Effect of catalysts on equilibria

Catalysts have no effect on either the equilibrium constant or the position of equilibrium of a reaction. A catalyst offers a separate pathway for a reaction to proceed with lower activation energy, E_a, and does not affect the energy levels of the reactants and products. As a result, a catalyst increases the rates of both the forward and reverse reactions by the same factor and allows a reaction to reach its dynamic equilibrium faster without affecting either the equilibrium constant or the position of equilibrium of the reaction.

Mathematically, let’s consider the following reversible reaction

$A\; \begin{matrix} k_f\\\rightleftharpoons \\ k_r \end{matrix}\: B$

where k_f is the rate constant of the forward reaction and k_r is that for the reverse reaction. In chemical kinetics, the rates of formation of B and A are:

$rate_f=k_f[A]\; \; \; and\; \; \;rate_r=k_r[B]$

At equilibrium, rate_f = rate_r

$k_f[A]=k_r[B]\; \; \; \; \; \; \; \; 22$

Substitute the equilibrium constant, K = [B]/[A] in eq22

$K=\frac{k_f}{k_r}\; \; \; \; \; \; \; \; 23$

Substitute the Arrhenius equation where $k=Ae^{-\frac{E_a}{RT}}$ in eq23

$K=\frac{A_fe^{-\frac{E_a}{RT}}}{A_re^{-\frac{E_a\, '}{RT}}}\; \; \; \; \; \; \; \; 24$

For a catalysed reaction where the activation energy decreases by an amount, $\left | \Delta E \right |$ ,

$K_{cat}=\frac{k_{f,cat}}{k_{r,cat}}=\frac{A_{f,cat}e^{-\frac{E_a-\left | \Delta E \right |}{RT}}}{A_{r,cat}e^{-\frac{E_a\, '-\left | \Delta E \right |}{RT}}}=\frac{A_{f,cat}e^{-\frac{E_a}{RT}}}{A_{r,cat}e^{-\frac{E_a\, '}{RT}}}\frac{e^{\frac{\left | \Delta E \right |}{RT}}}{e^{\frac{\left | \Delta E \right |}{RT}}} =\frac{A_{f,cat}e^{-\frac{E_a}{RT}}}{A_{r,cat}e^{-\frac{E_a\, '}{RT}}}\; \; \; \; \; \; \; \; 25$

Assuming the ratio of the forward pre-exponential factor and the reverse pre-exponential factor for the uncatalysed reaction (A_f/A_r) is the same as that for the catalysed one (A_f,cat/A_r,cat), the equilibrium constant for an uncatalysed reaction, eq24, is the same as that for a catalysed one, eq25. This means that the catalytic pathway does not affect the value of the equilibrium constant. Furthermore, both the numerator and denominator of eq24 are unchanged despite a change in the activation energy in the catalysed reaction, which means that the position of equilibrium remains the same. Lastly, the forward rate of reaction, k_f[A], and reverse rate of reaction, k_r[B], increase by the same factor of $e^{\frac{\left | \Delta E \right |}{RT}}$ for the catalysed reaction, allowing the reaction to achieve equilibrium faster.

Question

If a catalyst increases the rates of both the forward and reverse reactions of a reversible reaction, does it mean that we will not achieve a greater amount of products than an uncatalysed reaction (assuming we begin both the catalysed and uncatalysed reactions consisting of only the same amount of reactants under the same temperature and pressure, and that k_f> k_r for the uncatalysed reaction)?

Answer

Since a catalysed reaction does not alter either the equilibrium constant or the position of the equilibrium of the reaction, we will obtain the same amount of product as the uncatalysed reaction, but at a faster rate.

Effect of catalysts on equilibria

Question

Answer

Previous article: Effect of temperature on equilibrium

Content page of chemical equilibrium

Content page of intermediate chemistry

Main content page

Leave a Reply Cancel reply