Chemical Equilibrium Question 3

Question 3 - 29 January - Shift 1

Consider the following reaction approaching equilibrium at $27^{\circ} C$ and $1 atm$ pressure

$A+B \underset{K_r=10^{2}}{\stackrel{K_F=10^{3}}{\leftrightharpoons}} C+D$

The standard Gibb’s energy change $(\Delta_r G^{o})$ at $27^{\circ} C$ is $(-)$____________ $kJ mol^{-1}$

(Nearest integer).

Show Answer

Answer: (6)

Solution:

Formula: Relation between $\Delta G^{o}$ and $lnK _{eq}$

(Given : $R=8.3 J K^{-1} mol^{-1}$ and $\ln 10=2.3$ )

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

and $K _{eq}=\frac{K_f}{K_b}$

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

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

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

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

Ans $=6$