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Question

The value of ΔHΔU for the following reaction at 37o C will be:
2NH3(g)N2(g)+3H2(g)

A
51.54 kJ mol1
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B
51.54 J mol1
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C
51.54 J mol1
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D
5.154 kJ mol1
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Solution

The correct option is D 5.154 kJ mol1
We know, ΔH=ΔU+PΔV=ΔU+ΔngRT
where ΔH = Change in enthalpy
ΔU = Change in internal energy
Δng = nproductsnreactants
Δng = 42=2
So, ΔHΔU=2RT=2×8.314×310=5154.68 J mol1=5.154 kJ mol1

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