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Assume each reaction is carried out in an open container. For which reaction will $\Delta H = \Delta E$ ?
A
$H_2(g) + Br_2(g) \rightarrow 2HBr(g)$
B
$C(s) + 2H_2O(g) \rightarrow 2H_2(g) + CO_2(g)$
C
$PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)$
D
$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$
Detailed Solution
$\Delta H = \Delta E + \Delta n_g RT$, where $\Delta n_g$ = moles of gaseous products $-$ moles of gaseous reactants.
$\Delta H = \Delta E$ only when $\Delta n_g = 0$.
$H_2(g) + Br_2(g) \rightarrow 2HBr(g)$: $\Delta n_g = 2 - 2 = 0$, so $\Delta H = \Delta E$.
$C(s) + 2H_2O(g) \rightarrow 2H_2(g) + CO_2(g)$: $\Delta n_g = 3 - 2 = +1$ (the solid carbon is not counted).
$PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)$: $\Delta n_g = 2 - 1 = +1$.
$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$: $\Delta n_g = 2 - 3 = -1$.
So $\Delta H = \Delta E$ for $H_2(g) + Br_2(g) \rightarrow 2HBr(g)$.
$\Delta H = \Delta E$ only when $\Delta n_g = 0$.
$H_2(g) + Br_2(g) \rightarrow 2HBr(g)$: $\Delta n_g = 2 - 2 = 0$, so $\Delta H = \Delta E$.
$C(s) + 2H_2O(g) \rightarrow 2H_2(g) + CO_2(g)$: $\Delta n_g = 3 - 2 = +1$ (the solid carbon is not counted).
$PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)$: $\Delta n_g = 2 - 1 = +1$.
$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$: $\Delta n_g = 2 - 3 = -1$.
So $\Delta H = \Delta E$ for $H_2(g) + Br_2(g) \rightarrow 2HBr(g)$.