Question

The plot of log 𝑘𝑓 versus $\frac{1}{T}$ for a reversible reaction, A (g) ⇌ P (g) is shown.

Pre-exponential factors for the forward and backward reactions are 1015s-1 and 1011s-1 respectively. If the value of log K for the reaction at 500 K is 6, the value of |log kb| at 250K is_________
[K = equilibrium constant of the reaction,
𝑘𝑓 = rate constant of forward reaction,
𝑘𝑏 = rate constant of backward reaction]

Answer 1

log kb at 500 k:
log K =$log (\frac{k_f}{k_b})$, Since $\Rightarrow K= \frac{k_f}{k_b}$
$6=log\,k_f-log\,k_b$
$6=9-log\,k_b$
$log\,k_b=3 \,\,at \,\,500K$
$log (\frac{k_2}{k_1})= \frac{-E_a}{R}( \frac{1}{T_2}- \frac{1}{T_1})$
$k_b=A_be^{ \frac{-E_{ab}}{RT}}$
$Ink_b=InA_b- \frac{E_{ab}}{RT}$
$2.303\,log\,k_b=2.303\,log\,a_b- \frac{E_{ab}}{500R}$
$\frac{E_a}{500R}=2.303(log\,10^{11}-3)$
$E_a=2.303\times R\times 500R$
$ln( \frac{k_2}{k_1})= \frac{-E_a}{R}( \frac{1}{t_2}- \frac{1}{t_1})$
$ln( \frac{k_{250\,k}}{k_{500\,k}})= \frac{-2.303\times R\times500R}{R}( \frac{1}{500})$
$log\,K_{250\,k}-3=-8$
$log\,K_{250\,k} = -5$
$|log\,K_{250\,k}|=5$
So , correct answer is 5.