For the reaction,N_{2(g)} + 3 H_{2(g)} righta | vedclass

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(D) 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

$\Delta U - 2 RT$

Vedclass · Answer

(D) 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 $N_{2(g)} + 3 H_{2(g)} ightarrow 2 NH_{3(g)}$,the change in the number of moles of gaseous species $(\Delta n_g)$ is calculated as: $\Delta n_g = (	ext{moles of gaseous products}) - (	ext{moles of gaseous reactants}) = 2 - (1 + 3) = 2 - 4 = -2$. Substituting this value into the equation: $\Delta H = \Delta U + (-2) RT = \Delta U - 2 RT$. Therefore,the correct option is $D$.

For the reaction,$N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,$\Delta H$ is equal to

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