• # question_answer Determine $\Delta {{U}^{o}}$at 400 K for the following reaction using the listed enthalpies of reaction. $4CO(g)+8{{H}_{2}}(g)\xrightarrow[{}]{{}}$ $3C{{H}_{4}}(g)+C{{O}_{2}}(g)+2{{H}_{2}}O(l)$ $C\,\text{(graphite)+}\frac{1}{2}{{O}_{2}}\xrightarrow[{}]{{}}CO(g),$ $\Delta H_{1}^{o}=-110.5\,\text{kJ}$ $CO\,\text{(g)}\,\text{+}\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}CO(g),$ $\Delta H_{2}^{o}=-282.9\,\text{kJ}$ ${{H}_{2}}\,\text{(g)}\,\text{+}\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}{{H}_{2}}O(l),$ $\Delta H_{3}^{o}=-285.8\,\text{kJ}$ $C\,\text{(graphite)}\,+2{{H}_{2}}(g)\xrightarrow[{}]{{}}C{{H}_{4}}(g),$ $\Delta H_{4}^{o}=-74.8\,\text{kJ}$ A)  - 750.69 kJ         B)  - 820.79 kJ           C)  - 720.79 kJ          D)  - 1000.79 kJ

$\Delta {{H}^{o}}=-3\Delta H_{1}^{o}+\Delta H_{2}^{o}+2\Delta H_{3}^{o}+3\Delta H_{4}^{o}$ $=-747.4\,\text{kJ}$ $\Delta {{H}^{o}}=\Delta U_{{}}^{o}+\Delta ng\,RT$ $\because$$\Delta {{n}_{g}}=-8$ $-747.4=\Delta {{U}^{o}}-\frac{8\times 8.314\times 400}{1000}$ $-747.4+26.60=\Delta {{U}^{o}}$ $\Delta {{U}^{o}}=-720.79\,\text{kJ}$