Question

For a gaseous reaction 2A2(g)+5B2(g)→2A2B5(g) at 27∘C the heat change at constant pressure is found to be −50160J. Calculate the value of internal energy change (ΔU).(R=8.314J/Kmol)

Answer 1

C−38229J

ΔH=ΔU+ΔnRT

−50700=ΔU+(−5)×8.314×300

ΔU=−38229J

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