Page No 184: - Chapter 6 - Thermodynamics - Ncert Solutions class 11 - Chemistry

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Question 6.20:

The equilibrium constant for a reaction is 10. What will be the value of ΔG^θ? R = 8.314 JK^–1 mol^–1, T = 300 K.

Answer:

From the expression,

ΔG^θ = –2.303 RT logKeq

ΔG^θ for the reaction,

= (2.303) (8.314 JK^–1 mol^–1) (300 K) log10

= –5744.14 Jmol^–1

= –5.744 kJ mol^–1

Question 6.21:

Comment on the thermodynamic stability of NO₍g_), given

N₂₍g₎ +  O₂₍g₎ → NO₍g₎ ; ΔrH^θ = 90 kJ mol^–1

NO₍g₎ + O₂₍g₎ → NO₂₍g₎ : ΔrH^θ= –74 kJ mol^–1

Answer:

The positive value of ΔrH indicates that heat is absorbed during the formation of NO₍g₎. This means that NO₍g₎ has higher energy than the reactants (N₂ and O₂). Hence, NO₍g₎ is unstable.

The negative value of ΔrH indicates that heat is evolved during the formation of NO₂₍g₎ from NO₍g₎ and O₂₍g₎. The product, NO₂₍g₎ is stabilized with minimum energy.

Hence, unstable NO₍g₎ changes to stable NO₂₍g₎.

Question 6.22:

Calculate the entropy change in surroundings when 1.00 mol of H₂O₍l₎ is formed under standard conditions. ΔfH^θ = –286 kJ mol^–1.

Answer:

It is given that 286 kJ mol^–1 of heat is evolved on the formation of 1 mol of H₂O₍l₎. Thus, an equal amount of heat will be absorbed by the surroundings.

qsurr = +286 kJ mol^–1

Entropy change (ΔSsurr) for the surroundings = 

ΔSsurr = 959.73 J mol^–1 K^–1