And let's suppose the rate constant is k one and that means the rate law for this reaction is given by so the rate of the four reaction v forward is equal to k one multiplied by the concentration of A to the power of one, because the coefficient here is one. If this coefficient was two, then this exponent here would be two and so forth. Now, if we go in reverse, then the only thing that changes is this becomes the reactor, this becomes the product. And so we replace the concentration A with concentration B and we replace k one with K minus one. And remember that if this reaction achieves equilibrium, then the rate of the forward will be equal to the rate of the reverse and these two quantities will be equal. Now notice, k one is not necessarily equal to K minus one.