Question

The rate constant of a reaction is \(4.50 \times 10^{-5} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(195^{\circ} \mathrm{C}\) and \(3.20 \times 10^{-3} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(258^{\circ} \mathrm{C}\). What is the activation energy of the reaction?

Short answer

\(E_a = 139.37 \mathrm{kJ/mol}\)

Step-by-step solution

Identify Given Values

The rate constant at two different temperatures are given: \(k_1 = 4.50 \times 10^{-5} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(195^{\circ} \mathrm{C}\) and \(k_2 = 3.20 \times 10^{-3} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(258^{\circ} \mathrm{C}\).

Convert Temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin using the formula \(T(K) = T(^{\circ}C) + 273.15\).\(T_1 = 195 + 273.15 = 468.15 \mathrm{K}\).\(T_2 = 258 + 273.15 = 531.15 \mathrm{K}\).

Use Arrhenius Equation

The Arrhenius equation is \(k = A e^{-E_a / RT}\). The form used to compare two rate constants at different temperatures is:\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \].

Plug in the Values

Plug in the values into the equation from Step 3:\[ \ln \left( \frac{3.20 \times 10^{-3}}{4.50 \times 10^{-5}} \right) = \frac{-E_a}{8.314} \left( \frac{1}{531.15} - \frac{1}{468.15} \right) \]Simplify the values inside the logarithm:\[ \ln \left( \frac{3.20 \times 10^{-3}}{4.50 \times 10^{-5}} \right) = \ln(71.1111) \].

Calculate the Natural Logarithm

Calculate \( \ln(71.1111) \):\[ \ln(71.1111) = 4.2633 \].

Calculate the Temperature Difference in the Denominator

Calculate the difference in reciprocal temperatures:\[ \frac{1}{531.15} - \frac{1}{468.15} = 0.001883 \mathrm{K^{-1}} - 0.002137 \mathrm{K^{-1}} = -0.0002541 \mathrm{K^{-1}} \].

Solve for Activation Energy

Rearrange the equation to solve for activation energy (\(E_a\)):\[ 4.2633 = \frac{-E_a}{8.314} \times (-0.0002541) \]Thus, \(E_a = \frac{4.2633 \times 8.314}{0.0002541} = 139,365.4 \mathrm{J/mol} \).

Convert Activation Energy to kJ/mol

Since activation energy is commonly expressed in \(\mathrm{kJ/mol}\), convert \(139,365.4 \mathrm{J/mol}\) to \(\mathrm{kJ/mol}\):\[ 139,365.4 \mathrm{J/mol} = 139.37 \mathrm{kJ/mol} \].

Key concepts

Arrhenius equation

The Arrhenius equation is fundamental in understanding the temperature dependence of reaction rates. It is expressed as:
\[ k = A e^{-E_a / RT} \]
where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor or frequency factor
- \( E_a \) is the activation energy
- \( R \) is the gas constant, approximately 8.314 J/(mol·K)
- \( T \) is the temperature in Kelvin
The equation shows that as the temperature increases, the rate constant \( k \) also increases, provided all other factors remain constant. For comparison of rate constants at two different temperatures, a logarithmic form of the Arrhenius equation is used:
\[ \text{ln}\bigg(\frac{k_2}{k_1}\bigg) = \frac{-E_a}{R} \bigg(\frac{1}{T_2} - \frac{1}{T_1}\bigg) \]
This equation is useful for calculating the activation energy when you know the rate constants at two different temperatures.

Rate constant

The rate constant \(k\) is a proportionality factor in the rate equation of a reaction. It reflects the speed of a reaction under specific conditions. In the problem provided, the rate constants at two different temperatures are:
- \( k_1 = 4.50 \times 10^{-5} \text{ L/mol·s} \) at 195°C
- \( k_2 = 3.20 \times 10^{-3} \text{ L/mol·s} \) at 258°C
A higher rate constant means a faster reaction. The units of the rate constant can vary depending on the order of the reaction. In the context of the Arrhenius equation, knowing the rate constants at different temperatures allows us to determine the activation energy \(E_a\), which is crucial for understanding the temperature dependence of the reaction rate.

Temperature conversion

Temperature conversion, particularly from Celsius to Kelvin, is essential in such calculations. The formula to convert temperature from Celsius to Kelvin is:
\[ T(K) = T(^{\text{o}}C) + 273.15 \]
In the exercise, given temperatures in Celsius were converted to Kelvin:
- Temperature at 195°C to Kelvin:
\[ T_1 = 195 + 273.15 = 468.15 \text{ K} \]
- Temperature at 258°C to Kelvin:
\[ T_2 = 258 + 273.15 = 531.15 \text{ K} \]
These converted temperatures are then used in the logarithmic form of the Arrhenius equation to calculate the activation energy. Proper temperature conversion ensures accurate calculations, as the Kelvin scale is the standard unit when working with the Arrhenius equation and other thermodynamic equations.