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

The catalyst will increase the rate of reaction by approximately 211.89 times at \(25^{\circ} \mathrm{C}\) and 69.40 times at \(125^{\circ} \mathrm{C}\).

Step 1: Convert temperatures to Kelvin

To calculate the reaction rates, we must first convert the given temperatures to Kelvin. The conversion formula is: \(T_{K} = T_{C} + 273.15\) For the two given temperatures, the Kelvin values are: \(T_{25^{\circ}\mathrm{C}} = 25 + 273.15 = 298.15\,K\) \(T_{125^{\circ}\mathrm{C}} = 125 + 273.15 = 398.15\,K\)

Step 2: Calculate the reaction rate constants

Using the Arrhenius equation, we can calculate the reaction rate constants for both the catalyzed and uncatalyzed reactions. We have: \(k_{uncatalyzed} = Ae^{-E_a(uncatalyzed) / RT}\) \(k_{catalyzed} = Ae^{-E_a(catalyzed) / RT}\) Since the collision factor A remains the same in both equations, we can find the ratio of catalyzed reaction rate to uncatalyzed reaction rate as: \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-E_a(catalyzed) / RT}}{e^{-E_a(uncatalyzed) / RT}}\)

Step 3: Calculate the rate increase factors

Now, we can plug in the given activation energies and temperature values to calculate the rate increase factors for both temperatures. (a) For \(T = 298.15 K\): \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-(55 \times 10^3) / (8.314 \times 298.15)}}{e^{-(95 \times 10^3) / (8.314 \times 298.15)}} ≈ 211.89\) (b) For \(T = 398.15 K\): \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-(55 \times 10^3) / (8.314 \times 398.15)}}{e^{-(95 \times 10^3) / (8.314 \times 398.15)}} ≈ 69.40\) Therefore, the catalyst will increase the rate of reaction by approximately 211.89 times at \(25^{\circ} \mathrm{C}\) and 69.40 times at \(125^{\circ} \mathrm{C}\).

Key Concepts

Activation Energy

Activation energy is the minimum amount of energy that reacting particles need to undergo a chemical reaction. It acts like a barrier, and only particles with energy equal to or greater than this barrier will successfully react. Let's imagine activation energy as a hill that reactants have to climb. If you add a catalyst, it lowers the height of the hill, making it easier for reactions to happen.
An uncatalyzed reaction has a higher activation energy compared to a catalyzed one. In our example, the activation energy decreases from 95 kJ/mol to 55 kJ/mol when a catalyst is introduced. This significant reduction implies that more particles will have sufficient energy to overcome this barrier and participate in the reaction, making the reaction faster.

Arrhenius Equation

The Arrhenius equation is a formula used to express how reaction rates are affected by temperature and activation energy. It is given by:

• \( k = Ae^{-E_a/RT} \)

Here, \( k \) is the reaction rate constant, \( A \) is the pre-exponential factor or frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
The equation shows that the reaction rate \( k \) increases with a decrease in activation energy (as we see with catalysis) or with an increase in temperature. Exponent \(-E_a/RT\) is a critical factor; as \( E_a \) decreases or \( T \) increases, the exponent becomes less negative, which increases the value of \( k \). This equation helps explain why catalysis is so effective in increasing reaction rates.

Reaction Rate Constants

The reaction rate constant, \( k \), is a crucial part of the Arrhenius equation. It determines the speed of a chemical reaction at a given temperature and activation energy. With catalysis, even if the collision factor \( A \) remains the same, the lower activation energy reduces the negative impact on the exponent of the Arrhenius equation, leading to a higher \( k \).
In the context of our example, inserting the activation energies and temperature values into the Arrhenius equation for both catalyzed and uncatalyzed reactions helps us compute the reaction rate constants. The ratio of these constants gives us a quantitative understanding of how much faster (or slower) a reaction occurs—a key insight for optimizing chemical processes.

Temperature Conversion

Temperature conversion, especially from Celsius to Kelvin, is fundamental in calculations involving the Arrhenius equation. This is because reaction rates depend on the absolute temperature, not simply the temperature difference.
The formula for converting Celsius to Kelvin is:

• \( T_K = T_C + 273.15 \)

In our exercise, the temperatures 25°C and 125°C were converted to 298.15 K and 398.15 K, respectively. Kelvin is used because it ensures that temperature measurements start at absolute zero, providing an accurate scale for kinetic calculations as required by thermodynamic principles. Proper temperature conversion is key to correctly using the Arrhenius equation and assessing reaction kinetics.