Question

The equilibrium constant ${K_P}$ for the reaction $A(g) \rightleftharpoons B(g) + C(g)$ is 1 at 27C and 4 at 47. For the reaction calculate enthalpy change for the $B(g) \rightleftharpoons C(g) + A(g)$. (Given; R = 2cal/mol-K)
A.13.31 kcal/mol
B.14.31 kcal/mol
C.19.2 kcal/mol
D.55.63 kcal/mol

Answer 1

${K_P}$is an equilibrium constant at constant pressure. If it is given for two different temperatures, we can calculate the enthalpy change for the reaction using the Van’t Hoff Equation. It relates the change in equilibrium constant of a reaction to the change in temperature.
Formula used:
$\ln \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{R}[\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}]$

Complete step by step answer:
The Van’t Hoff equation relates the equilibrium constant of a reaction to change in temperature by assuming that the change in enthalpy of a reaction is constant as a function of temperature. Using the integrated form of the Van't Hoff equation shown below, we can determine the change in enthalpy.
$\ln \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{R}[\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}]$
We are provided the values such that
${K_1}$= 1, ${K_2}$= 4, ${T_1}$= 27C = 300K, ${T_2}$=47C = 320K
Substituting these values in the formula, we obtain
$\ln \dfrac{4}{1} = \dfrac{{\Delta H}}{2}[\dfrac{1}{{300}} - \dfrac{1}{{320}}]$
From this we get, $\Delta H$= 13.31 kcal/mol for $A(g) \rightleftharpoons B(g) + C(g)$
For reaction $B(g) + C(g) \rightleftharpoons A(g)$, the change in enthalpy will remain same but the sign will change is
the equation has been reversed. So, $\Delta H$= -13.31 kcal/mol for $B(g) + C(g) \rightleftharpoons A(g)$.

Hence, the correct option is (A).

Note:
As per Van’t Hoff equation, the reacting system of a chemical reaction behaves in an ideal manner and the change in enthalpy is independent of temperature for a small range of temperature change.