The enthalpy change (Delta H) for the reactio | vedclass

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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \De

Vedclass · Accepted Answer

$-87.42$

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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.For the reaction $N_{2(g)} + 3H_{2(g)} ightarrow 2NH_{3(g)}$,the change in the number of gaseous moles is $\Delta n_g = 2 - (1 + 3) = -2$.Given $\Delta H = -92.38 \ kJ = -92380 \ J$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.Substituting the values: $-92380 = \Delta U + (-2 	imes 8.314 	imes 298)$.$-92380 = \Delta U - 4955.144$.$\Delta U = -92380 + 4955.144 = -87424.856 \ J$.Converting to $kJ$: $\Delta U \approx -87.42 \ kJ$.

The enthalpy change $(\Delta H)$ for the reaction,$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$ is $-92.38 \ kJ$ at $298 \ K$. The internal energy change $\Delta U$ at $298 \ K$ is $... \ kJ$.

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