Thermodynamics

NCERT Solution

Part 2

Question 12: Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are -110, -393, 81 and 9.7 kJ mol^-1 respectively. Find the value of Δ_rH for the reaction:

N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

Answer: Δ_rH

= 9.7 + 3 × (-110) – (81 – 3 × 393)

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

Question 13: Given: N₂(g) + 3H₂(g) → 2NH₃

Δ_rH^Θ = -92.4 kJ mol^-1

What is the standard enthalpy of formation of NH₃ gas?

Answer: Standard enthalpy of formation of NH₃ = `(-92.4)/2=-46.2` kJ mol^-1

Question 14: Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:

CH₃OH(l) + `3/2`O₂(g) → CO_g + 2H₂O(l)

Δ_rH^θ = -726 kJ mol^-1

C (graphite) + O₂(g) → CO₂(g)

Δ_rH^θ = -393 kJ mol^-1

H₂(g) + `1/2`O₂(g) → H₂O(l)

Δ_rH^θ = -286 kJ mol^-1

Answer: The equation for formation of CH₃OH is as follows:

C(s) + 2H₂(g) + `1/2`O₂(g) → CH₃OH(l)

This equation can be achieved as follows:

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

Add: Equation (2) + 2 × equation (3)

C + O₂ + 2H₂ + 2O₂ → CO₂ + 2H₂

↠ C + 2H₂ + 2O₂ → CO₂ + 2H₂O

Subtract: Equation (1) from above equation

C + 2H₂ + 2O₂ - CH₃OH - 3/2 O₂ → CO₂ + 2H₂ - CO₂ - 2H₂O

↠ C + 2H₂ + ½ O₂ - CH₃OH = 0

↠ C + 2H₂ + ½ O₂ → CH₃OH

So, enthalpy of formation of CH₃OH 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

CCl₄(g) → C(g) + 4CL(g)

And calculate bond enthalpy of C-Cl in CCl₄(g)

Δ_vapH^Θ [CCl₄] = 30.5 kJ mol^-1

Δ_fH^Θ [CCl₄] = -135.5 kJ mol^-1

Δ_aH^Θ [C] = 715 kJ mol^-1

Δ_aH^Θ [Cl₂] = 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:

CCl₄(l) → CCl₄(g) [30.5 kJ mol^-1

C(s) + 2Cl₂ → CCl₄(l) [ - 135.5 kJ mol^-1]

C(s) → C(g) [715 kJ mol^-1]

Cl₂(g) → 2Cl(g) [242 kJ mol^-1]

The aimed equation is: CCl₄(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 + 2Cl₂ → C + 4Cl

Subtract equations (1) and (2) from above

C + 2Cl₂ - CCl₄ - C – 2Cl₂ → C + 4Cl – CCl₄ - CCl₄

↠ CCl₄ → 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) → Cl₂(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_gRT

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`N₂(g) + `1/2`O₂(g) → NO(g)

Δ_rH^Θ = 90 kJ mol^-1

NO(g) + `1/2`O₂(g) → NO₂(g)

Δ_rH^Θ = -74 kJ mol^-1

Answer: For NO (g): Δ_rH^Θ = +ve: unstable in nature

For NO₂ (g): Δ_rH^Θ = -ve: stable in nature

Question 22: Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. Δ_fH^Θ = -286 kJ mol^-1

Answer: q_rev = - Δ_fH^Θ

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

ΔS_surr `=(q_(re\v))/T`

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