Question

For a reversible reaction $A\rightleftharpoons B$, ${\log }_{10}K=1.737-\frac{2500}{T}$ (in litre, atm unit). The standard entropy, change to the closest value in $cal.K^{-1}$ is :

Answer 1

The Eyring equation is as given below.

$\log K=\frac{\Delta S}{2.303R}-\frac{\Delta H}{2.303RT}$

Thus,

$\frac{\Delta S}{2.303R}=1.737$

Hence, $\Delta S=2.303\times 1.737R=2.303\times 1.737\times 2\approx 8cal/(mol.K)$

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