Question

For the reaction at ${25}^{\circ }C,$ $N{H}_3\left(g\right)+\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right);\Delta H^{\circ }=11.04kcal.$ Calculate $\Delta U^{\circ }$ of the reaction at the given temperature

Answer 1

$N{H}_3\left(g\right)\to \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right);\Delta H^{\circ }=11.04$ kcal

$\Delta H^{\circ }=\Delta U^{\circ }+\Delta nRT$

$11.04=\Delta U^{\circ }+(\frac{1}{2}+\frac{3}{2}-1)\times 1.98\times {10}^{-3}\times 298$

$11.04=\Delta U^{\circ }+1\times 1.98\times 0.298$

$11.04=\Delta E^{\circ }+0.59$

$\Delta E^{\circ }=11.04-0.59=10.45$ kcal