Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIFree Energy and EquilibriumMedium2 minPYQ_2023
ChemistryMediummatching list

Consider the following reaction approaching equilibrium at 27°C and 1 atm pressure

A+B⇌kf=03kb=102C+D

The standard Gibb's energy change ΔrG° at 27°C is - _____ kJmol-1 (Nearest integer).

(Given : R=8.3J K-1 mol-1 and ln10=2.3)

Answer:
6
Solution:

We know,

 Keq=kfkb

∴ Keq=103102=10
Using gibbs free energy equation,

∴  ΔG°=-RTlnKeq  ⇒ΔG=-RTln10

⇒-8.3×300×2.3=-5.7 kJ mol-1≈6 kJ mole-1 (nearest integer)

Answer =6.

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium
⏱ 2mℹ️ Source: PYQ_2023

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