Consider the following reaction approaching equilibrium at 27^ ,C… - Vidyadip

MCQ

Consider the following reaction approaching equilibrium at $27^{\circ}\,C$ and $1\,atm$ pressure The standard Gibb's energy change $\left(\Delta_{ r } G ^{\circ}\right)$ at $27^{\circ}\,C$ is $(-)$ $..........kJ\,mol ^{-1}$ (Nearest integer). (Given : $R =8.3\,J\,K ^{-1}\, mol ^{-1}$ and $\ln 10=2.3$ )

✓
$6$
B
$3$
C
$12$
D
$9$

✓

Answer

Correct option: A.

$6$

a
$\because \Delta G ^{\circ}=- RT \ln K _{ eq }$

$\text { and } K _{ eq }=\frac{ K _{ f }}{ K _{ b }}$

$\therefore K _{ eq }=\frac{10^3}{10^2}=10$

$\therefore \Delta G =- RT \ln 10$

$\Rightarrow-(8.3 \times 300 \times 2.3)=-5.7\,kJ\,mole ^{-1} \approx 6\,kJ$

$\text { mole }{ }^{-1}(\text { nearest integer) }$

$\text { Ans }=6$

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