Question

The enthalpy change $(\Delta H)$ for the reaction ${N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)}$ is - 92.38 kJ at 298K. The internal energy change $\Delta U$ at 298K is :
(A) - 92.38 kJ
(B) - 87.42 kJ
(C) - 97.34 kJ
(D) - 89.9 kJ

Answer 1

Hint: We will use the relationship between Heat of reaction at constant pressure and that at constant volume. Heat of reaction is the energy that is released or absorbed when chemicals are transformed in a chemical reaction.

Formula used: (i) $\Delta H = \Delta U + \Delta {n_g}RT$
(ii) $\Delta {n_g} = {n_p} - {n_r}$

Complete step by step answer:
The reaction is
${N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)}$
In this reaction ${N_2}$ and ${H_2}$ are gaseous reactants. And $N{H_3}$ is a gaseous product. Reactants are usually the compounds written to the left side of the chemical equation and product is usually the compound written to the right side of the chemical equation. But it is not always necessary. You check the change from the reactant to product by observing the arrow used in the chemical equation. For this particular question, the flow of reaction is from left to right.
To observe the change in the number of molecules in the reaction, we will use the formula
$\therefore \Delta {n_g} = {n_p} - {n_r}$
Where, ${n_p}$ is gaseous product
${n_r}$ is gaseous Reactant
$ \Rightarrow \Delta {n_g} = 2 - (3 + 1)$
= 2 - 4
= - 2
We are given \[\Delta H = - 92.38\;kJmo{l^{ - 1}}\].
$ = - 92.38 \times 1000\;Jmol{e^{ - 1}}$
$ = - 92380\;Jmo{l^{ - 1}}$
We know $R = 8.314\;J{k^{ - 1}}$
The absolute temperature is T = 298K
Where,$\Delta H = \Delta U + \Delta {n_g}RT$
Where, $\Delta H$ = Change in enthalpy
$\Delta U$ = Change in internal energy
$\Delta {n_g}$ = Change in moles of gaseous reaction
R = Gas constant
T = Absolute temperature
Putting above values in the formula
We get $ - 92380\;Jmo{l^{ - 1}} = \Delta U + ( - 2) \times (8.314\;J{k^{ - 1}}mo{l^{ - 1}}) \times (298K)$
$ \Rightarrow - 92380\;Jmo{l^{ - 1}} = \Delta U + ( - 4953.95\;Jmo{l^{ - 1}})$
Rearranging the above equation, we get
$\Delta U = - 92380Jmo{l^{ - 1}} + 4953.95Jmo{l^{ - 1}}$
$ \Rightarrow \Delta U = - 87,426.048\;Jmo{l^{ - 1}}$
Divide it by 1000 to convert it into $kJmo{l^{ - 1}}$
$ \Rightarrow \Delta U = - 87.426\;kJmo{l^{ - 1}}$

Thus the correct answer is (B).

Note: A physical quantity should be used in the same unit. Here we use SI units for $\Delta H,\Delta U,R$ and T.
\[\Delta H\] is given in kJ so it should be converted into J for the sake of easy calculation and then converted back into kJ.