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Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.ΔG=ΔG∘+R T lnQP the condition for a system to be at equilibrium is thatΔG=0 and Qp=KPThus at equilibriumΔG∘=−R T lnKPNote: In the reaction, where all gaseous reactants and products; K represents KP, we may conclude that for standard reactions, i.e., at 1 M or 1 atmWhen ΔG∘=−ve or K>1: forward reaction is feasibleΔG∘=+ve or K<1: reverse reaction is feasibleΔG∘=0 or K=1: reaction is at equilibrium (very rare)Kc for reaction N2O4⇌2NO2 in chloroform at 291 K is 1.14. Calculate the free energy change of the reaction when the concentration of the two gases are 0.5 mol dm−3 each at the same temperature. (R=0.082 lit atm K−1mol−1)

Answer»

Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.

ΔG=ΔG∘+R T lnQP the condition for a system to be at equilibrium is that

ΔG=0 and Qp=KP

Thus at equilibrium

ΔG∘=−R T lnKP

Note: In the reaction, where all gaseous reactants and products; K represents KP, we may conclude that for standard reactions, i.e., at 1 M or 1 atm

When ΔG∘=−ve or K>1: forward reaction is feasible

ΔG∘=+ve or K<1: reverse reaction is feasible

ΔG∘=0 or K=1: reaction is at equilibrium (very rare)

Kc for reaction N2O4⇌2NO2 in chloroform at 291 K is 1.14. Calculate the free energy change of the reaction when the concentration of the two gases are 0.5 mol dm−3 each at the same temperature. (R=0.082 lit atm K−1mol−1)


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