What will be the value of logarithm of equilibrium constant K_P i… - Vidyadip

MCQ

What will be the value of logarithm of equilibrium constant $\ce{K_P}$ if the standard free energy change of a reaction is $\Delta\text{G}^\circ=-115\text{kJ}$ at $298K$ will be :

A
$2.303$
B
$13.83$
C
$2.015$
✓
$20.15$

✓

Answer

Correct option: D.

$20.15$

$\Delta\text{G}^\circ=-115\times10^3\text{J}$
$\text{T}=298\text{K, R}=8.314\text{JK}^{-1}\text{mol}^{-1}$
$-\Delta\text{G}^\circ=2.303\text{RT}\log_{10}\text{K}_\text{p}$
$-(-115\times10^3)=2.303\times8.314$
$\times298\log_{10}\text{K}_\text{p}$
$\log_{10}\text{K}_\text{p}=\frac{115000}{2.303\times8.314\times298}=20.15$

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