Question

The activation energy for the decomposition of hydrogen peroxide: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ is \(42 \mathrm{~kJ} / \mathrm{mol}\), whereas when the reaction is catalyzed by the enzyme catalase, it is \(7.0 \mathrm{~kJ} / \mathrm{mol}\). Calculate the temperature that would cause the uncatalyzed decomposition to proceed as rapidly as the enzyme-catalyzed decomposition at \(20^{\circ} \mathrm{C}\). Assume the frequency factor A to be the same in both cases.

Short answer

The uncatalyzed reaction proceeds as rapidly at approximately 1486 °C.

Step-by-step solution

Understand the Arrhenius Equation

The rate constant of a reaction is given by the Arrhenius equation: \(k = Ae^{-E_a/(RT)}\), where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the absolute temperature in Kelvin. For the catalyzed and uncatalyzed reactions to proceed at the same rate, their rate constants must be equal.

Set Up the Equation for Equal Rates

Since the frequency factor \(A\) is the same for both reactions, set the Arrhenius expressions equal: \(e^{-E_{a, ext{cat}}/(RT_1)} = e^{-E_{a, ext{uncat}}/(RT_2)}\). This simplifies to \(E_{a, ext{cat}}/(RT_1) = E_{a, ext{uncat}}/(RT_2)\), where \(T_1\) is the initial temperature (\(20^{\circ}C = 293.15 \text{K}\) for the catalyzed reaction) and \(T_2\) is the temperature we need to find for the uncatalyzed reaction.

Solve for the Uncatalyzed Reaction Temperature

Rearrange the equation from Step 2 to solve for \(T_2\): \(T_2 = (E_{a, ext{uncat}}/E_{a, ext{cat}}) \times T_1\). Substitute the given \(E_{a, ext{cat}} = 7.0 \text{ kJ/mol}\) and \(E_{a, ext{uncat}} = 42 \text{ kJ/mol}\) along with \(T_1 = 293.15 \text{ K}\) into the equation. Thus, \(T_2 = (42/7.0) \times 293.15\, \text{K}\).

Calculate and Convert Temperature to Celsius

Calculate \(T_2\) using the expression from Step 3: \(T_2 = 6 \times 293.15 = 1758.9 \text{ K}\). Convert \(T_2\) from Kelvin to Celsius: \(T_2\, (^{\circ}C) = 1758.9 - 273.15 = 1485.75 \text{ °C}\).

Key concepts

Arrhenius Equation

The Arrhenius Equation is a key formula in chemistry used to determine the rate of a chemical reaction. The equation is: \[ k = Ae^{-E_a/(RT)} \]Here,

• \( k \): Rate constant of the reaction
• \( A \): Frequency factor, which is related to the rate at which molecules collide
• \( E_a \): Activation energy, the minimum energy required to start a reaction
• \( R \): Universal gas constant, approximately \(8.314 \, \text{J/mol K}\)
• \( T \): Absolute temperature in Kelvin

This equation shows the dependence of the reaction rate on the temperature and activation energy. By increasing the temperature or lowering the activation energy, the rate of the reaction increases, allowing reactions to proceed faster.

Catalysis

Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst. Catalysts work by lowering the activation energy \( E_a \) required for the reaction to proceed.This means reactions can occur more rapidly or at lower temperatures. In the context of hydrogen peroxide decomposition, the enzyme catalase acts as a natural catalyst. It significantly lowers the activation energy from \( 42 \, \text{kJ/mol} \) to \( 7.0 \, \text{kJ/mol} \), making the reaction much faster.Effective catalysts are crucial in many industrial and biological processes, ensuring that they are efficient and require less energy. Importantly, catalysts are not consumed or altered permanently in reactions, allowing them to function repeatedly.

Hydrogen Peroxide Decomposition

Hydrogen peroxide \((\text{H}_2\text{O}_2)\) decomposes into water \((\text{H}_2\text{O})\) and oxygen gas \((\text{O}_2)\). The reaction proceeds more quickly in the presence of a catalyst like catalase.The decomposition can be represented by:\[ 2 \text{H}_2\text{O}_2(aq) \longrightarrow 2 \text{H}_2\text{O}(l) + \text{O}_2(g) \]Without catalysis, hydrogen peroxide decomposes slowly under normal conditions. The significant drop in activation energy when catalase is present makes the process suitable for biological systems like human cells, where hydrogen peroxide is commonly broken down to avoid damage.

Temperature Calculation

To find the temperature at which the uncatalyzed decomposition of hydrogen peroxide proceeds at the same rate as the enzyme-catalyzed reaction at \(20^\circ \text{C}\), we can use the relationship derived from the Arrhenius Equation.For reactions to proceed at the same rate, their rate constants must be equal:\[ e^{-E_{a, \text{cat}}/(RT_1)} = e^{-E_{a, \text{uncat}}/(RT_2)} \]After simplification, it can be rearranged to:\[ T_2 = \left( \frac{E_{a, \text{uncat}}}{E_{a, \text{cat}}} \right) \times T_1 \]Substituting the given values:\( E_{a, \text{cat}} = 7.0 \, \text{kJ/mol} \), \( E_{a, \text{uncat}} = 42 \, \text{kJ/mol} \), and \( T_1 = 293.15 \, \text{K} \) (which corresponds to \(20^\circ \text{C}\)), the calculation results in:\[ T_2 = 6 \times 293.15 \, \text{K} = 1758.9 \, \text{K} \]Converting this temperature from Kelvin to Celsius gives:\[ T_2 \, (^\circ \text{C}) = 1758.9 - 273.15 = 1485.75 \, ^\circ \text{C} \]This reflects the significantly higher temperature needed for the uncatalyzed reaction to match the catalyzed reaction's rate.