# Thermodynamics

## NCERT Solution

### Part 2

Question 12: Enthalpies of formation of CO(g), CO_{2}(g), N_{2}O(g) and N_{2}O_{4}(g) are -110, -393, 81 and 9.7 kJ mol^{-1} respectively. Find the value of Δ_{r}H for the reaction:

N_{2}O_{4}(g) + 3CO(g) → N_{2}O(g) + 3CO_{2}(g)

**Answer:** Δ_{r}H

= 9.7 + 3 × (-110) – (81 – 3 × 393)

= 9.7 – 330 – 81 + 1179 = 777.7 = 778 kJ mol^{-1}

Question 13: Given: N_{2}(g) + 3H_{2}(g) → 2NH_{3}

Δ_{r}H^{Θ} = -92.4 kJ mol^{-1}

What is the standard enthalpy of formation of NH_{3} gas?

**Answer:** Standard enthalpy of formation of NH_{3} = `(-92.4)/2=-46.2` kJ mol^{-1}

Question 14: Calculate the standard enthalpy of formation of CH_{3}OH(l) from the following data:

CH_{3}OH(l) + `3/2`O_{2}(g) → CO_{g} + 2H_{2}O(l)

Δ_{r}H^{θ} = -726 kJ mol^{-1}

C (graphite) + O_{2}(g) → CO_{2}(g)

Δ_{r}H^{θ} = -393 kJ mol^{-1}

H_{2}(g) + `1/2`O_{2}(g) → H_{2}O(l)

Δ_{r}H^{θ} = -286 kJ mol^{-1}

**Answer:** The equation for formation of CH_{3}OH is as follows:

C(s) + 2H_{2}(g) + `1/2`O_{2}(g) → CH_{3}OH(l)

This equation can be achieved as follows:

Equation (2) + 2 × equation (3) – equation (1)

Add: Equation (2) + 2 × equation (3)

C + O_{2} + 2H_{2} + 2O_{2} → CO_{2} + 2H_{2}

↠ C + 2H_{2} + 2O_{2} → CO_{2} + 2H_{2}O

Subtract: Equation (1) from above equation

C + 2H_{2} + 2O_{2} - CH_{3}OH - 3/2 O_{2} → CO_{2} + 2H_{2} - CO_{2} - 2H_{2}O

↠ C + 2H_{2} + ½ O_{2} - CH_{3}OH = 0

↠ C + 2H_{2} + ½ O_{2} → CH_{3}OH

So, enthalpy of formation of CH_{3}OH can be calculated as follows:

Equation (2) + 2 × equation (3) – equation (1)

= -393 – 2 × 286 + 726 = -239 kJ mol^{-1}

Question 15: Calculate the enthalpy change for the process

CCl_{4}(g) → C(g) + 4CL(g)

And calculate bond enthalpy of C-Cl in CCl_{4}(g)

Δ_{vap}H^{Θ} [CCl_{4}] = 30.5 kJ mol^{-1}

Δ_{f}H^{Θ} [CCl_{4}] = -135.5 kJ mol^{-1}

Δ_{a}H^{Θ} [C] = 715 kJ mol^{-1}

Δ_{a}H^{Θ} [Cl_{2}] = 242 kJ mol^{-1}

Here, Δ_{a} H^{Θ} is enthalpy of atomization

**Answer:** Following equations can be written for various enthalpies given in question:

CCl_{4}(l) → CCl_{4}(g) [30.5 kJ mol^{-1}

C(s) + 2Cl_{2} → CCl_{4}(l) [ - 135.5 kJ mol^{-1}]

C(s) → C(g) [715 kJ mol^{-1}]

Cl_{2}(g) → 2Cl(g) [242 kJ mol^{-1}]

The aimed equation is: CCl_{4}(g) → C(g) + 4CL(g)

This can be achieved as follows:

Equation (3) + 2 × equation (4) – equation (1) - equation (2)

Add equation 3 and 4 as follows:

C + 2Cl_{2} → C + 4Cl

Subtract equations (1) and (2) from above

C + 2Cl_{2} - CCl_{4} - C – 2Cl_{2} → C + 4Cl – CCl_{4} - CCl_{4}

↠ CCl_{4} → C + 4Cl

So, required enthalpy can be calculated in the same order of additions:

Equation (3) + 2 × equation (4) – equation (1) - equation (2)

= 715 + 2 × 242 – 30.5 + 135.5 = 1304 kJ

So, bond enthalpy in each C-Cl bond `=(1304)/4=326` kJ mol^{-1}

Question 16: For an isolated system, ΔU = 0, what will be ΔS?

**Answer:** For an isolated system, ΔS > 0 because entropy tends to increase.

Question 17: For the reaction at 298 K

2A + B → C

ΔH = 400 kJ mol^{-1} and ΔS = 0.2 kJ K^{-1} mol^{-1}

At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature ranage.

**Answer:** As per Gibbs equation,

ΔG = ΔH – TΔS

For ΔG = 0

ΔH = TΔS

Or, `T=(ΔH)/(ΔS)`

`=(400)/(0.2)=2000` K

Question 18: For the reaction, 2Cl(g) → Cl_{2}(g) what are the signs of ΔH and ΔS?

**Answer:** Energy is released in bond formation, so ΔH is negative. Entropy decreases in formation of molecule, so ΔS will be negative.

Question 19: For the reaction:

2A(g) + B(g) → 2D(g)

ΔU^{Θ} = -10.5 kJ and ΔS^{Θ} = -44.1 JK^{-1}

Calculate ΔG^{Θ} for the reaction and predict whether the reaction may occur spontaneously.

**Answer:** ΔH = ΔU + Δn_{g}RT

Given; ΔU = -10.5 kJ

Δn_{g} = 2 – 3 = - 1 mol

R = 8.3145 × 10^{-3} kJ mol^{-1}

T = 298 K

So, ΔH = -10.5 – 1 × 8.314 × 10^{-3} × 298

= - 10.5 – 8.314 × 0.298 = 12.978 kJ

According to Gibbs equation:

ΔG = ΔH – TΔS

= - 12.978 – 298 × - 0.0441

= - 12.978 + 13.142 =0.164 kJ

As ΔG is positive, the reaction will be non-spontaneous

Question 20: The equilibrium constant for a reaction is 10. What will the value of ΔG^{Θ}? R = 8.314 JK^{-1} mol^{-1}, T = 300 K .

**Answer:** ΔG = - RT ln K = - 2.303 RT log K

Given: R = 8.314 JK^{-1} mol^{-1}, T = 300 K, K = 10

So, ΔG = - 2.303 × 8.314 × 300 × log 10

= - 5527 J mol^{-1} = - 5.527 kJ mol^{-1}

Question 21: Comment on the thermodynamic stability of NO(g), given

`1/2`N_{2}(g) + `1/2`O_{2}(g) → NO(g)

Δ_{r}H^{Θ} = 90 kJ mol^{-1}

NO(g) + `1/2`O_{2}(g) → NO_{2}(g)

Δ_{r}H^{Θ} = -74 kJ mol^{-1}

**Answer:** For NO (g): Δ_{r}H^{Θ} = +ve: unstable in nature

For NO_{2} (g): Δ_{r}H^{Θ} = -ve: stable in nature

Question 22: Calculate the entropy change in surroundings when 1.00 mol of H_{2}O(l) is formed under standard conditions. Δ_{f}H^{Θ} = -286 kJ mol^{-1}

**Answer:** q_{rev} = - Δ_{f}H^{Θ}

= - 286 kJ mol^{-1} = 286000 J mol^{-1}

ΔS_{surr} `=(q_(re\v))/T`

`=(286000)/(298)=959` J K^{-1} mol^{-1}