Equation 1 relates the activation energy of a reverse process to the activation energy of the forward process, which are always different by the reaction energy, ΔEr. Therefore, given a constant horizontal displacement in fA,-1, a vertical displacement of the same magnitude would appear in fA. The first and second derivatives (Equations 2 and 3) show that the gradients of the forward and reverse reactions are complementary, adding up to one. The curvature of the backward reaction, as seen in the behaviour of the second derivative, is the mirror image of that of the forward reaction about the y axis (as -∆𝐸 𝑟,-1 = ∆𝐸 𝑟 ).