Question

Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\) 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

At equilibrium, we can express the equilibrium constant (K) as the ratio of the forward and reverse reaction rate constants, k_f and k_r: \(K = \frac{k_f}{k_r}\). We apply the Arrhenius equation to find k_f and k_r as \(k_f = A_{f}e^{-E_{f}/RT}\) and \(k_r = A_{r}e^{-E_{r}/RT}\), respectively. Expressing K as a ratio of k_f and k_r, we obtain \(K = e^{\frac{E_{r}-E_{f}}{RT}}\). Using the relationship between ΔG° and K, we derive the equilibrium constant K as a function of the forward and reverse reaction rate constants, proving the statement. For part b, the statement, "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction," is false as a catalyst lowers the activation energy for both the forward and reverse reactions, thereby increasing the rate of both reactions without altering the overall equilibrium constant K.

Step-by-step solution

Understand and represent equilibrium constant K

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. Using the given relationship, we can represent the equilibrium constant, K, as the ratio of the forward and reverse reaction rate constants: \(K = \frac{k_f}{k_r}\). Now, our goal is to derive K using the provided relationships.

Write the relationship between ΔG° and K

First, recall that the relationship between standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) is given by: \(\Delta G^\circ = -R T \ln(K)\), where R is the gas constant and T is the temperature, in Kelvin.

Apply the Arrhenius equation to find k_f and k_r

The Arrhenius equation defines the relationship between the rate constant k and the activation energy E, given by: \(k = Ae^{-E/RT}\), where A is the pre-exponential factor. We define \(E_{f}\) as the activation energy of the forward reaction and \(E_{r}\) as the activation energy of the reverse reaction. Then, for the forward reaction, we can write: \(k_f = A_{f}e^{-E_{f}/RT}\), and for the reverse reaction: \(k_r = A_{r}e^{-E_{r}/RT}\).

Express K as the ratio of k_f and k_r

Express the equilibrium constant (K) as the ratio of the forward and reverse reaction rate constants, k_f and k_r, using the Arrhenius equation: \(K = \frac{k_f}{k_r} = \frac{A_{f}e^{-E_{f}/RT}}{A_{r}e^{-E_{r}/RT}}\)

Simplify and find the relationship between ΔG°, K, and E_r - E_f

Simplify the expression: \(K = e^{\frac{-E_{f}+E_{r}}{RT}} = e^{\frac{E_{r}-E_{f}}{RT}}\) Using the relationship between ΔG° and K, we have: \(\Delta G^\circ = -R T \ln(K) = -RT(\ln e^{(E_{r}-E_{f})/RT}) = - (E_{r}-E_{f})\) Now, we have derived the equilibrium constant K as a function of the forward and reverse reaction rate constants, k_f and k_r, proving the statement in the exercise. For part b:

Evaluate the statement

The statement given is: "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

Evaluate the effect of a catalyst on forward and reverse reactions

A catalyst lowers the activation energy for both the forward and reverse reactions, thereby increasing the rate of both reactions. It does not change the overall equilibrium constant K and thus does not alter the relative concentrations of the reactants and products at equilibrium.

Conclude about the falsity of the statement

Since a catalyst increases the rate of both the forward and reverse reactions and does not discriminate between them, the statement, "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction," is false.

Key concepts

Gibbs free energy

Gibbs free energy is a crucial concept in understanding chemical reactions and their spontaneity. It is denoted by the symbol \( \Delta G \). This thermodynamic property helps us determine whether a reaction can occur without any additional energy.

The equation \( \Delta G = \Delta H - T\Delta S \) provides insight into how enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) affect a reaction. A negative \( \Delta G \) implies that the reaction proceeds spontaneously. On the contrary, a positive value indicates non-spontaneity.

In equilibrium, the Gibbs free energy change (\(\Delta G^{\circ}\)) is directly related to the equilibrium constant (\(K\)). The relationship \( \Delta G^{\circ} = -RT\ln K \) showcases how the standard Gibbs free energy change influences the position of equilibrium. The reaction will favor products if the equilibrium constant is greater than one, and reactants if it is less than one.

Understanding this relationship is pivotal in calculating and determining chemical equilibrium, providing a comprehensive picture of reaction possibilities.

Arrhenius equation

The Arrhenius equation is fundamental to predicting reaction rates and understanding temperature's impact.

The formula is expressed as \( k = Ae^{-E_a/RT} \), where:

• \(k\) is the rate constant
• \(A\) is the pre-exponential factor, representing the frequency of collisions
• \(E_a\) is the activation energy
• \(R\) is the universal gas constant
• \(T\) is the temperature in Kelvin

The equation shows that reactions occur faster when the temperature increases or the activation energy decreases. A higher temperature causes particles to move faster, leading to more effective collisions.

Using the Arrhenius equation helps chemists understand and manipulate reaction conditions, thereby controlling the rates of chemical processes. It also allows for the calculation of activation energy by plotting the natural logarithm of the rate constant against the inverse of temperature.

Catalysis

Catalysis plays an instrumental role in enhancing reaction rates without being consumed by the reaction. Catalysts work by providing an alternate route with a lower activation energy compared to the uncatalyzed pathway.

This reduction in activation energy doesn't affect the thermodynamics of the reaction, like Gibbs free energy, but it allows the reaction to occur more swiftly and efficiently.

Catalysts increase the rates of both forward and reverse reactions equally, essentially speeding up the attainment of equilibrium without shifting it. They are widely used in industrial processes to achieve desired results faster, making them cost-effective and practical solutions in chemical manufacturing.

Understanding how catalysts work helps us improve reaction efficiency while maintaining the integrity and balance of the chemical system.

Activation energy

Activation energy is the minimum amount of energy required for a chemical reaction to occur. This concept is fundamental as it represents the energy barrier that reactants must overcome to transform into products.

In the energy profile of a reaction, activation energy (\(E_a\)) is the peak barrier that needs surmounting. It determines the reaction's speed; higher activation energy usually results in slower reactions since fewer molecules have sufficient energy to react.

The relationship between activation energy and temperature is well-illustrated by the Arrhenius equation, where a lower activation energy or higher temperature results in increased reaction rates.

Reducing activation energy, often achieved through catalysis, allows reactions to proceed at practical rates at lower temperatures. This understanding is vital in various applications, from manufacturing to biochemistry, where reaction conditions must align with desired outcomes.