Problem 87 - Chemistry: The Central Science

Question

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C}\), (b) \(125^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

At 25°C (298.15 K), the catalyst increases the rate of the reaction by a factor of about 713.95. At 125°C (398.15 K), the catalyst increases the rate of the reaction by a factor of about 49.65.

Step 1: Understand the Arrhenius equation

The Arrhenius equation is used to calculate the rate constant (k) of a reaction, given the activation energy (Ea) and temperature (T): \[k = Ae^{\frac{-Ea}{RT}}\] where A is the pre-exponential factor (also known as the collision factor), R is the gas constant (8.314 J/(mol K)), and T is the temperature in Kelvin. Since we are only interested in the ratio of the rate constants for the uncatalyzed and catalyzed reactions, the pre-exponential factor will cancel out. Therefore, we only need to compare the exponents for the two reactions.

Step 2: Convert temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin: (a) \(T_1 = 25 + 273.15 = 298.15 K\) (b) \(T_2 = 125 + 273.15 = 398.15 K\)

Step 3: Calculate the ratio of the rate constants

For both temperatures, calculate the ratio of the rate constants for the catalyzed reaction (Ea = 55 kJ/mol) to the uncatalyzed reaction (Ea = 95 kJ/mol): (a) At 298.15 K: \[\frac{k_{cat}}{k_{uncat}} = \frac{e^{\frac{-55 \times 10^3}{8.314 \times 298.15}}}{e^{\frac{-95 \times 10^3}{8.314 \times 298.15}}} = e^{\frac{40 \times 10^3}{8.314 \times 298.15}} \approx 713.95\] (b) At 398.15 K: \[\frac{k_{cat}}{k_{uncat}} = \frac{e^{\frac{-55 \times 10^3}{8.314 \times 398.15}}}{e^{\frac{-95 \times 10^3}{8.314 \times 398.15}}} = e^{\frac{40 \times 10^3}{8.314 \times 398.15}} \approx 49.65\]

Step 4: Interpret the results

The factor by which the catalyst increases the rate of the reaction at each temperature is: (a) At 25°C (298.15 K): The catalyst increases the rate of the reaction by a factor of about 713.95 (b) At 125°C (398.15 K): The catalyst increases the rate of the reaction by a factor of about 49.65