What is the significance of Gibbs free energy?

Gibbs Energy question?

• Here's the problem: A chemical reaction at 1.0 bar and 298 K has ΔrG° = -335 kJ/mol. The Gibbs function of the reactants changes by +7.5 kJ/mol when the pressure is changed from 1.0 bar to some other pressure P2. The products change by +8.4 kJ/mol over the same pressure change, 1.0 bar to P2. What is ΔrG° at P2 and 298 K? Relevant equations (Maybe?): ΔG = ΔG° + RTln(P2/P1) I'm not sure what to do with the Gibbs functions of the reactants and products. Can I subtract reactants from products (8.4 - 7.5) to get ΔrG? I did this and plugged all of the numbers into the above formula, solving for P2 = 7.52 bar. What now? Am I on the right track at all?

• Answer:

  Under both sets of conditions, ΔrG = 0 because they are at equilibrium. What changes is ΔrG° because the standard conditions for pressure change. Typically standard pressure is 1 bar. ΔrG = ΔrG° + RT ln (K) 0 = ΔrG° + RT ln (Keq) ΔrG° = - RT ln (Keq) where Keq is different for each set of pressure conditions They tell you what happens to ΔrG°. Under the second set of conditions ΔrG°(new conditions) = ΔrG°(typical standard conditions) + Gibbs function product changes - Gibbs function reactant changes = - 335 kJ/mol + (+8.4 kJ/mol) - (+7.5 kJ/mol) = - 334.1 kJ/mol So ΔrG° at P2 and 298 K = - 334.1 kJ/mol