How to prove that the forward and reverse reactions have the same rate at equilibrium?

I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that the rate constant K is related to the forward and reverse reaction constants: $$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$

In answering the question it is given that: $$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$

And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to $E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

• Are $K_\mathrm{f/r}$ rate constants? – Martin - マーチン Jan 14 '16 at 7:56
• You can find your answer here, chemed.chem.purdue.edu/genchem/topicreview/bp/ch22/react.html (take care with your second equation, above). – user1945827 Jan 14 '16 at 9:56
• In the reverse reaction, the products and reactants and switched over from the forward reaction. Products are just what is made from a chemical reaction, not necessarily your desired or intended 'product'. – Spontification Jan 14 '16 at 14:20