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Q.

Given at 300 K :

NH3(g)⇌12N2(g)+32H2(g); ΔrH∘=193.27kJmol−1

The value of ΔrU∘ for this reaction would be:

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a

188.29 KJ mol-1

b

198.27 KJ mol-1

c

190.78 KJ mol-1

d

19.576 KJ mol-1

answer is A.

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Detailed Solution

The value of ΔrU∘ for this reaction would be:

ΔU∘=ΔH∘−RT =193.27−8.314×3001000=190.78 kJ mol−1

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