Question

The plot of $\ln K_{c}vs\cdot \frac{1}{T}$ shows an intercept of $\ln A=4$. The change in standard entropy for the reaction at equilibrium in cal.

Answer 1

$\ln K_c=lnA-\frac{E_a}{RT}$ ......(1)

The Eyring equation is as given below.

$\ln K_c=\frac{\Delta S}{R}-\frac{\Delta H}{RT}$......(2)

From equations (1) and (2), it can be observed that

$lnA=\frac{\Delta S}{R}=4$

Hence, $\Delta S=4R=4\times 2cal/mol/K=8cal/mol/K$.

The change in standard entropy for the reaction at equilibrium is $8cal/mol/K$