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

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Question

(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

$-87.42$

Vedclass · Answer

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.Rearranging for $\Delta U$: $\Delta U = \Delta H - \Delta n_g RT$.For the reaction $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$,the change in the number of gaseous moles is $\Delta n_g = n_p - n_r = 2 - (1 + 3) = -2$.Given $\Delta H = -92.38 \ kJ$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.Substituting the values: $\Delta U = -92.38 - (-2 	imes 8.314 	imes 10^{-3} 	imes 298)$.$\Delta U = -92.38 - (-4.957) = -92.38 + 4.957 = -87.423 \ kJ$.Thus,$\Delta U \approx -87.42 \ kJ$.

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

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