Problem 92 - Chemistry An Atoms First Approach

Question

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \space\mathrm{kJ} / \mathrm{mol} .\) In the presence of a catalyst at \(37^{\circ} \mathrm{C}\) the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Step-by-Step Solution

Answer

The activation energy for the catalyzed reaction is approximately \(29.10\space\mathrm{kJ/mol}\).

Step 1: Write the Arrhenius equation for catalyzed and uncatalyzed reactions

We have the Arrhenius equation for the rate constant (k) for both the uncatalyzed (k1) and the catalyzed (k2) reactions: \[k_1=A\space e^{\frac{-E_a}{RT}}\] \[k_2=A\space e^{\frac{-E_{a^{'}}}{RT}}\] Where: \(k_1\) - rate constant for uncatalyzed reaction \(k_2\) - rate constant for catalyzed reaction A - frequency factor (same for both reactions) \(-E_a\) - activation energy for uncatalyzed reaction \(-E_{a^{'}}\) - activation energy for catalyzed reaction R - gas constant (\(8.314\space\mathrm{J} / \mathrm{mol \cdot K}\)) T - temperature in Kelvin

Step 2: Calculate the temperature in Kelvin and set the rate constants ratio

First, convert the given temperature from Celsius to Kelvin by adding 273.15: \[T = 37^\circ C + 273.15 = 310.15 K\] Then, set the ratio for the rate constants based on the given rate constant increase factor: \[\frac{k_2}{k_1} = 2.50 \times 10^{3}\]

Step 3: Divide one Arrhenius equation by the other and eliminate A

Divide the Arrhenius equation for the catalyzed reaction by the one for the uncatalyzed reaction: \[\frac{k_2}{k_1} = \frac{A\space e^{\frac{-E_{a^{'}}}{RT}}}{A\space e^{\frac{-E_a}{RT}}}\] A will cancel out: \[\frac{k_2}{k_1} = \frac{e^{\frac{-E_{a^{'}}}{RT}}}{e^{\frac{-E_a}{RT}}}\] Now substitute the calculated ratio and activation energy of the uncatalyzed reaction: \[\frac{1}{2.50 \times 10^{3}} = \frac{e^{\frac{-E_{a^{'}}}{(8.314)(310.15)}}}{e^{\frac{-50\,000}{(8.314)(310.15)}}}\]

Step 4: Solve for the activation energy of the catalyzed reaction

Rearrange the equation so that expressions with \(-E_{a^{'}}\) and \(-E_{a}\) are on the same side, and then take the natural logarithm of both sides: \[\ln\left(\frac{1}{2.50 \times 10^{3}}\right) = \frac{-E_{a^{'}}}{(8.314)(310.15)} - \frac{-50\,000}{(8.314)(310.15)}\] Next, solve for \(-E_{a^{'}}\): \[-E_{a^{'}} = [(8.314)(310.15)]\left[\ln\left(\frac{1}{2.50 \times 10^{3}}\right) + \frac{50\,000}{(8.314)(310.15)}\right]\] Now calculate \(E_{a^{'}}\): \[E_{a^{'}} = -\left[(8.314)(310.15)\left(\ln\left(\frac{1}{2.50 \times 10^{3}}\right) - \frac{50\,000}{(8.314)(310.15)}\right)\right] \approx 29.10 \space\mathrm{kJ/mol}\]

Step 5: Express the answer

The activation energy for the catalyzed reaction, rounding to two decimal points, is approximately \(29.10\space\mathrm{kJ/mol}\).

Key Concepts

Arrhenius Equation

The Arrhenius equation is fundamental in chemistry for understanding how temperature affects the rate of a chemical reaction. It is expressed as:\[k = A e^{-E_a/RT}\]Where:

• \(k\) is the rate constant of the reaction.
• \(A\) represents the frequency factor, which is related to the frequency of collisions with correct orientation.
• \(E_a\) is the activation energy, or the energy barrier that must be overcome for the reaction to occur.
• \(R\) is the gas constant \(8.314\, \text{J/mol}\cdot\text{K}\).
• \(T\) is the absolute temperature in Kelvin.

The equation shows that as the temperature or the frequency factor increases, the rate constant \(k\) also increases, thereby speeding up the reaction. Furthermore, a lower activation energy \(E_a\) makes the reaction easier to occur, enhancing the rate constant and thus the reaction rate itself.

Catalysis

Catalysis is a powerful concept in chemistry that involves the use of catalysts to increase the rate of a chemical reaction. A catalyst achieves this by lowering the activation energy \(E_a\) required for the reaction.- Catalysts are not consumed during the reaction, meaning they can be used repeatedly.- They provide an alternative reaction pathway with a lower energy barrier, which speeds up the reaction.In a biochemical setting, enzymes often serve as catalysts. For example, in the biochemical reaction described, a catalyst decreased the activation energy from 50.0 \(\text{kJ/mol}\) to approximately 29.10 \(\text{kJ/mol}\), making the process more efficient.Using catalysts is essential for many industrial processes as well as biological functions to ensure reactions happen at a reasonable rate without the need for high temperatures.

Biochemical Reaction

Biochemical reactions are chemical processes that occur in living organisms. These reactions are crucial for sustaining life, allowing complex biological processes to take place.- They often involve enzymes, which are biological catalysts that speed up reactions.- Conditions such as temperature and pH are critical for optimal enzyme activity and, by extension, for the fidelity of biochemical reactions.For instance, the exercise illustrates a biochemical reaction at body temperature \(37^\circ \text{C}\). The catalyst helps maintain the reaction rate at this physiological temperature, which is energetically more favorable than at higher temperatures that might damage cells.

Rate Constant

The rate constant, denoted as \(k\), is a scaling factor that relates the rate of a chemical reaction to the concentrations of reactants.- In the Arrhenius equation, \(k\) increases as temperature increases or when the activation energy \(E_a\) is lowered by a catalyst.- For physical and biochemical reactions, the rate constant helps predict how fast the reaction will occur under different conditions.In the given problem, the presence of a catalyst significantly increases the rate constant \(k_2\) by a factor of \(2.50 \times 10^3\), compared to the uncatalyzed reaction \(k_1\). This increase demonstrates how effective catalysts are in improving the efficiency of chemical processes.