Question

Consider the reaction $A \rightleftharpoons B$ at $1000\, K$. At time '$t$', the temperature of the system was increased to $2000 \,K$ and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure of $A$ was maintained at $1$ bar. Given below is the plot of the partial pressure of $B$ with time. What is the ratio of the standard Gibbs energy of the reaction at $1000\, K$ to that at $2000\, K$ ?

Answer 1

$K _{1}$ at $1000\, K =\frac{10}{1}=10$
$K _{2}$ at $2000\, K =\frac{100}{1}=100$
$\Delta G _{1}^{\circ}=- RT _{1} \ln 10=-2.303\, RT _{1}$
$\Delta G _{2}^{\circ}=-2.303 \,RT _{2} \log 100$
$=-2 \times 2.303 \,RT _{2}$
$\frac{\Delta G _{1}^{\circ}}{\Delta G _{2}^{\circ}}=\frac{1 \times 1000}{2 \times 2000}=\frac{1}{4}=0.25$