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

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Question

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

Vedclass · Accepted Answer

$-87.42$

Vedclass · Answer

(B) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$. Here,$\Delta n_g$ is the change in the number of moles of gaseous species: $\Delta n_g = \sum n_p(g) - \sum n_R(g) = 2 - (1 + 3) = -2 \ mol$. Rearranging the formula for $\Delta U$: $\Delta U = \Delta H - \Delta n_g RT$. Substituting the given values: $\Delta U = (-92.38 \ kJ) - (-2 \ mol) 	imes (8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) 	imes (298 \ K)$. $\Delta U = -92.38 \ kJ + 4.955 \ kJ \approx -87.42 \ kJ$.

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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