Class 11 Chemistry

# 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 ΔrH for the reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

= 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

ΔrHΘ = -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) + 3/2O2(g) → COg + 2H2O(l)

ΔrHθ = -726 kJ mol-1

C (graphite) + O2(g) → CO2(g)

ΔrHθ = -393 kJ mol-1

H2(g) + 1/2O2(g) → H2O(l)

ΔrHθ = -286 kJ mol-1

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

C(s) + 2H2(g) + 1/2O2(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)

ΔvapHΘ [CCl4] = 30.5 kJ mol-1

ΔfHΘ [CCl4] = -135.5 kJ mol-1

ΔaHΘ [C] = 715 kJ mol-1

ΔaHΘ [Cl2] = 242 kJ mol-1

Here, Δa HΘ 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.

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)

ΔUΘ = -10.5 kJ and ΔSΘ = -44.1 JK-1

Calculate ΔGΘ for the reaction and predict whether the reaction may occur spontaneously.

Answer: ΔH = ΔU + ΔngRT

Given; ΔU = -10.5 kJ

Δng = 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 ΔGΘ? 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

1/2N2(g) + 1/2O2(g) → NO(g)

ΔrHΘ = 90 kJ mol-1

NO(g) + 1/2O2(g) → NO2(g)

ΔrHΘ = -74 kJ mol-1

Answer: For NO (g): ΔrHΘ = +ve: unstable in nature

For NO2 (g): ΔrHΘ = -ve: stable in nature

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

ΔSsurr =(q_(re\v))/T
=(286000)/(298)=959 J K-1 mol-1