Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium.Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g)=-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g)=-300kcal/mole DeltaG_(f)^(@)D(l)=-224.606 kcal/mole DeltaG_(f)^(@)D(g)= -226.9.9 kcal/ mole, All values at 500K Calculate rate constant of the backward reaction for the following reaction at 500K: A(g)+B(g)iffC(g)+D(l) " if " K_(f)=10"bar"^(-1) sec^(-1)

Equilibrium constant for a reaction can be obtained by kinetics approa

1.	Equilibrium constant for a reaction can be obtained by kinetics approach or by thermodynamics approch. While from kinetics approach at equililbrium, the rate of forward and backward will be same, from thermodynamics approach Gibbs free energy will be minimized at equilibrium.Using this information and following thermodynamics values, answer the question that follow: DeltaG_(f)^(@)A(g)=-200kcal/mole DeltaG_(f)^(@)B(g) =-320 kcal/mole DeltaG_(f)^(@)C(g)=-300kcal/mole DeltaG_(f)^(@)D(l)=-224.606 kcal/mole DeltaG_(f)^(@)D(g)= -226.9.9 kcal/ mole, All values at 500K Calculate rate constant of the backward reaction for the following reaction at 500K: A(g)+B(g)iffC(g)+D(l) " if " K_(f)=10"bar"^(-1) sec^(-1)
Answer» `10 "BAR"^(-1) SEC^(-1)` `0.1 "bar"^(-1) sec^(-1)` `0.1 sec^(-1)` `10SEC^(-1)` Answer :C