Consider the equilibrium A ⇌B in which both the forward and reverse reactions are elementary (si

Consider the equilibrium A ⇌B in which both the forward and...

Consider the equilibrium $\mathrm{A} \rightleftharpoons \mathrm{B}$ in which both the forward and reverse reactions are elementary (single-step) reactions. Assume that the only effect of a catalyst on the reaction is to lower the activation energies of the forward and reverse reactions, as shown in Figure $15.15 .$ Using the Arrhenius equation (Section 14.5$)$ , prove that the equilibrium constant is the same for the catalyzed reaction as for the uncatalyzed one.

Key Concepts

Equilibrium Constant

The equilibrium constant (K) is a fundamental parameter in chemical thermodynamics that quantifies the ratio of the concentrations (or pressures) of products to reactants at equilibrium. In the context of both catalyzed and uncatalyzed reactions, K depends only on the free energy change of the overall reaction and is independent of the reaction pathway or the rate constants. Since catalysts only affect the speeds of the forward and reverse reactions without changing the thermodynamic free energy change, the equilibrium constant remains the same.

Catalysts

Catalysts are substances that increase the rates of both the forward and reverse reactions by providing an alternative reaction pathway with lower activation energies. They do not alter the position of equilibrium, as they equally enhance the formation of products and the regeneration of reactants. This fundamental property ensures that while catalysts speed up the attainment of equilibrium, they have no effect on the equilibrium constant.

Arrhenius Equation

The Arrhenius equation relates the rate constant of a reaction to the activation energy and temperature, providing a mathematical description of how reaction rates depend on these parameters. It shows that even when the activation energies are lowered by a catalyst, both the forward and reverse reaction rate constants are modified in such a way that their ratio—and hence the equilibrium constant—remains unchanged. This equation underpins the quantitative understanding of how catalysts speed up chemical reactions without affecting equilibrium.

Activation Energy

Activation energy is the energy barrier that must be overcome for reactants to be converted into products. In an uncatalyzed reaction, higher activation energies correspond to slower rates. A catalyst lowers these energy barriers for both the forward and reverse reactions, thereby accelerating the reaction without changing the net energy change of the reaction. This selective lowering of the activation energies is crucial for understanding why catalysts do not alter the equilibrium constant.

Elementary Reaction

An elementary reaction is one that occurs in a single step and has a rate law that can be directly inferred from its stoichiometry. In such reactions, the elementary nature means every molecular event is straightforwardly related to its rate constant. This assumption simplifies the analysis because the effects of a catalyst on reaction kinetics can be directly linked to the changes in activation energies for the forward and reverse elementary steps, reinforcing that the equilibrium constant, defined by the ratio of these rate constants, remains unaffected.

Transcript

00:02 In this problem, we are considering the equilibrium between species a and species b.

00:10 We are told that a catalyst is used which lowers the activation energy of the reaction.

00:18 So here in this plot with energy on the y -axis and reaction progress on the x -axis, the top curve here is an uncatalized reaction with a large activation energy, and the catalyzed reaction has a smaller activation energy.

00:34 And the question asks us to use the rnas equation to prove that the equilibrium constant is unchanged.

00:46 All that's happening is that equilibrium is achieved faster.

00:52 So let's remember the rnas equation.

00:56 That's some rate constant little k equals a pre -exponential factor a times e to the negative ea for activation energy over rt, where r is the ideal gas constant and t is the temperature.

01:19 Let's also remember how we are going to compute kc, that's products over reactants, at equilibrium, which i'll denote with the subscript e.

01:38 So we are interested in finding whether kc will be the same or different when we're using this catalyst.

01:49 What we can do next is consider the forward reaction.

01:52 So this is going to be kf.

02:00 Kf, written as the rnae equation, is going to be a.

02:05 I'm going to put a subscript a subscript of f because that's the pre -exponential factor for the forward direction, times e to the negative ea, f for the forward direction, divided by rt, as we're assuming constant temperature here.

02:32 When we employ the use of a catalyst, the term ea itself decreases, so that gives us a less negative exponent, and this term as a whole increases...

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