Chemical KineticsClass 12 Chemistry Notes

Introduction to Chemical Kinetics

Chemical kinetics is the branch of chemistry that helps us understand the speed, or rate, of chemical reactions and the step-by-step processes, or mechanisms, by which they occur.

While other branches of chemistry answer different questions, kinetics focuses specifically on "how fast?" and "how?":

• Thermodynamics tells us if a reaction is feasible (possible). A reaction is feasible if the change in Gibbs free energy is negative ($\Delta G < 0$).
• Chemical Equilibrium tells us the extent to which a reaction will proceed before stopping.
• Chemical Kinetics tells us the speed at which a reaction reaches equilibrium.

Kinetic studies are crucial for practical applications, like understanding how quickly food spoils, designing fast-setting dental fillings, or controlling how fuel burns in an engine.

Rate of a Chemical Reaction

The rate of a reaction is defined as the change in the concentration of a reactant or a product per unit of time.

We can express this in two ways:

1. Rate of disappearance of a reactant: Since reactants are used up, their concentration decreases over time.
2. Rate of appearance of a product: Since products are formed, their concentration increases over time.

Average Rate of Reaction

Consider a simple reaction where a reactant (R) turns into a product (P): $R \rightarrow P$

If we measure the concentration at two different times, $t_1$ and $t_2$, we can calculate the average rate ($r_{av}$) over that time interval ($\Delta t = t_2 - t_1$).

• In terms of reactant R: $r_{av} = -\frac{\Delta[R]}{\Delta t} = -\frac{[R]_2 - [R]_1}{t_2 - t_1}$ [!note] A negative sign is used because the concentration of the reactant decreases ($\Delta[R]$ is negative), and reaction rates must always be positive.

• In terms of product P: $r_{av} = +\frac{\Delta[P]}{\Delta t} = +\frac{[P]_2 - [P]_1}{t_2 - t_1}$

Instantaneous Rate of Reaction

The average rate gives us the speed over a period, but the rate of a reaction usually slows down as reactants are consumed. To find the rate at a specific moment, we use the instantaneous rate ($r_{inst}$). This is the average rate over an infinitesimally small time interval ($\Delta t$ approaches zero, written as $dt$).

Graphically, the instantaneous rate at any time $t$ is the slope of the tangent to the concentration-time curve at that point.

• In terms of reactant R: $r_{inst} = -\frac{d[R]}{dt}$
• In terms of product P: $r_{inst} = +\frac{d[P]}{dt}$

Units of Rate of a Reaction

The units for the rate of reaction are typically concentration time⁻¹.

• If concentration is in moles per litre ($\text{mol L}^{-1}$) and time is in seconds (s), the units are $\text{mol L}^{-1} \text{s}^{-1}$.
• For reactions involving gases, concentration is often expressed as partial pressure. In this case, the units might be $\text{atm s}^{-1}$.

Rate Expression and Stoichiometry

When the stoichiometric coefficients in a balanced chemical equation are not all 1, the rate of disappearance or appearance of each substance will be different. To define a single, unique rate for the entire reaction, we divide the rate of change of each substance by its stoichiometric coefficient.

For a general reaction: $aA + bB \rightarrow cC + dD$

The rate of reaction is given by: $\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$

Given

• Reaction: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$
• Initial concentration $[N_2O_5]_1 = 2.33 \text{ mol L}^{-1}$
• Final concentration $[N_2O_5]_2 = 2.08 \text{ mol L}^{-1}$
• Time interval $\Delta t = 184 \text{ min}$

To Find

• Average rate in $\text{mol L}^{-1} \text{min}^{-1}$, $\text{mol L}^{-1} \text{h}^{-1}$, and $\text{mol L}^{-1} \text{s}^{-1}$
• Rate of production of $NO_2$

Formula

$\text{Average Rate} = -\frac{1}{2}\frac{\Delta[N_2O_5]}{\Delta t}$ $\text{Rate} = \frac{1}{4}\frac{\Delta[NO_2]}{\Delta t}$

Solution

First, calculate the average rate of reaction in $\text{mol L}^{-1} \text{min}^{-1}$. $\text{Average Rate} = -\frac{1}{2} \frac{(2.08 - 2.33) \text{ mol L}^{-1}}{184 \text{ min}} = -\frac{1}{2} \frac{-0.25 \text{ mol L}^{-1}}{184 \text{ min}}$ $\text{Average Rate} = 6.79 \times 10^{-4} \text{ mol L}^{-1} \text{min}^{-1}$

Now, convert this rate to other units:

• In hours: $6.79 \times 10^{-4} \text{ mol L}^{-1} \text{min}^{-1} \times \frac{60 \text{ min}}{1 \text{ h}} = 4.07 \times 10^{-2} \text{ mol L}^{-1} \text{h}^{-1}$
• In seconds: $6.79 \times 10^{-4} \text{ mol L}^{-1} \text{min}^{-1} \times \frac{1 \text{ min}}{60 \text{ s}} = 1.13 \times 10^{-5} \text{ mol L}^{-1} \text{s}^{-1}$

Next, find the rate of production of $NO_2$. From the rate expression: $\text{Average Rate} = \frac{1}{4}\frac{\Delta[NO_2]}{\Delta t}$ Therefore, the rate of production of $NO_2$ is: $\frac{\Delta[NO_2]}{\Delta t} = 4 \times \text{Average Rate}$ $\frac{\Delta[NO_2]}{\Delta t} = 4 \times (6.79 \times 10^{-4} \text{ mol L}^{-1} \text{min}^{-1}) = 2.72 \times 10^{-3} \text{ mol L}^{-1} \text{min}^{-1}$

Final Answer The average rate is $6.79 \times 10^{-4} \text{ mol L}^{-1} \text{min}^{-1}$, $4.07 \times 10^{-2} \text{ mol L}^{-1} \text{h}^{-1}$, or $1.13 \times 10^{-5} \text{ mol L}^{-1} \text{s}^{-1}$. The rate of production of $NO_2$ is $2.72 \times 10^{-3} \text{ mol L}^{-1} \text{min}^{-1}$.

Factors Influencing Rate of a Reaction

The rate of a reaction is influenced by several factors:

• Concentration of reactants (or pressure for gases)
• Temperature
• Presence of a catalyst

Dependence of Rate on Concentration

Generally, the rate of a reaction increases as the concentration of reactants increases. This relationship is described by the rate law (also called the rate equation or rate expression).

Rate Expression and Rate Constant

For a general reaction: $aA + bB \rightarrow cC + dD$ The rate law is expressed as: $\text{Rate} = k[A]^x[B]^y$

Here:

• k is the rate constant, a proportionality constant that is specific to the reaction and depends on temperature.
• [A] and [B] are the molar concentrations of the reactants.
• x and y are exponents that define how the rate depends on the concentration of each reactant.

This form of the rate law is also known as the differential rate equation: $-\frac{d[R]}{dt} = k[A]^x[B]^y$

Order of a Reaction

The order of a reaction describes how sensitive the reaction rate is to changes in reactant concentrations.

• The order with respect to a reactant is the exponent of its concentration term in the rate law. For example, the reaction is of order 'x' with respect to reactant A.
• The overall order of the reaction is the sum of the exponents of all concentration terms in the rate law (overall order = x + y).

Reaction order can be a whole number (0, 1, 2, 3) or even a fraction.

• A zero-order reaction means the rate is independent of the concentration of reactants (Rate = k).

To Find

(a) Overall order for Rate = $k[A]^{1/2}[B]^{3/2}$ (b) Overall order for Rate = $k[A]^{3/2}[B]^{-1}$

Formula

Overall order = sum of exponents in the rate law.

Solution

(a) For Rate = $k[A]^{1/2}[B]^{3/2}$ The exponents are 1/2 and 3/2. $\text{Order} = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$ This is a second-order reaction.

(b) For Rate = $k[A]^{3/2}[B]^{-1}$ The exponents are 3/2 and -1. $\text{Order} = \frac{3}{2} + (-1) = \frac{3}{2} - \frac{2}{2} = \frac{1}{2}$ This is a half-order reaction.

Units of Rate Constant

The units of the rate constant, $k$, depend on the overall order of the reaction (n). The general formula for the units of $k$ is: $k = \frac{\text{Rate}}{(\text{concentration})^n} = \frac{\text{concentration}}{\text{time}} \times \frac{1}{(\text{concentration})^n} = (\text{concentration})^{1-n} (\text{time})^{-1}$

Using concentration in $\text{mol L}^{-1}$ and time in s:

• Zero order (n=0): Units of $k$ are $(\text{mol L}^{-1})^{1-0} \text{s}^{-1} = \text{mol L}^{-1} \text{s}^{-1}$.
• First order (n=1): Units of $k$ are $(\text{mol L}^{-1})^{1-1} \text{s}^{-1} = \text{s}^{-1}$.
• Second order (n=2): Units of $k$ are $(\text{mol L}^{-1})^{1-2} \text{s}^{-1} = \text{mol}^{-1} \text{L s}^{-1}$.

Solution

(i) The units are $\text{L mol}^{-1} \text{s}^{-1}$, which can be written as $\text{mol}^{-1} \text{L s}^{-1}$. This corresponds to the units for a second-order reaction.

(ii) The unit is $\text{s}^{-1}$. This corresponds to the unit for a first-order reaction.

Molecularity of a Reaction

While order is an experimental concept, molecularity is a theoretical concept that applies only to elementary reactions (reactions that occur in a single step).

Molecularity is the number of reacting species (atoms, ions, or molecules) that must collide simultaneously to bring about the chemical reaction in an elementary step.

• Unimolecular: One reacting species is involved. Example: $\text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}$
• Bimolecular: Two species collide. Example: $2\text{HI} \rightarrow \text{H}_2 + \text{I}_2$
• Trimolecular (or Termolecular): Three species collide. Example: $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$

Reactions with molecularity greater than three are very rare because the probability of more than three molecules colliding simultaneously with the correct orientation and energy is extremely low.

Elementary vs. Complex Reactions

• Elementary Reactions: Occur in a single step.
• Complex Reactions: Occur through a sequence of elementary reactions, called the reaction mechanism.

For complex reactions, the overall rate is determined by the slowest step in the mechanism, known as the rate-determining step.

Comparing Order and Molecularity

Integrated Rate Equations

The differential rate law shows how rate depends on concentration. By integrating this equation, we get an integrated rate equation, which directly relates the concentration of reactants to time. This is more convenient for analyzing experimental data.

Zero-Order Reactions

A zero-order reaction is one where the rate is independent of the reactant's concentration.

• Differential Rate Law: $\text{Rate} = -\frac{d[R]}{dt} = k[R]^0 = k$

• Integrated Rate Law: $[R] = -kt + [R]_0$ where $[R]_0$ is the initial concentration of the reactant R. This equation is in the form of a straight line ($y = mx + c$). A plot of $[R]$ versus time ($t$) gives a straight line with a slope of $-k$ and a y-intercept of $[R]_0$.

• Examples: Zero-order reactions are uncommon but can occur on surfaces, such as the decomposition of gaseous ammonia on a hot platinum surface at high pressure. $2NH_3(g) \xrightarrow{1130 K, \text{Pt catalyst}} N_2(g) + 3H_2(g)$ $\text{Rate} = k[NH_3]^0 = k$

First-Order Reactions

In a first-order reaction, the rate is directly proportional to the first power of the reactant's concentration.

• Differential Rate Law: $\text{Rate} = -\frac{d[R]}{dt} = k[R]$

• Integrated Rate Law (Natural Logarithm): $\ln[R] = -kt + \ln[R]_0$ A plot of $\ln[R]$ versus time ($t$) gives a straight line with a slope of $-k$ and a y-intercept of $\ln[R]_0$.

• Integrated Rate Law (Base-10 Logarithm): $k = \frac{2.303}{t} \log\frac{[R]_0}{[R]}$ This is the most commonly used form for calculations.

• Exponential Form: $[R] = [R]_0 e^{-kt}$ This shows that the concentration of the reactant decreases exponentially with time.

• Examples: All radioactive decay processes are first-order reactions. The decomposition of $N_2O_5$ is also a first-order reaction.

Given

• Order = First order
• Initial concentration, $[R]_0 = 1.24 \times 10^{-2} \text{ mol L}^{-1}$
• Concentration at time t, $[R] = 0.20 \times 10^{-2} \text{ mol L}^{-1}$
• Time, $t = 60 \text{ min}$

To Find

Rate constant, $k$

Formula

For a first-order reaction: $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ (Note: The text uses $[R]_1$ for initial and $[R]_2$ for final, which is equivalent to $[R]_0$ and $[R]$).

Solution

Substitute the given values into the formula: $k = \frac{2.303}{60 \text{ min}} \log \frac{1.24 \times 10^{-2} \text{ mol L}^{-1}}{0.20 \times 10^{-2} \text{ mol L}^{-1}}$ $k = \frac{2.303}{60} \log(6.2) \text{ min}^{-1}$ Using $\log(6.2) \approx 0.7924$: $k = \frac{2.303}{60} \times 0.7924 \text{ min}^{-1}$ $k = 0.0304 \text{ min}^{-1}$

Final Answer The rate constant of the reaction is $0.0304 \text{ min}^{-1}$.

First-Order Gas-Phase Reactions

For a gas-phase reaction like $A(g) \rightarrow B(g) + C(g)$, it's often easier to measure pressure than concentration. The first-order rate law can be expressed in terms of partial pressures.

If $p_i$ is the initial pressure of A and $p_t$ is the total pressure at time $t$, the rate constant is given by: $k = \frac{2.303}{t} \log \frac{p_i}{(2p_i - p_t)}$

Given

• Initial total pressure (which is the initial pressure of $N_2O_5$), $p_i = 0.5 \text{ atm}$
• Total pressure at $t=100$ s, $p_t = 0.512 \text{ atm}$
• Time, $t = 100 \text{ s}$

To Find

Rate constant, $k$

Formula

First, we need to relate the partial pressure of $N_2O_5$ at time $t$, $p_{N_2O_5}$, to the total pressure $p_t$. Let the decrease in pressure of $N_2O_5$ be $2x$. $p_{N_2O_5} = p_i - 2x$ $p_{N_2O_4} = 2x$ $p_{O_2} = x$ Total pressure $p_t = (p_i - 2x) + 2x + x = p_i + x \implies x = p_t - p_i$ So, $p_{N_2O_5} = p_i - 2(p_t - p_i) = 3p_i - 2p_t$. Wait, the source text calculation is different. Let's re-derive based on the source. Source derivation: $p_t = (0.5 - 2x) + 2x + x = 0.5 + x \implies x = p_t - 0.5$ $p_{N_2O_5} = 0.5 - 2x = 0.5 - 2(p_t - 0.5) = 0.5 - 2p_t + 1 = 1.5 - 2p_t$. This is correct. The rate law for a first-order reaction in terms of pressure is: $k = \frac{2.303}{t} \log \frac{p_i}{p_A}$ where $p_A$ is the partial pressure of the reactant at time $t$.

Solution

1. Find the partial pressure of $N_2O_5$ at t = 100 s: The initial pressure of $N_2O_5$ is $p_i = 0.5 \text{ atm}$. Using the derived expression: $p_{N_2O_5} = 1.5 - 2p_t$ At $t = 100$ s, $p_t = 0.512$ atm. $p_{N_2O_5} = 1.5 - 2(0.512) = 1.5 - 1.024 = 0.476 \text{ atm}$

2. Calculate the rate constant k: $k = \frac{2.303}{100 \text{ s}} \log \frac{0.5 \text{ atm}}{0.476 \text{ atm}}$ $k = \frac{2.303}{100} \log(1.0504) \text{ s}^{-1}$ Using $\log(1.0504) \approx 0.0216$: $k = \frac{2.303}{100} \times 0.0216 \text{ s}^{-1} = 4.98 \times 10^{-4} \text{ s}^{-1}$

Final Answer The rate constant is $4.98 \times 10^{-4} \text{ s}^{-1}$.

Half-Life of a Reaction

The half-life ($t_{1/2}$) of a reaction is the time required for the concentration of a reactant to decrease to one half of its initial value.

• For a Zero-Order Reaction: $t_{1/2} = \frac{[R]_0}{2k}$ The half-life is directly proportional to the initial concentration.

• For a First-Order Reaction: $t_{1/2} = \frac{2.303}{k} \log(2) = \frac{0.693}{k}$ The half-life is constant and does not depend on the initial concentration.

Given

• Rate constant, $k = 5.5 \times 10^{-14} s^{-1}$
• Reaction order = First order

To Find

Half-life, $t_{1/2}$

Formula

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

Solution

$t_{1/2} = \frac{0.693}{5.5 \times 10^{-14} s^{-1}} = 1.26 \times 10^{13} s$

Final Answer The half-life of the reaction is $1.26 \times 10^{13} s$.

Given

• Reaction completion = 99.9%
• Reaction order = First order

To Find

Show that $t_{99.9\%} = 10 \times t_{1/2}$

Formula

$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$ $t_{1/2} = \frac{0.693}{k}$

Solution

Step 1: Calculate the time for 99.9% completion ($t_{99.9\%}$) If the reaction is 99.9% complete, the amount of reactant remaining is: $[R] = [R]_0 - 0.999[R]_0 = 0.001[R]_0$ Now, substitute this into the integrated rate law: $t_{99.9\%} = \frac{2.303}{k} \log \frac{[R]_0}{0.001[R]_0} = \frac{2.303}{k} \log \frac{1}{0.001}$ $t_{99.9\%} = \frac{2.303}{k} \log(1000) = \frac{2.303}{k} \log(10^3)$ Since $\log(10^3) = 3$: $t_{99.9\%} = \frac{2.303 \times 3}{k} = \frac{6.909}{k}$

Step 2: Compare $t_{99.9\%}$ with $t_{1/2}$ Take the ratio of the two times: $\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909/k}{0.693/k} = \frac{6.909}{0.693} \approx 10$ Therefore, $t_{99.9\%} = 10 \times t_{1/2}$.

Final Answer The time required for 99.9% completion of a first-order reaction is indeed 10 times its half-life.

Pseudo First-Order Reactions

Some reactions that are chemically of a higher order behave like first-order reactions under certain conditions. These are called pseudo first-order reactions.

This typically happens when one of the reactants is present in a very large excess (like the solvent). Its concentration remains almost constant throughout the reaction, so the reaction rate appears to depend only on the concentration of the other reactant.

Temperature Dependence of the Rate of a Reaction

Most chemical reaction rates increase significantly with an increase in temperature. A common rule of thumb is that for many reactions, the rate constant nearly doubles for every 10°C (or 10 K) rise in temperature.

This relationship is described quantitatively by the Arrhenius equation. $k = A e^{-E_a/RT}$ Where:

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

Activation Energy and the Activated Complex

For a reaction to happen, reactant molecules must collide with enough energy to overcome an energy barrier. This minimum energy is the activation energy ($E_a$).

When molecules collide with sufficient energy, they form a temporary, unstable, high-energy species called the activated complex (or transition state). This complex then breaks apart to form the products.

The activation energy is the energy difference between the reactants and the activated complex.

The Maxwell-Boltzmann Distribution

Molecules in a sample do not all have the same kinetic energy. The Maxwell-Boltzmann distribution curve shows the fraction of molecules possessing a certain kinetic energy.

• When temperature increases, the curve flattens and shifts to the right.
• This means that at a higher temperature, a larger fraction of molecules have kinetic energy equal to or greater than the activation energy ($E_a$).
• Because more molecules can overcome the energy barrier, the reaction rate increases. The factor $e^{-E_a/RT}$ in the Arrhenius equation represents this fraction of effective molecules.

Determining Activation Energy

By taking the natural logarithm of the Arrhenius equation, we get a linear equation: $\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A$ This is in the form $y = mx + c$. A plot of $\ln k$ versus $1/T$ gives a straight line with:

• Slope = $-E_a/R$
• Y-intercept = $\ln A$

If we know the rate constants ($k_1$ and $k_2$) at two different temperatures ($T_1$ and $T_2$), we can calculate $E_a$ using the following equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$

Given

• $T_1 = 500 \text{ K}$, $k_1 = 0.02 \text{ s}^{-1}$
• $T_2 = 700 \text{ K}$, $k_2 = 0.07 \text{ s}^{-1}$
• $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$

To Find

• Activation energy, $E_a$
• Arrhenius factor, A

Formula

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$ $k = A e^{-E_a/RT}$

Solution

(i) Calculate Activation Energy, $E_a$ $\log \frac{0.07}{0.02} = \frac{E_a}{2.303 \times 8.314 \text{ J K}^{-1} \text{mol}^{-1}} \left[ \frac{700 - 500}{700 \times 500} \right]$ $\log(3.5) = \frac{E_a}{19.147} \left[ \frac{200}{350000} \right]$ $0.544 = \frac{E_a}{19.147} (5.714 \times 10^{-4})$ $E_a = \frac{0.544 \times 19.147}{5.714 \times 10^{-4}} = 18230.8 \text{ J mol}^{-1}$

(ii) Calculate Arrhenius Factor, A Rearrange the Arrhenius equation: $A = k e^{E_a/RT}$. Use the data from either temperature (e.g., $T_1 = 500$ K). $A = (0.02 \text{ s}^{-1}) \times e^{(18230.8 / (8.314 \times 500))}$ $A = 0.02 \times e^{4.385}$ $A = 0.02 \times 80.25 \approx 1.61 \text{ s}^{-1}$ (The source text calculation $A = 0.02/0.012 = 1.61$ seems to have calculated $e^{-Ea/RT}$ as 0.012, which is correct.)

Final Answer The activation energy $E_a$ is $18230.8 \text{ J mol}^{-1}$ (or 18.23 kJ mol⁻¹). The Arrhenius factor A is approximately $1.61 \text{ s}^{-1}$.

Given

• $T_1 = 600 \text{ K}$, $k_1 = 1.60 \times 10^{-5} \text{s}^{-1}$
• $T_2 = 700 \text{ K}$
• $E_a = 209 \text{ kJ/mol} = 209000 \text{ J mol}^{-1}$
• $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$

To Find

Rate constant at 700 K, $k_2$

Formula

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$ Which can be rearranged to: $\log k_2 = \log k_1 + \frac{E_a}{2.303 R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$

Solution

$\log k_2 = \log(1.60 \times 10^{-5}) + \frac{209000 \text{ J mol}^{-1}}{2.303 \times 8.314 \text{ J K}^{-1} \text{mol}^{-1}} \left[ \frac{1}{600 \text{ K}} - \frac{1}{700 \text{ K}} \right]$ $\log k_2 = -4.796 + \frac{209000}{19.147} \left[ 0.001667 - 0.001429 \right]$ $\log k_2 = -4.796 + 10915.5 \times (0.000238)$ $\log k_2 = -4.796 + 2.599 = -2.197$ To find $k_2$, we take the antilog: $k_2 = 10^{-2.197} = 6.36 \times 10^{-3} \text{ s}^{-1}$

Final Answer The rate constant at 700 K is $6.36 \times 10^{-3} \text{ s}^{-1}$.

Effect of Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. A substance that reduces the rate is called an inhibitor.

A catalyst works by providing an alternative reaction pathway with a lower activation energy ($E_a$). By lowering the energy barrier, a larger fraction of reactant molecules have enough energy to react, thus increasing the reaction rate.

Key properties of a catalyst:

• It does not alter the Gibbs energy ($\Delta G$) of a reaction; it cannot make a non-spontaneous reaction spontaneous.
• It does not change the equilibrium constant of a reaction.
• It helps a reaction reach equilibrium faster by speeding up both the forward and reverse reactions to the same extent.

Collision Theory of Chemical Reactions

The collision theory provides a more detailed picture of how reactions occur at the molecular level. It is based on the kinetic theory of gases and has two main ideas:

1. For a reaction to occur, reactant molecules must collide with each other.
2. Not all collisions result in a reaction.

For a collision to be an effective collision (one that leads to the formation of products), two conditions must be met:

1. Energy Factor: The colliding molecules must possess a minimum amount of kinetic energy, called the threshold energy. This energy is needed to break existing bonds and is related to the activation energy.
2. Orientation Factor: The molecules must collide in a proper orientation that allows for the formation of new bonds. If the orientation is improper, the molecules will simply bounce off each other, even if they have sufficient energy.

The rate of reaction can be expressed by the equation: $\text{Rate} = P Z_{AB} e^{-E_a/RT}$ Where:

• $Z_{AB}$ is the collision frequency (the number of collisions between reactants A and B per second per unit volume).
• $e^{-E_a/RT}$ is the fraction of molecules with energy greater than or equal to $E_a$.
• P is the probability or steric factor, which accounts for the fact that collisions must have the correct orientation.

In essence, collision theory states that the rate of a reaction depends on the frequency of collisions that are both sufficiently energetic and correctly oriented.