathrm{~s}^{-1}60 s−1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th 1 / 16^{\text {th }}1/16th value?

Solution

Solution

For a first order reaction, show that time required for 99%99 \%99% completion is twice the time required for the completion of 90%90 \%90% of reaction.

Solution

A first order reaction takes 40 min for 30%30 \%30% decomposition. Calculate t1/2t_{1 / 2}t1/2​.

Solution

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t(sec)t(\mathrm{sec})t(sec)P(mm of Hg)\mathrm{P}(\mathrm{mm} \text { of } \mathrm{Hg})P(mm of Hg)035.036054.072063.0 Calculate the rate constant.

Solution

Solution

Solution

The rate constant for the decomposition of hydrocarbons is 2.418×10−5 s−12.418 \times 10^{-5} \mathrm{~s}^{-1}2.418×10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol179.9 \mathrm{~kJ} / \mathrm{mol}179.9 kJ/mol, what will be the value of pre-exponential factor.

Solution

Solution

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2=3.00t_{1/2}=3.00t1/2​=3.00 hours. What fraction of sample of sucrose remains after 8 hours ?

Solution

The decomposition of hydrocarbon follows the equation k=(4.5×1011 s−1)e−28000K/Tk=\left(4.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{-28000 K / T}k=(4.5×1011 s−1)e−28000K/T Calculate EaE_{\mathrm{a}}Ea​.

Solution

Solution

Solution

The time required for 10%10 \%10% completion of a first order reaction at 298 K is equal to that required for its 25%25 \%25% completion at 308 K. If the value of AAA is 4×1010 s−14 \times 10^{10} \mathrm{~s}^{-1}4×1010 s−1. Calculate kkk at 318 K and EaE_{\mathrm{a}}Ea​.

Solution

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Solution

$3 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ Rate $=k[\mathrm{NO}]^{2}$

$\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$ Rate $=k\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2}$

$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})$ Rate $=k\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right]$

In a pseudo first order reaction in water, the following results were obtained:
$\mathrm{t} / \mathrm{s}$ 0 30 60 90
$[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}$ 0.55 0.31 0.17 0.085
Calculate the average rate of reaction between the time interval 30 to 60 seconds.

$200 \mathrm{~s}^{-1}$

$2 \mathrm{~min}^{-1}$

4 years $^{-1}$

The half-life for radioactive decay of $^{14} \mathrm{C}$ is 5730 years. An archaeological artifact containing wood had only $80 \%$ of the $^{14} \mathrm{C}$ found in a living tree. Estimate the age of the sample.

Plot $\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$ against $t$ .

Draw a graph between $\log \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$ and $t$ .

Calculate the half-life period from $k$ and compare it with (ii).

The rate constant for a first order reaction is $60 \mathrm{~s}^{-1}$ . How much time will it take to reduce the initial concentration of the reactant to its $1 / 16^{\text {th }}$ value?

For a first order reaction, show that time required for $99 \%$ completion is twice the time required for the completion of $90 \%$ of reaction.

A first order reaction takes 40 min for $30 \%$ decomposition. Calculate $t_{1 / 2}$ .

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
$t(\mathrm{sec})$ $\mathrm{P}(\mathrm{mm} \text { of } \mathrm{Hg})$
0 35.0
360 54.0
720 63.0
Calculate the rate constant.

The rate constant for the decomposition of hydrocarbons is $2.418 \times 10^{-5} \mathrm{~s}^{-1}$ at 546 K. If the energy of activation is $179.9 \mathrm{~kJ} / \mathrm{mol}$ , what will be the value of pre-exponential factor.

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t_{1/2}=3.00$ hours. What fraction of sample of sucrose remains after 8 hours ?

The decomposition of hydrocarbon follows the equation $k=\left(4.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{-28000 K / T}$ Calculate $E_{\mathrm{a}}$ .

The time required for $10 \%$ completion of a first order reaction at 298 K is equal to that required for its $25 \%$ completion at 308 K. If the value of $A$ is $4 \times 10^{10} \mathrm{~s}^{-1}$ . Calculate $k$ at 318 K and $E_{\mathrm{a}}$ .