We know that increasing the temperature increases the rate of reaction by increasing the rate constant \( k \). The relationship between the rate constant, temperature, and activation energy is quantified by the Arrhenius equation.
Key Principle
The rate constant \( k \) increases exponentially as temperature increases. This is because a higher temperature increases the fraction of molecules that possess kinetic energy equal to or greater than the activation energy (\( E_a \)).
The Arrhenius Equation
The equation that relates the rate constant \( k \) to the absolute temperature \( T \):
\( k = Ae^{-\frac{E_a}{RT}} \)
Where:
- \( k \) is the rate constant (units vary).
- \( A \) is the pre-exponential factor (same units as \( k \)). It is also called the Arrhenius constant. It represents the frequency of collisions with correct orientation.
- \( E_a \) is the activation energy in J mol⁻¹.
- \( R \) is the gas constant (8.314 J K⁻¹mol⁻¹).
- \( T \) is the absolute temperature in Kelvin (K).
- \( e \) is the mathematical base of natural logarithms (approx. 2.718).
Units are the most common source of calculation errors here! Activation energy is usually given in kJ mol⁻¹, but the gas constant \( R \) is in J K⁻¹mol⁻¹. You must multiply \( E_a \) by 1000 to convert it to J mol⁻¹ before using it in the Arrhenius equation.
The Logarithmic Form of the Arrhenius Equation
To determine \( E_a \) and \( A \) from experimental data, we take the natural logarithm (\( \ln \)) of both sides of the Arrhenius equation. This yields the straight-line equation:
This matches the mathematical equation for a straight line, \( y = mx + c \):
- \( y \) corresponds to \( \ln k \)
- \( x \) corresponds to \( \frac{1}{T} \) (measured in K⁻¹)
- \( m \) (gradient) corresponds to \( -\frac{E_a}{R} \)
- \( c \) (y-intercept) corresponds to \( \ln A \)
Plotting the Arrhenius Graph
If we run a reaction at multiple temperatures, calculate the rate constant \( k \) for each, and plot \( \ln k \) against \( 1/T \), we get a straight line with a negative gradient:
How to calculate \( E_a \) and \( A \) from experimental data
- Draw a line of best fit on your Arrhenius plot.
- Choose two points on the line that are far apart to calculate the gradient:
\( m = \frac{\Delta y}{\Delta x} = \frac{\ln k_2 - \ln k_1}{\frac{1}{T_2} - \frac{1}{T_1}} \)
- The gradient \( m \) will be negative. Equate it to \( -E_a/R \) and solve for \( E_a \):
\( E_a = -\text{Gradient} \times R \)
Multiply the negative gradient by 8.314 J K⁻¹mol⁻¹. This gives \( E_a \) in J mol⁻¹. Divide by 1000 to convert to kJ mol⁻¹.
- To find the pre-exponential factor \( A \), either extrapolate the line to find the y-intercept (\( c = \ln A \)), or select a point (\( 1/T, \ln k \)) on the line of best fit and solve:
\( \ln A = \ln k - m \left(\frac{1}{T}\right) \implies A = e^{\ln A} \)
Step 1: Convert units
\( E_a = 75.0 \times 1000 = 75,000\text{ J mol}^{-1} \)
Step 2: Rearrange the Arrhenius equation for \( A \):
\( A = \frac{k}{e^{-\frac{E_a}{RT}}} = k \cdot e^{\frac{E_a}{RT}} \)
Step 3: Calculate the exponent value:
\( \frac{E_a}{RT} = \frac{75,000}{8.314 \times 300} = \frac{75,000}{2494.2} = 30.0697 \)
Step 4: Solve for \( A \):
\( A = (2.45 \times 10^{-4}) \times e^{30.0697} \)
\( A = (2.45 \times 10^{-4}) \times (1.1458 \times 10^{13}) = 2.81 \times 10^9\text{ s}^{-1} \)
Step 1: Set up the gradient equation
\( \text{Gradient} = -\frac{E_a}{R} \)
Step 2: Solve for \( E_a \) in J mol⁻¹
\( E_a = -\text{Gradient} \times R \)
\( E_a = -(-1.20 \times 10^4\text{ K}) \times 8.314\text{ J K}^{-1}\text{mol}^{-1} \)
\( E_a = 1.20 \times 10^4 \times 8.314 = 99,768\text{ J mol}^{-1} \)
Step 3: Convert to kJ mol⁻¹
\( E_a = \frac{99,768}{1000} = 99.8\text{ kJ mol}^{-1} \)
Therefore, the activation energy of the reaction is 99.8 kJ mol⁻¹.
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