Question

Consider a reaction:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
Given that ,
$K_{eq}=7\times {10}^5$
$\Delta S_{N{H}_3}^{\circ }=-23.66calmo{l}^{-1}{K}^{-1}$
Both the values are at 300K
Find the value of $\Delta H_{N{H}_3}^{\circ }$ ( in $kcalmo{l}^{-1}$) ( upto two decimal) at 300 K
Take $ln\left(7\right)=1.95$

Answer 1

At temperature $T$, equilibrium constant is $K_{eq}$ ,we know:

$ln\left(K_{eq}\right)=-\frac{\Delta H^o}{RT}+\frac{\Delta S^o}{R}$
$R=2calmo{l}^{-1}{K}^{-1}$
Putting the given values
$ln(7\times {10}^5)=\frac{-\Delta H^{\circ }}{2\times 300}+\frac{(-23.66)}{2}(1.95+5\times 2.3)=\frac{-\Delta H^{\circ }}{600}-11.83(13.45+11.83)\times 600=-\Delta H^{\circ }\Delta H^{\circ }=-15168calmo{l}^{-1}=-15.16kcalmo{l}^{-1}$