Question

a) Using the free energy profile for a simple one-step reaction, we show that at equilibrium K=k_f/k_r, where kf and kr are the rate constants for the forward and reverse reactions, respectively. Hint: Use the relationship∆G°= -RT In(K), and represent kf and kr using the Arrhenius equation (k = Ae^-E₂^/RT).

b. Why is the following statement false? “A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction.”

Short answer

a)In this problem, we are given a free energy profile for a simple one-step reaction and are asked to show that at equilibrium.

$K=\frac{K_t}{K_f}$

b) Catalysts increase the forward rate, while reducing the reverse rate. Because A catalyst is a substance that increases the rate of a reaction without being altered or used up in the reaction. Both the forward and reverse rates of the reaction are accelerated by a catalyst.

Step-by-step solution

Determine the equation (k = Ae-E2/RT)

$\Delta G°=-RT\ln \left(K\right)$

In addition, we know the Arrhenius equation, which relates K to the activation energy for forward and reverse reactions:

$$\begin{gathered} k_t=Ae\frac{E_f}{RT}and \\ {k}_r=Ae\frac{E_f}{RT} \end{gathered}$$

Also, if we look at the free energy profile, we that$\Delta G=E_r-E_f$

Determine to solve for K in terms of the activation energies:

$$\begin{gathered} k_f=A{e}^{\frac{E_f}{RT}} \\ \frac{k_f}{A}=e{}^{\frac{E_f}{RT}} \\ \ln \left(\frac{k_f}{A}\right)=-\frac{E_f}{RT} \\ {E}_f=-RT\ln \left(\frac{k_r}{A}\right) \end{gathered}$$

And if we replace the f subscripts with r.

$E_r=-RT\ln \left(\frac{k_r}{A}\right)$

Determine to how catalyst increases the rate of reaction

A catalyst increases the rate of reaction in a slightly unconventional way compared to other means of increasing the reaction rate. The role of a catalyst is to lower the activation energy so that a greater proportion of the particles have enough energy to react.