Question
For which change $$\Delta H \ne \Delta E\,:$$
A.
$${H_{2\left( g \right)}} + {I_{2\left( g \right)}} \to 2HI\left( g \right)$$
B.
$$HC{\text{l}} + NaOH \to NaC{\text{l}}$$
C.
$${C_{\left( s \right)}} + {O_{{2_{\left( g \right)}}}} \to C{o_{{2_{\left( g \right)}}}}$$
D.
$${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$$
Answer :
$${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$$
Solution :
$$\eqalign{
& \Delta H = \Delta E + \Delta nRT\,\,{\text{For}}\,\Delta H \ne \Delta E,\Delta n \ne 0 \cr
& {\text{Where}}\,\Delta n = {\text{no}}{\text{.}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{gaseous}}\,{\text{products}} - {\text{no}}{\text{.}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{gaseous}}\,{\text{reactants}} \cr
& \left( {\text{a}} \right)\,\,\Delta n = 2 - 2 = 0 \cr
& \left( {\text{b}} \right)\,\,\Delta n = 0\,\,\,\left( {\because \,\,{\text{they}}\,{\text{are either in solid or liquid state}}} \right) \cr
& \left( {\text{c}} \right)\,\;\Delta n = 1 - 1 = 0\,\,\,\left( {\because \,\,C\,{\text{is}}\,{\text{in}}\,{\text{solid}}\,{\text{state}}} \right) \cr
& \left( {\text{d}} \right)\,\,\Delta n = 2 - 4 = - 2 \cr} $$