Chemistry - Chemical Equilibrium Question with Solution | TestHub
ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minQB
ChemistryEasynumerical
At 27°C, the rate of forward reaction at time, t, is e5 times greater than that of reverse reaction for the elementary reaction: A + B ⇌ P. The magnitude of free energy change involved at that time (in cal) is (take R = 2 cal/mol/K)
Answer:
3000.00
Solution:
Forward rate/reverse rate = kf[A][B]/(kr[P]) = K/Q = e5 ΔG = RT ln(Q/K) = -RT ln(K/Q)
ΔG = -(2)(300)(5) = -3000 cal
Magnitude = 3000 cal
Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed
⏱ 2mℹ️ Source: QB
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