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Question: The enthalpy change \((\Delta H)\) for the reaction \({N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)}\)...

The enthalpy change (ΔH)(\Delta H) for the reaction N2(g)+3H2(g)→2NH3(g){N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)} is - 92.38 kJ at 298K. The internal energy change ΔU\Delta U at 298K is :
(A) - 92.38 kJ
(B) - 87.42 kJ
(C) - 97.34 kJ
(D) - 89.9 kJ

Explanation

Solution

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) ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
(ii) Δng=np−nr\Delta {n_g} = {n_p} - {n_r}

Complete step by step answer:
The reaction is
N2(g)+3H2(g)→2NH3(g){N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)}
In this reaction N2{N_2} and H2{H_2} are gaseous reactants. And NH3N{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
∴Δng=np−nr\therefore \Delta {n_g} = {n_p} - {n_r}
Where, np{n_p} is gaseous product
nr{n_r} is gaseous Reactant
⇒Δng=2−(3+1)\Rightarrow \Delta {n_g} = 2 - (3 + 1)
= 2 - 4
= - 2
We are given ΔH=−92.38  kJmol−1\Delta H = - 92.38\;kJmo{l^{ - 1}}.
=−92.38×1000  Jmole−1= - 92.38 \times 1000\;Jmol{e^{ - 1}}
=−92380  Jmol−1= - 92380\;Jmo{l^{ - 1}}
We know R=8.314  Jk−1R = 8.314\;J{k^{ - 1}}
The absolute temperature is T = 298K
Where,ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
Where, ΔH\Delta H = Change in enthalpy
ΔU\Delta U = Change in internal energy
Δng\Delta {n_g} = Change in moles of gaseous reaction
R = Gas constant
T = Absolute temperature
Putting above values in the formula
We get −92380  Jmol−1=ΔU+(−2)×(8.314  Jk−1mol−1)×(298K) - 92380\;Jmo{l^{ - 1}} = \Delta U + ( - 2) \times (8.314\;J{k^{ - 1}}mo{l^{ - 1}}) \times (298K)
⇒−92380  Jmol−1=ΔU+(−4953.95  Jmol−1)\Rightarrow - 92380\;Jmo{l^{ - 1}} = \Delta U + ( - 4953.95\;Jmo{l^{ - 1}})
Rearranging the above equation, we get
ΔU=−92380Jmol−1+4953.95Jmol−1\Delta U = - 92380Jmo{l^{ - 1}} + 4953.95Jmo{l^{ - 1}}
⇒ΔU=−87,426.048  Jmol−1\Rightarrow \Delta U = - 87,426.048\;Jmo{l^{ - 1}}
Divide it by 1000 to convert it into kJmol−1kJmo{l^{ - 1}}
⇒ΔU=−87.426  kJmol−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 ΔH,ΔU,R\Delta H,\Delta U,R and T.
ΔH\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.