Question

The value of $\Delta H-\Delta U$ for the following reaction at ${37}^o$ C will be:
$2N{H}_3\left(g\right)\to N_2\left(g\right)+3H_2\left(g\right)$

• A. $51.54{\text{kJ mol}}^{-1}$
• B. $51.54{\text{J mol}}^{-1}$
• C. $-51.54{\text{J mol}}^{-1}$
• D. $5.154{\text{kJ mol}}^{-1}$

Answer 1

The correct option is D $5.154{\text{kJ mol}}^{-1}$

We know, $\Delta H=\Delta U+P\Delta V=\Delta U+\Delta n_{g}RT$
where $\Delta H$ = Change in enthalpy
$\Delta U$ = Change in internal energy
$\Delta n_g$ = $n_{products}-n_{reactants}$
$\Delta n_g$ = $4-2=2$
So, $\Delta H-\Delta U=2RT=2\times 8.314\times 310=5154.68{\text{J mol}}^{-1}=5.154{\text{kJ mol}}^{-1}$