Chemical Equilibrium Question 3
Question 3 - 29 January - Shift 1
Consider the following reaction approaching equilibrium at $27^{\circ} C$ and $1 atm$ pressure
$A+B \underset{K_r=10^{2}}{\stackrel{K_F=10^{3}}{\rightarrow}} C+D$
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 $=6$
