For the reaction N_{2(g)} + 3H_{2(g)} to 2NH_ | vedclass

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Question

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

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$\Delta H = \Delta U - 2RT$

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(A) 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 $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$,the number of moles of gaseous products is $n_p = 2$ and the number of moles of gaseous reactants is $n_r = 1 + 3 = 4$. Therefore,the change in the number of gaseous moles is $\Delta n_g = n_p - n_r = 2 - 4 = -2$. Substituting this value into the equation: $\Delta H = \Delta U + (-2)RT = \Delta U - 2RT$.

For the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$,the relationship between $\Delta H$ and $\Delta U$ is:

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