Question

For the reaction at $25^{\circ }C$ ,
$N{H}_3(g)\to \frac{1}{2}{N}_2(g)+\frac{3}{2}{H}_2(g);\Delta H^{\circ }=11.04kcal$
Calculate $\Delta U^{{}^{\circ }}$ for the reaction at the given temperature.
(Note: Report your answer after multiplying with 100 and rounding up to the nearest integer value.)

Answer 1

Answer: 1044

Chemical reaction:
$N{H}_3(g)\to \frac{1}{2}{N}_2(g)+\frac{3}{2}{H}_2(g)$
$\Delta n=\left(\frac{1}{2}+\frac{3}{2}\right)-1=2-1=1$
$\Delta H^{\circ }=\Delta U^{\circ }+\Delta n_{g}RT$
Enthalpy and internal energy are extensive properties.

Enthalpy and internal energy are state functions.
$\Delta U^{\circ }=\Delta H^{\circ }-\Delta n_{g}RT$
$=11.04-\frac{(1)(1.98)(298)}{1000}$
$=11.04-0.59$
$=10.44kcal$
Final answer $=100\times 10.44=1044$