CHEMISTRY COMMUNITY

Question 11.111 in the textbook asks "A certain enzyme-catalyzed reaction in a biochemical cycle has an equilibrium constant that is 10 times the equilibrium constant of the next step in the cycle. If the standard Gibbs free energy of the first reaction is -200. kJ/mol, what is the standard Gibbs free energy of the second reaction?"

I assume that you would use the equation G=-RT lN(K) setting -200 equal to -RT lN(K) to get the K value that would be multiplied by 10 to re-solve for G. However without a given temperature this is not possible. Is there a method of factoring out temperature and if not how is this problem solved?

If K is increased 10 times (K' = 10K) then
ln(K')=ln(10*K)=ln(10)+ln(K)

So the gibbs free energy which is now RTln(K') = RTln(10) + RTln(K) has increased from the previous amount by RTln(10)