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Question 8.4: Calculating ΔE for a Reaction The reaction of nitrogen with ......

Calculating ΔE for a Reaction

The reaction of nitrogen with hydrogen to make ammonia has ΔH° = -92.2 kJ. What is the value of ΔE in kilojoules if the reaction is carried out at a constant pressure of 40.0 atm and the volume change is -1.12 L?

N_2(g) + 3 H_2(g) → 2 NH_3(g) ΔH° = -92.2 kJ

STRATEGY

We are given an enthalpy change ΔH, a volume change ΔV, and a pressure P and asked to find an energy change ΔE. Rearrange the equation ΔH = ΔE + PΔV to the form ΔE = ΔH – PΔV, and substitute the appropriate values for ΔH, P, and ΔV:

Step-by-Step
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\begin{aligned}& \Delta E=\Delta H-P \Delta V \quad \text { where } \Delta H=-92.2 \mathrm{~kJ} \\& \begin{aligned}& P \Delta V=(40.0 \mathrm{~atm})(-1.12 \mathrm{~L})=-44.8 \mathrm{~L} \cdot \text { atm } \\&=(-44.8 \mathrm{~L} \cdot \text { atm })\left(101 \frac{\mathrm{J}}{\mathrm{L} \cdot \mathrm{atm}}\right)=-4520 \mathrm{~J}=-4.52 \mathrm{~kJ} \\& \Delta E=(-92.2 \mathrm{~kJ})-(-4.52 \mathrm{~kJ})=-87.7 \mathrm{~kJ}\end{aligned}\end{aligned}

Note that ΔE is smaller (less negative) than ΔH for this reaction because the volume change is negative. The products have less volume than the reactants, so a contraction occurs and a small amount of PV work is gained by the system.

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