Answer: (iii) <Δ U^ominus Explanation: The relationship between enthalpy change ( Δ H ) and internal energy change ( Δ U ) is given by: Δ H = Δ U + Δ n_g RT where Δ n_g is the change in the number of moles of gaseous substances. The combustion reaction for methane is: CH_4(g) + 2O_2(g) → CO_2(g) +…

Q4 (ii): >Δ U^ominus

Answer: (iii) <Δ U^ominus Explanation: The relationship between enthalpy change ( Δ H ) and internal energy change ( Δ U ) is given by: Δ H = Δ U + Δ n_g RT where Δ n_g is the change in the number of moles of gaseous substances. The combustion reaction for methane is: CH_4(g) + 2O_2(g) → CO_2(g) +…

Q4 (iii): <Δ U^ominus

Answer: (iii) <Δ U^ominus Explanation: The relationship between enthalpy change ( Δ H ) and internal energy change ( Δ U ) is given by: Δ H = Δ U + Δ n_g RT where Δ n_g is the change in the number of moles of gaseous substances. The combustion reaction for methane is: CH_4(g) + 2O_2(g) → CO_2(g) +…

Q5 (i): -74.8 ~kJ ~mol^-1

Answer: (i) -74.8 ~kJ ~mol^-1 Explanation: To Find: The enthalpy of formation of methane, Δ_f H^ominus(CH_4). Target Reaction: C(graphite) + 2H_2(g) → CH_4(g) Given Reactions (Combustion Enthalpies): 1. CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(l); Δ_c H^ominus = -890.3 ~kJ mol^-1 2. C(graphite) + O_2(g)…

Q5 (ii): -52.27 ~kJ ~mol^-1

Answer: (i) -74.8 ~kJ ~mol^-1 Explanation: To Find: The enthalpy of formation of methane, Δ_f H^ominus(CH_4). Target Reaction: C(graphite) + 2H_2(g) → CH_4(g) Given Reactions (Combustion Enthalpies): 1. CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(l); Δ_c H^ominus = -890.3 ~kJ mol^-1 2. C(graphite) + O_2(g)…

Q5 (iii): +74.8 ~kJ ~mol^-1

Answer: (i) -74.8 ~kJ ~mol^-1 Explanation: To Find: The enthalpy of formation of methane, Δ_f H^ominus(CH_4). Target Reaction: C(graphite) + 2H_2(g) → CH_4(g) Given Reactions (Combustion Enthalpies): 1. CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(l); Δ_c H^ominus = -890.3 ~kJ mol^-1 2. C(graphite) + O_2(g)…

Q5 (iv): +52.26 ~kJ ~mol^-1.

Answer: (i) -74.8 ~kJ ~mol^-1 Explanation: To Find: The enthalpy of formation of methane, Δ_f H^ominus(CH_4). Target Reaction: C(graphite) + 2H_2(g) → CH_4(g) Given Reactions (Combustion Enthalpies): 1. CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(l); Δ_c H^ominus = -890.3 ~kJ mol^-1 2. C(graphite) + O_2(g)…

Thermodynamics Class 11 Chemistry NCERT Solutions

Q1EXERCISES

Choose the correct answer. A thermodynamic state function is a quantity

(i)

used to determine heat changes

(ii)

whose value is independent of path

(iii)

used to determine pressure volume work

(iv)

whose value depends on temperature only.

Solution

Answer: (ii) whose value is independent of path

Explanation: A thermodynamic state function is a property of a system that depends only on its current state, not on the path taken to reach that state. Examples include internal energy (U), enthalpy (H), entropy (S), and Gibbs free energy (G). In contrast, quantities like heat (q) and work (w) are path functions because their values depend on the specific process followed.

Q2EXERCISES

For the process to occur under adiabatic conditions, the correct condition is:

(i)

$\Delta T=0$

(ii)

$\Delta p=0$

(iii)

$q=0$

(iv)

$\mathrm{w}=0$

Solution

Answer: (iii)

q=0

Explanation: An adiabatic process is defined as one in which there is no heat transfer between the system and its surroundings. Therefore, the heat change,

q

, is equal to zero. For an adiabatic process, the first law of thermodynamics becomes

\Delta U = w

.

Q3EXERCISES

The enthalpies of all elements in their standard states are:

(i)

unity

(ii)

zero

(iii)

< 0

(iv)

different for each element

Solution

Answer: (ii) zero

Explanation: By convention, the standard enthalpy of formation (

\Delta_f H^{\ominus}

) of an element in its most stable form (its standard state) at 298 K and 1 bar pressure is defined as zero. This serves as a reference point for calculating the enthalpies of formation of compounds.

Q4EXERCISES

$\Delta U^{\ominus}$ of combustion of methane is $-\mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}$ . The value of $\Delta H^{\ominus}$ is

(i)

$=\Delta U^{\ominus}$

(ii)

$>\Delta U^{\ominus}$

(iii)

$<\Delta U^{\ominus}$

(iv)

$=0$

Solution

Answer: (iii)

<\Delta U^{\ominus}

Explanation: The relationship between enthalpy change (

\Delta H

) and internal energy change (

\Delta U

) is given by:

\Delta H = \Delta U + \Delta n_g RT

where

\Delta n_g

is the change in the number of moles of gaseous substances.

The combustion reaction for methane is:

\mathrm{CH}_{4}(\mathrm{g}) + 2\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{l})

Now, we calculate

\Delta n_g

:

\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})

\Delta n_g = (1) - (1 + 2) = -2

Substituting this into the equation:

\Delta H^{\ominus} = \Delta U^{\ominus} + (-2)RT

\Delta H^{\ominus} = \Delta U^{\ominus} - 2RT

Since R (gas constant) and T (temperature) are positive values, the term

2RT

is positive. Therefore,

\Delta H^{\ominus}

is less than

\Delta U^{\ominus}

.

Q5EXERCISES

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1} -393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ , and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Enthalpy of formation of $\mathrm{CH}_{4}(\mathrm{g})$ will be

(i)

$\quad -74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii)

$\quad -52.27 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii)

$\quad +74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iv)

$\quad +52.26 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Solution

Answer: (i)

\quad -74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}

Explanation: To Find: The enthalpy of formation of methane,

\Delta_f H^{\ominus}(\mathrm{CH}_4)

.

Target Reaction:

\mathrm{C}(\text{graphite}) + 2\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})

Given Reactions (Combustion Enthalpies):

$\mathrm{CH}_{4}(\mathrm{g}) + 2\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{l}); \quad \Delta_c H^{\ominus} = -890.3 \mathrm{~kJ} \mathrm{mol}^{-1}$
$\mathrm{C}(\text{graphite}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}); \quad \Delta_c H^{\ominus} = -393.5 \mathrm{~kJ} \mathrm{mol}^{-1}$
$\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2}\mathrm{O}(\mathrm{l}); \quad \Delta_c H^{\ominus} = -285.8 \mathrm{~kJ} \mathrm{mol}^{-1}$

Applying Hess's Law: We need to manipulate the given equations to obtain the target reaction.

Keep equation (2) as it is to get $\mathrm{C}(\text{graphite})$ on the reactant side.
Multiply equation (3) by 2 to get $2\mathrm{H}_{2}(\mathrm{g})$ on the reactant side.
Reverse equation (1) to get $\mathrm{CH}_{4}(\mathrm{g})$ on the product side.

Step-by-step calculation:

Eq (2): $\mathrm{C}(\text{graphite}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}); \quad \Delta H = -393.5 \mathrm{~kJ}$
2 x Eq (3): $2\mathrm{H}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{H}_{2}\mathrm{O}(\mathrm{l}); \quad \Delta H = 2 \times (-285.8) = -571.6 \mathrm{~kJ}$
Reverse Eq (1): $\mathrm{CO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_{4}(\mathrm{g}) + 2\mathrm{O}_{2}(\mathrm{g}); \quad \Delta H = +890.3 \mathrm{~kJ}$

Summing the manipulated equations:

(\mathrm{C} + \mathrm{O}_2) + (2\mathrm{H}_2 + \mathrm{O}_2) + (\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}) \rightarrow (\mathrm{CO}_2) + (2\mathrm{H}_2\mathrm{O}) + (\mathrm{CH}_4 + 2\mathrm{O}_2)

Canceling common species on both sides (

\mathrm{CO}_2, 2\mathrm{H}_2\mathrm{O}, 2\mathrm{O}_2

):

\mathrm{C}(\text{graphite}) + 2\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{4}(\mathrm{g})

Calculating the total enthalpy change:

\Delta_f H^{\ominus}(\mathrm{CH}_4) = (-393.5) + (-571.6) + (890.3)

\Delta_f H^{\ominus}(\mathrm{CH}_4) = -965.1 + 890.3 = -74.8 \mathrm{~kJ} \mathrm{mol}^{-1}

Q6EXERCISES

A reaction, $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}+\mathrm{q}$ is found to have a positive entropy change. The reaction will be

(i)

possible at high temperature

(ii)

possible only at low temperature

(iii)

not possible at any temperature

(iv)

possible at any temperature

Solution

Answer: (iv) possible at any temperature

Explanation: The spontaneity of a reaction is determined by the Gibbs free energy change (

\Delta G

):

\Delta G = \Delta H - T\Delta S

A reaction is spontaneous if

\Delta G < 0

.

From the given information:

The reaction is $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}+\mathrm{q}$ . The term '+q' indicates that heat is released, so the reaction is exothermic. Therefore, the enthalpy change $\Delta H$ is negative ( $\Delta H < 0$ ).
The reaction has a positive entropy change. Therefore, $\Delta S$ is positive ( $\Delta S > 0$ ).

Substituting these signs into the Gibbs equation:

\Delta G = (\text{negative value}) - T(\text{positive value})

Since temperature (T) is always positive in Kelvin, the term

T\Delta S

is always positive. The equation becomes:

\Delta G = (\text{negative}) - (\text{positive}) = (\text{negative})

Since both terms on the right side are negative,

\Delta G

will always be negative, regardless of the temperature. Therefore, the reaction is spontaneous at all temperatures.

Q7EXERCISES

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Solution

Given: Heat absorbed by the system,

q = +701 \mathrm{~J}

Work done by the system,

w = -394 \mathrm{~J}

To Find: The change in internal energy,

\Delta U

.

Formula: According to the First Law of Thermodynamics:

\Delta U = q + w

Solution: Substitute the given values into the formula:

\Delta U = (+701 \mathrm{~J}) + (-394 \mathrm{~J})

\Delta U = 701 - 394

\Delta U = 307 \mathrm{~J}

Final Answer: The change in internal energy for the process is

307 \mathrm{~J}

.

Q8EXERCISES

The reaction of cyanamide, $\mathrm{NH}_{2} \mathrm{CN}(\mathrm{s})$ , with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at 298 K . Calculate enthalpy change for the reaction at 298 K . $\mathrm{NH}_{2} \mathrm{CN}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Solution

Given: Internal energy change,

\Delta U = -742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}

Temperature,

T = 298 \mathrm{~K}

Gas constant,

R = 8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} = 8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

Reaction:

\mathrm{NH}_{2} \mathrm{CN}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

To Find: Enthalpy change,

\Delta H

.

Formula:

\Delta H = \Delta U + \Delta n_g RT

Solution: First, calculate the change in the number of moles of gas,

\Delta n_g

.

\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})

From the balanced equation:

Moles of gaseous products = Moles of $\mathrm{N}_{2}(\mathrm{g})$ + Moles of $\mathrm{CO}_{2}(\mathrm{g}) = 1 + 1 = 2 \mathrm{~mol}$
Moles of gaseous reactants = Moles of $\mathrm{O}_{2}(\mathrm{g}) = \frac{3}{2} = 1.5 \mathrm{~mol}$

Therefore,

\Delta n_g = 2 - 1.5 = 0.5 \mathrm{~mol}

Now, substitute the values into the formula for

\Delta H

:

\Delta H = -742.7 \mathrm{~kJ} \mathrm{~mol}^{-1} + (0.5 \mathrm{~mol}) \times (8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}) \times (298 \mathrm{~K})

\Delta H = -742.7 + (0.5 \times 8.314 \times 10^{-3} \times 298)

\Delta H = -742.7 + 1.2387

\Delta H \approx -741.46 \mathrm{~kJ} \mathrm{~mol}^{-1}

Final Answer: The enthalpy change for the reaction is approximately

-741.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

Q9EXERCISES

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from $35^{\circ} \mathrm{C}$ to $55^{\circ} \mathrm{C}$ . Molar heat capacity of Al is $24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ .

Solution

Given: Mass of aluminium,

m = 60.0 \mathrm{~g}

Initial temperature,

T_1 = 35^{\circ} \mathrm{C}

Final temperature,

T_2 = 55^{\circ} \mathrm{C}

Molar heat capacity of Al,

C_m = 24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

Molar mass of Al,

M = 27.0 \mathrm{~g} \mathrm{~mol}^{-1}

To Find: Heat required,

q

, in kJ.

Formula:

q = n \times C_m \times \Delta T

where

n

is the number of moles and

\Delta T

is the change in temperature.

Solution:

Calculate the number of moles (n): $n = \frac{\text{mass}}{\text{molar mass}} = \frac{60.0 \mathrm{~g}}{27.0 \mathrm{~g} \mathrm{mol}^{-1}} = 2.222 \mathrm{~mol}$
Calculate the change in temperature ( $\Delta T$ ): $\Delta T = T_2 - T_1 = 55^{\circ} \mathrm{C} - 35^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}$ A change of $20^{\circ} \mathrm{C}$ is equal to a change of $20 \mathrm{~K}$ . So, $\Delta T = 20 \mathrm{~K}$ .
Calculate the heat required (q): $q = (2.222 \mathrm{~mol}) \times (24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}) \times (20 \mathrm{~K})$ $q = 1066.56 \mathrm{~J}$
Convert the heat from J to kJ: $q = \frac{1066.56 \mathrm{~J}}{1000 \mathrm{~J/kJ}} = 1.06656 \mathrm{~kJ}$

Final Answer: The heat necessary is approximately

1.07 \mathrm{~kJ}

.

Q10EXERCISES

Calculate the enthalpy change on freezing of 1.0 mol of water at $10.0^{\circ} \mathrm{C}$ to ice at $-10.0^{\circ} \mathrm{C}$ . $\Delta_{\text{fus}} H=6.03 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $0^{\circ} \mathrm{C}$ . $C_{p}[\mathrm{H}_{2} \mathrm{O}(\mathrm{l})]=75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ $C_{p}[\mathrm{H}_{2} \mathrm{O}(\mathrm{s})]=36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Solution

Given: Amount of water,

n = 1.0 \mathrm{~mol}

Initial temperature =

10.0^{\circ} \mathrm{C} = 283.15 \mathrm{~K}

Final temperature =

-10.0^{\circ} \mathrm{C} = 263.15 \mathrm{~K}

Enthalpy of fusion,

\Delta_{\text{fus}} H = 6.03 \mathrm{~kJ} \mathrm{~mol}^{-1}

Molar heat capacity of liquid water,

C_{p,l} = 75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

Molar heat capacity of ice,

C_{p,s} = 36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

To Find: Total enthalpy change,

\Delta H_{total}

.

Solution: The process occurs in three steps:

Step 1: Cooling liquid water from $10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ . $\Delta T_1 = (0^{\circ} \mathrm{C} - 10^{\circ} \mathrm{C}) = -10 \mathrm{~K}$ $\Delta H_1 = n \times C_{p,l} \times \Delta T_1 = (1.0 \mathrm{~mol}) \times (75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}) \times (-10 \mathrm{~K}) = -753 \mathrm{~J}$
Step 2: Freezing water at $0^{\circ} \mathrm{C}$ . Freezing is the reverse of fusion, so the enthalpy change is the negative of the enthalpy of fusion. $\Delta H_2 = -\Delta_{\text{fus}} H = -6.03 \mathrm{~kJ} \mathrm{~mol}^{-1} = -6030 \mathrm{~J} \mathrm{~mol}^{-1}$ For 1 mole, $\Delta H_2 = -6030 \mathrm{~J}$ .
Step 3: Cooling ice from $0^{\circ} \mathrm{C}$ to $-10^{\circ} \mathrm{C}$ . $\Delta T_3 = (-10^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C}) = -10 \mathrm{~K}$ $\Delta H_3 = n \times C_{p,s} \times \Delta T_3 = (1.0 \mathrm{~mol}) \times (36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}) \times (-10 \mathrm{~K}) = -368 \mathrm{~J}$

Total Enthalpy Change: The total enthalpy change is the sum of the enthalpy changes of the three steps.

\Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3

\Delta H_{total} = (-753 \mathrm{~J}) + (-6030 \mathrm{~J}) + (-368 \mathrm{~J})

\Delta H_{total} = -7151 \mathrm{~J}

Converting to kJ:

\Delta H_{total} = -7.151 \mathrm{~kJ}

Final Answer: The enthalpy change on freezing 1.0 mol of water from

10.0^{\circ} \mathrm{C}

to

-10.0^{\circ} \mathrm{C}

is

-7.151 \mathrm{~kJ}

.

Q11EXERCISES

Enthalpy of combustion of carbon to $\mathrm{CO}_{2}$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . Calculate the heat released upon formation of 35.2 g of $\mathrm{CO}_{2}$ from carbon and dioxygen gas.

Solution

Given: Enthalpy of combustion of C to

\mathrm{CO}_{2}

,

\Delta_c H = -393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

Mass of

\mathrm{CO}_{2}

formed =

35.2 \mathrm{~g}

To Find: Heat released for the formation of 35.2 g of

\mathrm{CO}_{2}

.

Solution: The combustion reaction is:

\mathrm{C}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})

The given enthalpy change corresponds to the formation of 1 mole of

\mathrm{CO}_{2}

.

Calculate the molar mass of $\mathrm{CO}_{2}$ : Molar mass = $12.01 + 2 \times 16.00 = 44.01 \mathrm{~g} \mathrm{~mol}^{-1}$
Calculate the number of moles of $\mathrm{CO}_{2}$ in 35.2 g: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{35.2 \mathrm{~g}}{44.01 \mathrm{~g} \mathrm{mol}^{-1}} \approx 0.7998 \mathrm{~mol}$
Calculate the total heat released: Heat released = (moles of $\mathrm{CO}_{2}$ ) $\times$ (heat released per mole) Heat released = $0.7998 \mathrm{~mol} \times 393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Heat released $\approx 314.52 \mathrm{~kJ}$

Final Answer: The heat released upon the formation of 35.2 g of

\mathrm{CO}_{2}

is approximately

314.5 \mathrm{~kJ}

.

Q12EXERCISES

Enthalpies of formation of $\mathrm{CO}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{g}), \mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ and $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})$ are $-110, -393, 81$ and $9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Find the value of $\Delta_{r} H$ for the reaction: $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+3 \mathrm{CO}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{~g})+3 \mathrm{CO}_{2}(\mathrm{~g})$

Solution

Given: Reaction:

\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+3 \mathrm{CO}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{~g})+3 \mathrm{CO}_{2}(\mathrm{~g})

Standard enthalpies of formation (

\Delta_f H^{\ominus}

):

$\Delta_f H^{\ominus}(\mathrm{CO}, \mathrm{g}) = -110 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta_f H^{\ominus}(\mathrm{CO}_2, \mathrm{g}) = -393 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta_f H^{\ominus}(\mathrm{N}_2\mathrm{O}, \mathrm{g}) = 81 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta_f H^{\ominus}(\mathrm{N}_2\mathrm{O}_4, \mathrm{g}) = 9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

To Find: The enthalpy change for the reaction,

\Delta_r H

.

Formula:

\Delta_r H^{\ominus} = \sum \Delta_f H^{\ominus}(\text{products}) - \sum \Delta_f H^{\ominus}(\text{reactants})

Solution: Applying the formula to the given reaction:

\Delta_r H = [\Delta_f H^{\ominus}(\mathrm{N}_{2} \mathrm{O}) + 3 \times \Delta_f H^{\ominus}(\mathrm{CO}_{2})] - [\Delta_f H^{\ominus}(\mathrm{N}_{2} \mathrm{O}_{4}) + 3 \times \Delta_f H^{\ominus}(\mathrm{CO})]

Substitute the given values:

\Delta_r H = [(81) + 3 \times (-393)] - [(9.7) + 3 \times (-110)]

\Delta_r H = [81 - 1179] - [9.7 - 330]

\Delta_r H = [-1098] - [-320.3]

\Delta_r H = -1098 + 320.3

\Delta_r H = -777.7 \mathrm{~kJ} \mathrm{~mol}^{-1}

Final Answer: The value of

\Delta_r H

for the reaction is

-777.7 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

Q13EXERCISES

Given $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta_{r} H^{\ominus}=-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ What is the standard enthalpy of formation of $\mathrm{NH}_{3}$ gas?

Solution

Given: The reaction for the formation of 2 moles of ammonia:

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})

Standard enthalpy change for this reaction,

\Delta_{r} H^{\ominus}=-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

To Find: The standard enthalpy of formation of

\mathrm{NH}_{3}

gas,

\Delta_{f} H^{\ominus}(\mathrm{NH}_3)

.

Solution: The standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. The given reaction produces two moles of

\mathrm{NH}_{3}

.

The reaction for the formation of one mole of

\mathrm{NH}_{3}

is:

\frac{1}{2}\mathrm{N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{NH}_{3}(\mathrm{~g})

To find the enthalpy change for this reaction, we divide the given enthalpy change by 2:

\Delta_{f} H^{\ominus}(\mathrm{NH}_3) = \frac{\Delta_{r} H^{\ominus}}{2}

\Delta_{f} H^{\ominus}(\mathrm{NH}_3) = \frac{-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}}{2}

\Delta_{f} H^{\ominus}(\mathrm{NH}_3) = -46.2 \mathrm{~kJ} \mathrm{~mol}^{-1}

Final Answer: The standard enthalpy of formation of

\mathrm{NH}_{3}

gas is

-46.2 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

Q14EXERCISES

Calculate the standard enthalpy of formation of $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})$ from the following data: $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{r} H^{\ominus}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}$ C (graphite) $+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{c} H^{\ominus}=-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{f} H^{\ominus}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Solution

To Find: The standard enthalpy of formation of methanol,

\Delta_f H^{\ominus}(\mathrm{CH}_3\mathrm{OH})

.

Target Reaction:

\mathrm{C}(\text{graphite}) + 2\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{3}\mathrm{OH}(\mathrm{l})

Given Reactions:

$\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}); \quad \Delta H_1 = -726 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}(\text{graphite}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}); \quad \Delta H_2 = -393 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}); \quad \Delta H_3 = -286 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Applying Hess's Law: We need to combine the given equations to obtain the target reaction.

Reverse equation (1) to get $\mathrm{CH}_{3}\mathrm{OH}(\mathrm{l})$ on the product side.
Keep equation (2) as it is for $\mathrm{C}(\text{graphite})$ on the reactant side.
Multiply equation (3) by 2 for $2\mathrm{H}_{2}(\mathrm{g})$ on the reactant side.

Step-by-step manipulation:

Reverse Eq (1): $\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}); \quad \Delta H = -(-726) = +726 \mathrm{~kJ}$
Eq (2): $\mathrm{C}(\text{graphite}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}); \quad \Delta H = -393 \mathrm{~kJ}$
2 x Eq (3): $2\mathrm{H}_{2}(\mathrm{~g}) + \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}); \quad \Delta H = 2 \times (-286) = -572 \mathrm{~kJ}$

Summing the manipulated equations:

(\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}) + (\mathrm{C} + \mathrm{O}_2) + (2\mathrm{H}_2 + \mathrm{O}_2) \rightarrow (\mathrm{CH}_3\mathrm{OH} + \frac{3}{2}\mathrm{O}_2) + (\mathrm{CO}_2) + (2\mathrm{H}_2\mathrm{O})

Canceling common species (

\mathrm{CO}_2, 2\mathrm{H}_2\mathrm{O}, \frac{3}{2}\mathrm{O}_2

from the total

2\mathrm{O}_2

on reactant side):

\mathrm{C}(\text{graphite}) + 2\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{3}\mathrm{OH}(\mathrm{l})

Calculating the total enthalpy change:

\Delta_f H^{\ominus}(\mathrm{CH}_3\mathrm{OH}) = (+726) + (-393) + (-572)

\Delta_f H^{\ominus}(\mathrm{CH}_3\mathrm{OH}) = 726 - 965 = -239 \mathrm{~kJ} \mathrm{~mol}^{-1}

Final Answer: The standard enthalpy of formation of

\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})

is

-239 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

Q15EXERCISES

Calculate the enthalpy change for the process $\mathrm{CCl}_{4}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})$ and calculate bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ in $\mathrm{CCl}_{4}(\mathrm{~g})$ . $\Delta_{\text{vap}} H^{\ominus}(\mathrm{CCl}_{4})=30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . $\Delta_{f} H^{\ominus}(\mathrm{CCl}_{4})=-135.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . $\Delta_{a} H^{\ominus}(\mathrm{C})=715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ , where $\Delta_{a} H^{\ominus}$ is enthalpy of atomisation $\Delta_{a} H^{\ominus}(\mathrm{Cl}_{2})=242 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

Part 1: Calculate enthalpy change for the process $\mathrm{CCl}_{4}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})$

This process represents the enthalpy of atomization of gaseous

\mathrm{CCl}_4

. We can calculate it using a Hess's Law cycle with the following reactions:

Formation of $\mathrm{CCl}_4(\mathrm{l})$ : $\mathrm{C}(\text{graphite}) + 2\mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{CCl}_4(\mathrm{l}); \quad \Delta_f H^{\ominus} = -135.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Vaporization of $\mathrm{CCl}_4(\mathrm{l})$ : $\mathrm{CCl}_4(\mathrm{l}) \rightarrow \mathrm{CCl}_4(\mathrm{g}); \quad \Delta_{\text{vap}} H^{\ominus} = 30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Atomization of Carbon: $\mathrm{C}(\text{graphite}) \rightarrow \mathrm{C}(\mathrm{g}); \quad \Delta_a H^{\ominus} = 715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Atomization of Chlorine: $\mathrm{Cl}_2(\mathrm{g}) \rightarrow 2\mathrm{Cl}(\mathrm{g}); \quad \Delta_a H^{\ominus} = 242 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Target Reaction:

\mathrm{CCl}_{4}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})

Applying Hess's Law: We can express the enthalpy of atomization (

\Delta_a H^{\ominus}

) as:

\Delta_a H^{\ominus}(\mathrm{CCl}_4, \mathrm{g}) = \Delta_f H^{\ominus}(\mathrm{C}, \mathrm{g}) + 4\Delta_f H^{\ominus}(\mathrm{Cl}, \mathrm{g}) - \Delta_f H^{\ominus}(\mathrm{CCl}_4, \mathrm{g})

First, we need to find the values for each term:

$\Delta_f H^{\ominus}(\mathrm{C}, \mathrm{g}) = \Delta_a H^{\ominus}(\mathrm{C}) = 715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
From reaction 4, the enthalpy to form 2 moles of Cl(g) is 242 kJ. So, for 1 mole: $\Delta_f H^{\ominus}(\mathrm{Cl}, \mathrm{g}) = \frac{1}{2} \Delta_a H^{\ominus}(\mathrm{Cl}_2) = \frac{242}{2} = 121 \mathrm{~kJ} \mathrm{~mol}^{-1}$
We need $\Delta_f H^{\ominus}(\mathrm{CCl}_4, \mathrm{g})$ . We can get this by adding reactions 1 and 2: $(\mathrm{C} + 2\mathrm{Cl}_2 \rightarrow \mathrm{CCl}_4(\mathrm{l})) + (\mathrm{CCl}_4(\mathrm{l}) \rightarrow \mathrm{CCl}_4(\mathrm{g})) \Rightarrow \mathrm{C}(\text{graphite}) + 2\mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{CCl}_4(\mathrm{g})$ $\Delta_f H^{\ominus}(\mathrm{CCl}_4, \mathrm{g}) = \Delta_f H^{\ominus}(\mathrm{CCl}_4, \mathrm{l}) + \Delta_{\text{vap}} H^{\ominus}(\mathrm{CCl}_4) = -135.5 + 30.5 = -105.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Now, calculate the enthalpy change for the target process:

\Delta H = (715.0) + 4(121) - (-105.0)

\Delta H = 715.0 + 484 + 105.0

\Delta H = 1304.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

Part 2: Calculate bond enthalpy of C-Cl in $\mathrm{CCl}_{4}(\mathrm{~g})$

The enthalpy change calculated in Part 1 is the energy required to break all four C-Cl bonds in one mole of

\mathrm{CCl}_4

. The C-Cl bond enthalpy is the average energy per bond.

\text{Bond Enthalpy (C-Cl)} = \frac{\text{Total Atomization Enthalpy}}{4}

\text{Bond Enthalpy (C-Cl)} = \frac{1304.0 \mathrm{~kJ} \mathrm{~mol}^{-1}}{4}

\text{Bond Enthalpy (C-Cl)} = 326.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

Final Answer: The enthalpy change for the process is

1304.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

. The bond enthalpy of C-Cl in

\mathrm{CCl}_{4}

is

326.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

Q16EXERCISES

For an isolated system, $\Delta U=0$ , what will be $\Delta S$ ?

Solution

Explanation: An isolated system is one that cannot exchange energy (heat or work) or matter with its surroundings. For any process occurring in such a system, the change in internal energy is zero, i.e.,

\Delta U = 0

.

The Second Law of Thermodynamics states that for any spontaneous process, the total entropy of the universe (or an isolated system) must increase. For a process at equilibrium, the entropy is maximum and does not change.

Therefore, for an isolated system:

If a process is spontaneous, the entropy will increase: $\Delta S > 0$ .
If the system is at equilibrium, the entropy will not change: $\Delta S = 0$ .

Combining these, for any process (spontaneous or at equilibrium) in an isolated system, the change in entropy must be greater than or equal to zero.

Final Answer: For a spontaneous process in an isolated system,

\Delta S > 0

. For a system at equilibrium,

\Delta S = 0

. In general, for any process in an isolated system,

\Delta S \ge 0

.

Q17EXERCISES

For the reaction at 298 K , $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}$ $\Delta H=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range.

Solution

Given: Enthalpy change,

\Delta H = 400 \mathrm{~kJ} \mathrm{~mol}^{-1}

Entropy change,

\Delta S = 0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

To Find: The temperature (T) at which the reaction becomes spontaneous.

Formula: The spontaneity of a reaction is determined by the Gibbs free energy change:

\Delta G = \Delta H - T\Delta S

A reaction is spontaneous when

\Delta G < 0

.

Solution: For the reaction to be spontaneous, we must have:

\Delta H - T\Delta S < 0

Substitute the given values:

400 \mathrm{~kJ} \mathrm{~mol}^{-1} - T \times (0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}) < 0

Rearrange the inequality to solve for T:

400 < T \times 0.2

T > \frac{400}{0.2}

T > 2000 \mathrm{~K}

Final Answer: The reaction will become spontaneous at temperatures above

2000 \mathrm{~K}

.

Q18EXERCISES

For the reaction, $2 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g})$ , what are the signs of $\Delta H$ and $\Delta S$ ?

Solution

Explanation: Consider the reaction:

2 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g})

Sign of $\Delta H$ (Enthalpy Change): The reaction involves the formation of a chemical bond (a Cl-Cl covalent bond) from two separate chlorine atoms. Bond formation is an exothermic process, meaning it releases energy. Therefore, the enthalpy change, $\Delta H$ , is negative. $\Delta H < 0$
Sign of $\Delta S$ (Entropy Change): Entropy is a measure of disorder or randomness. In this reaction, two moles of gaseous atoms combine to form one mole of gaseous molecules. The number of independent particles decreases, leading to a more ordered and less random system. Therefore, the entropy change, $\Delta S$ , is negative. $\Delta S < 0$

Final Answer: For the given reaction, the sign of

\Delta H

is negative, and the sign of

\Delta S

is negative.

Q19EXERCISES

For the reaction $2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow 2 \mathrm{D}(\mathrm{~g})$ $\Delta U^{\ominus}=-10.5 \mathrm{~kJ}$ and $\Delta S^{\ominus}=-44.1 \mathrm{JK}^{-1}$ . Calculate $\Delta G^{\ominus}$ for the reaction, and predict whether the reaction may occur spontaneously.

Solution

Given: Reaction:

2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightarrow 2 \mathrm{D}(\mathrm{~g})

Standard internal energy change,

\Delta U^{\ominus}=-10.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

Standard entropy change,

\Delta S^{\ominus}=-44.1 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} = -0.0441 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

Standard temperature,

T = 298 \mathrm{~K}

Gas constant,

R = 8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} = 8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

To Find: Standard Gibbs energy change,

\Delta G^{\ominus}

, and predict spontaneity.

Solution: Step 1: Calculate the standard enthalpy change ( $\Delta H^{\ominus}$ ). We use the relationship

\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT

. First, find

\Delta n_g

:

\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})

\Delta n_g = (2) - (2 + 1) = -1 \mathrm{~mol}

Now, calculate

\Delta H^{\ominus}

:

\Delta H^{\ominus} = -10.5 \mathrm{~kJ} + (-1 \mathrm{~mol}) \times (8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}) \times (298 \mathrm{~K})

\Delta H^{\ominus} = -10.5 - 2.4776

\Delta H^{\ominus} = -12.9776 \mathrm{~kJ} \mathrm{~mol}^{-1}

Step 2: Calculate the standard Gibbs energy change ( $\Delta G^{\ominus}$ ). We use the formula

\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

.

\Delta G^{\ominus} = -12.9776 \mathrm{~kJ} \mathrm{~mol}^{-1} - (298 \mathrm{~K}) \times (-0.0441 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1})

\Delta G^{\ominus} = -12.9776 + 13.1418

\Delta G^{\ominus} = +0.1642 \mathrm{~kJ} \mathrm{~mol}^{-1}

Step 3: Predict spontaneity. Since the value of

\Delta G^{\ominus}

is positive, the reaction is not spontaneous under standard conditions.

Final Answer:

\Delta G^{\ominus} = +0.1642 \mathrm{~kJ} \mathrm{~mol}^{-1}

. Since

\Delta G^{\ominus} > 0

, the reaction is non-spontaneous.

Q20EXERCISES

The equilibrium constant for a reaction is 10. What will be the value of $\Delta G^{\ominus}$ ? $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~T}=300 \mathrm{~K}$ .

Solution

Given: Equilibrium constant,

K = 10

Gas constant,

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

Temperature,

T = 300 \mathrm{~K}

To Find: Standard Gibbs energy change,

\Delta G^{\ominus}

.

Formula: The relationship between standard Gibbs energy change and the equilibrium constant is:

\Delta G^{\ominus} = -RT \ln K

This can also be written using the base-10 logarithm:

\Delta G^{\ominus} = -2.303 RT \log K

Solution: Substitute the given values into the formula:

\Delta G^{\ominus} = -2.303 \times (8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}) \times (300 \mathrm{~K}) \times \log(10)

Since

\log(10) = 1

:

\Delta G^{\ominus} = -2.303 \times 8.314 \times 300

\Delta G^{\ominus} = -5744.14 \mathrm{~J} \mathrm{~mol}^{-1}

To express the answer in kJ/mol:

\Delta G^{\ominus} = -5.744 \mathrm{~kJ} \mathrm{~mol}^{-1}

Final Answer: The value of

\Delta G^{\ominus}

is

-5.744 \mathrm{~kJ} \mathrm{~mol}^{-1}

.

Q21EXERCISES

Comment on the thermodynamic stability of $\mathrm{NO}(\mathrm{g})$ , given $\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{NO}(\mathrm{g}) ; \quad \Delta_{r} H^{\ominus}=90 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{NO}_{2}(\mathrm{~g}): \quad \Delta_{r} H^{\ominus}=-74 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

Analysis: Thermodynamic stability refers to whether a substance is in its lowest energy state under a given set of conditions. A compound is considered thermodynamically unstable if it has a tendency to decompose into its elements or react to form more stable compounds.

Stability with respect to its elements: The first reaction represents the formation of nitric oxide (NO) from its elements, nitrogen and oxygen. $\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{NO}(\mathrm{g}) ; \quad \Delta_{f} H^{\ominus}=+90 \mathrm{~kJ} \mathrm{~mol}^{-1}$ The standard enthalpy of formation ( $\Delta_{f} H^{\ominus}$ ) is large and positive. This means the formation of NO from N $_2$ and O $_2$ is a highly endothermic process. Consequently, the reverse reaction (decomposition of NO into its elements) is highly exothermic ( $\Delta H = -90 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ). A substance with a positive enthalpy of formation is thermodynamically unstable relative to its constituent elements. Thus, NO is unstable and will tend to decompose back into N $_2$ and O $_2$ .
Stability with respect to oxidation: The second reaction shows the oxidation of NO to nitrogen dioxide (NO $_2$ ). $\mathrm{NO}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{NO}_{2}(\mathrm{~g}): \quad \Delta_{r} H^{\ominus}=-74 \mathrm{~kJ} \mathrm{~mol}^{-1}$ This reaction is exothermic, meaning that NO readily reacts with oxygen to form a more stable oxide, NO $_2$ . This indicates that NO is also unstable with respect to oxidation.

Conclusion: Based on both reactions, nitric oxide (NO) is a thermodynamically unstable compound. It has a high positive enthalpy of formation, making it unstable relative to N

_2

and O

_2

, and it readily undergoes an exothermic reaction to form NO

_2

, making it unstable with respect to oxidation.

Q22EXERCISES

Calculate the entropy change in surroundings when 1.00 mol of $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ is formed under standard conditions. $\Delta_{f} H^{\ominus}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

Solution

Given: Formation of 1.00 mol of

\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

. Standard enthalpy of formation,

\Delta_{f} H^{\ominus}_{\text{sys}} = -286 \mathrm{~kJ} \mathrm{~mol}^{-1}

Standard conditions imply a temperature,

T = 298 \mathrm{~K}

.

To Find: The entropy change in the surroundings,

\Delta S_{\text{surr}}

.

Formula: The entropy change of the surroundings is related to the heat exchanged with the system at constant pressure:

\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T}

where

q_{\text{surr}}

is the heat absorbed by the surroundings. The heat absorbed by the surroundings is equal to the negative of the enthalpy change of the system:

q_{\text{surr}} = -\Delta H_{\text{sys}}

Therefore,

\Delta S_{\text{surr}} = \frac{-\Delta H_{\text{sys}}}{T}

Solution: Substitute the given values into the formula:

\Delta H_{\text{sys}} = -286 \mathrm{~kJ} \mathrm{~mol}^{-1} = -286000 \mathrm{~J} \mathrm{~mol}^{-1}

\Delta S_{\text{surr}} = \frac{-(-286000 \mathrm{~J} \mathrm{~mol}^{-1})}{298 \mathrm{~K}}

\Delta S_{\text{surr}} = \frac{+286000}{298} \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

\Delta S_{\text{surr}} \approx 959.73 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

Final Answer: The entropy change in the surroundings is approximately

+959.7 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

.

ThermodynamicsClass 11 Chemistry NCERT Solutions

Choose the correct answer. A thermodynamic state function is a quantity

used to determine heat changes

whose value is independent of path

used to determine pressure volume work

whose value depends on temperature only.

Solution

For the process to occur under adiabatic conditions, the correct condition is:

ΔT=0\Delta T=0ΔT=0

Δp=0\Delta p=0Δp=0

q=0q=0q=0

w=0\mathrm{w}=0w=0

Solution

The enthalpies of all elements in their standard states are:

unity

zero

< 0

different for each element

Solution

ΔU⊖\Delta U^{\ominus}ΔU⊖ of combustion of methane is −XkJmol−1-\mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}−XkJmol−1. The value of ΔH⊖\Delta H^{\ominus}ΔH⊖ is

=ΔU⊖=\Delta U^{\ominus}=ΔU⊖

>ΔU⊖>\Delta U^{\ominus}>ΔU⊖

<ΔU⊖<\Delta U^{\ominus}<ΔU⊖

=0=0=0

Solution

−74.8 kJ mol−1\quad -74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}−74.8 kJ mol−1

−52.27 kJ mol−1\quad -52.27 \mathrm{~kJ} \mathrm{~mol}^{-1}−52.27 kJ mol−1

+74.8 kJ mol−1\quad +74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}+74.8 kJ mol−1

+52.26 kJ mol−1\quad +52.26 \mathrm{~kJ} \mathrm{~mol}^{-1}+52.26 kJ mol−1.

Solution

A reaction, A+B→C+D+q\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}+\mathrm{q}A+B→C+D+q is found to have a positive entropy change. The reaction will be

possible at high temperature

possible only at low temperature

not possible at any temperature

possible at any temperature

Solution

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Solution

Solution

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35∘C35^{\circ} \mathrm{C}35∘C to 55∘C55^{\circ} \mathrm{C}55∘C. Molar heat capacity of Al is 24 J mol−1 K−124 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}24 J mol−1 K−1.

Solution

Solution

Enthalpy of combustion of carbon to CO2\mathrm{CO}_{2}CO2​ is −393.5 kJ mol−1-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}−393.5 kJ mol−1. Calculate the heat released upon formation of 35.2 g of CO2\mathrm{CO}_{2}CO2​ from carbon and dioxygen gas.

Solution

Solution

Solution

Solution

Solution

For an isolated system, ΔU=0\Delta U=0ΔU=0, what will be ΔS\Delta SΔS ?

Solution

Solution

For the reaction, 2Cl(g)→Cl2(g)2 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g})2Cl(g)→Cl2​(g), what are the signs of ΔH\Delta HΔH and ΔS\Delta SΔS ?

Solution

Solution

The equilibrium constant for a reaction is 10. What will be the value of ΔG⊖\Delta G^{\ominus}ΔG⊖ ? R=8.314JK−1 mol−1, T=300 K\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~T}=300 \mathrm{~K}R=8.314JK−1 mol−1, T=300 K.

Solution

Solution

Calculate the entropy change in surroundings when 1.00 mol of H2O(l)\mathrm{H}_{2} \mathrm{O}(\mathrm{l})H2​O(l) is formed under standard conditions. ΔfH⊖=−286 kJ mol−1\Delta_{f} H^{\ominus}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}Δf​H⊖=−286 kJ mol−1.

Solution

$\Delta T=0$

$\Delta p=0$

$q=0$

$\mathrm{w}=0$

$\Delta U^{\ominus}$ of combustion of methane is $-\mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}$ . The value of $\Delta H^{\ominus}$ is

$=\Delta U^{\ominus}$

$>\Delta U^{\ominus}$

$<\Delta U^{\ominus}$

$=0$

$\quad -74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\quad -52.27 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\quad +74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\quad +52.26 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

A reaction, $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}+\mathrm{q}$ is found to have a positive entropy change. The reaction will be

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from $35^{\circ} \mathrm{C}$ to $55^{\circ} \mathrm{C}$ . Molar heat capacity of Al is $24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ .

Enthalpy of combustion of carbon to $\mathrm{CO}_{2}$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ . Calculate the heat released upon formation of 35.2 g of $\mathrm{CO}_{2}$ from carbon and dioxygen gas.

For an isolated system, $\Delta U=0$ , what will be $\Delta S$ ?

For the reaction, $2 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g})$ , what are the signs of $\Delta H$ and $\Delta S$ ?

The equilibrium constant for a reaction is 10. What will be the value of $\Delta G^{\ominus}$ ? $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~T}=300 \mathrm{~K}$ .

Calculate the entropy change in surroundings when 1.00 mol of $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ is formed under standard conditions. $\Delta_{f} H^{\ominus}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .