Q.
For the reaction at $25^\circ C$ ,
$NH _{3}( g ) \rightarrow \frac{1}{2} N _{2}( g )+\frac{3}{2} H _{2}( g ) ; \quad \Delta H ^{\circ}=11.04 kcal$
Calculate $ΔU^{^\circ }$ for the reaction at the given temperature.
(Note: Report your answer after multiplying with 100 and rounding up to the nearest integer value.)
NTA AbhyasNTA Abhyas 2022
Solution:
Chemical reaction:
$NH _{3}( g ) \rightarrow \frac{1}{2} N _{2}( g )+\frac{3}{2} H _{2}( g )$
$\Delta n=\left(\frac{1}{2} + \frac{3}{2}\right)-1=2-1=1$
$\Delta H^\circ =\Delta U^\circ +\Delta n_{g}RT$
Enthalpy and internal energy are extensive properties.
Enthalpy and internal energy are state functions.
$\Delta U^\circ =\Delta H^\circ -\Delta n_{g}RT$
$=11.04-\frac{\left(\right. 1 \left.\right) \left(\right. 1 . 98 \left.\right) \left(\right. 298 \left.\right)}{1000}$
$=11.04-0.59$
$=10.44kcal$
Final answer $=100\times 10.44=1044$
$NH _{3}( g ) \rightarrow \frac{1}{2} N _{2}( g )+\frac{3}{2} H _{2}( g )$
$\Delta n=\left(\frac{1}{2} + \frac{3}{2}\right)-1=2-1=1$
$\Delta H^\circ =\Delta U^\circ +\Delta n_{g}RT$
Enthalpy and internal energy are extensive properties.
$\Delta U^\circ =\Delta H^\circ -\Delta n_{g}RT$
$=11.04-\frac{\left(\right. 1 \left.\right) \left(\right. 1 . 98 \left.\right) \left(\right. 298 \left.\right)}{1000}$
$=11.04-0.59$
$=10.44kcal$
Final answer $=100\times 10.44=1044$