Identify the reaction for which Delta H neq D | vedclass

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Question

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by $\Delta H = \Delta E + \Delta n_g RT$. F

Vedclass · Accepted Answer

$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}$

Vedclass · Answer

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by $\Delta H = \Delta E + \Delta n_g RT$. For $\Delta H 
eq \Delta E$,the change in the number of gaseous moles $(\Delta n_g)$ must not be equal to $0$. $(A)$ $\Delta n_g = 1 - 1 = 0$. $(B)$ $\Delta n_g = 2 - (1 + 1) = 0$. $(C)$ $\Delta n_g = 2 - (1 + 1) = 0$. $(D)$ $\Delta n_g = 1 - (1 + 0.5) = -0.5$. Since $\Delta n_g 
eq 0$ for option $D$,$\Delta H 
eq \Delta E$ for this reaction.

Identify the reaction for which $\Delta H \neq \Delta E$ :

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