For which of the following reaction, $\Delta H=\Delta E+2RT$ ?

In thermodynamic reaction,

$\Delta H=\Delta E+\Delta nRT$

Where, $\Delta H$ is the enthalpy change, $\Delta E$ is the internal energy change, $\Delta n$ is the number of gaseous moles of product - number of gaseous moles of reactant. And $R$ and $T$ are the gas constant and temperature respectively.

We have to find the reaction which justifies the thermodynamic system, $\Delta H=\Delta E+2RT$ i.e., which will have $\Delta n=2$.

Let's find the value of $\Delta n$ for the given equations.

$2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\to 2S{O}_{3\left(g\right)}=\Delta n=2-3=-1$

${\mathrm{NH}}_4{\mathrm{HS}}_{\left(s\right)}\to {\mathrm{NH}}_{3\left(g\right)}+{\mathrm{H}}_2{S}_{\left(g\right)}=\Delta n=2-0=+2$

$N_{2\left(g\right)}+O_{2\left(g\right)}\to 2N{O}_{\left(g\right)}=\Delta n=2-2=0$

${\mathrm{PCl}}_{5\left(g\right)}\to {\mathrm{PCl}}_{3\left(g\right)}+{\mathrm{Cl}}_{2\left(g\right)}=\Delta n=2-1=+1$

So, the reaction ${\mathrm{NH}}_4{\mathrm{HS}}_{\left(s\right)}\to {\mathrm{NH}}_{3\left(g\right)}+{\mathrm{H}}_2{S}_{\left(g\right)}$ has, $\Delta H=\Delta E+2RT$ because the value of $\Delta n$ is $+2$ here.

Therefore, option (B) is correct.