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)
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)
The correct option is B −19.67 lit atm
From the given data
T=291 K;R=0.082 lit atm K−1 mol−1
Kc=1.14
The reaction quotient Qc for the reaction
Since QP=Qc(RT)Δng and Δng=2−1=1
⇒QP=0.5×(0.082×291)=11.93
⇒KP=Kc(RT)Δng
⇒KP=1.14×(0.082×291)=27.1
Substituting these values in the equation
ΔG=ΔG∘+RT ln QP=−RT ln KP+RT ln QP
⇒−2.303RT(log KP−log QP)
⇒ΔG=−(2.303×0.082×291) [log 27.2−log 11.93]=19.67 lit atm
From the given data
T=291 K;R=0.082 lit atm K−1 mol−1
Kc=1.14
The reaction quotient Qc for the reaction
Since QP=Qc(RT)Δng and Δng=2−1=1
⇒QP=0.5×(0.082×291)=11.93
⇒KP=Kc(RT)Δng
⇒KP=1.14×(0.082×291)=27.1
Substituting these values in the equation
ΔG=ΔG∘+RT ln QP=−RT ln KP+RT ln QP
⇒−2.303RT(log KP−log QP)
⇒ΔG=−(2.303×0.082×291) [log 27.2−log 11.93]=19.67 lit atm
