The Arrhenius equation links reaction rate to temperature and activation energy

The Arrhenius equation is a mathematical formula that connects the rate constant (k) of a chemical reaction with the temperature (T) and the activation energy (E_a). This equation helps us understand how these factors affect the rate of a chemical reaction.

The Arrhenius equation is given by:

$\text{k }=\text{ A}{e}^{(\frac{-{\text{E}}_{\text{a}}}{\text{RT}})}$

Where:

• k = rate constant.
• A = pre-exponential factor (also known as the Arrhenius constant).
• e = mathematical constant (the base of natural logarithms).
• E_a = activation energy (J mol^-1).
• R = gas constant (8.31 J K^-1 mol^-1).
• T = temperature (K).

Influence of activation energy and temperature on reaction rate

The Arrhenius equation reveals important relationships between the rate constant, activation energy, and temperature:

1. Activation energy (E_a):

• The higher the E_a, the lower the rate constant (k) becomes.
• A high activation energy means that fewer reactant molecules have sufficient kinetic energy to overcome the energy barrier and react successfully.
• Consequently, a reaction with a high E_a will have a slower rate.

2. Temperature (T):

• An increase in temperature leads to an exponential increase in the rate constant (k).
• At higher temperatures, reactant molecules move faster, resulting in more frequent collisions.
• Additionally, higher temperatures provide greater kinetic energy to reactant molecules, allowing a higher proportion to exceed the activation energy barrier.
• Both of these factors contribute to the increased reaction rate when temperature increases.

Notably, even a small temperature increase causes a large increase in rate due to the significantly higher fraction of molecules that have energies greater than, or equal to, E_a. Consequently, a much higher proportion of collisions become successful and reaction rate increases considerably.

Calculating rate constants and activation energies

To calculate either the rate constant (k) or activation energy (E_a) when other variables are known, we often use the logarithmic form of the Arrhenius equation:

$\ln (\text{k})=-\frac{{\text{E}}_{\text{a}}}{\text{RT}}+\ln (\text{A})$

This equation represents a straight line where:

• $\text{y }=\ln (\text{k})$
• $\text{x }=\frac{1}{\text{T}}$
• $\text{Gradient }=-\frac{{\text{E}}_{\text{a}}}{\text{R}}$
• y-intercept = ln(A)

Worked example 1 - Calculating activation energy

The decomposition of dinitrogen pentoxide (N₂O₅) at 310 K has a rate constant of $2.50\times {10}^{-3}{\text{ s}}^{-1}$. The Arrhenius constant for this reaction is $6.90\times {10}^{12}{\text{ s}}^{-1}$.

Calculate the activation energy, in kJ mol^-1, for this reaction, given R = 8.31 J K^-1 mol^-1. Give your answer to 3 significant figures.

Step 1: Logarithmic form of the Arrhenius equation

$\ln (\text{k})=\frac{-{\text{E}}_{\text{a}}}{\text{RT}}+\ln (\text{A})$

Step 2: Rearrange equation to solve for E_a_

${\text{E}}_{\text{a}}=(\ln (\text{A})-\ln (\text{k}))\times \text{ RT}$

Step 3: Substitution and correct evaluation

${\text{E}}_{\text{a}}=(\ln (6.90\times {10}^{12})-\ln (2.50\times {10}^{-3}))\times 8.31\times 310$

$=35.6\times 8.31\times 310=\mathrm{91,600}{\text{ J mol}}^{-1}$

Step 4: Conversion from J mol^-1 to kJ mol^-1

To convert from J mol^-1 to kJ mol^-1, divide by 1,000.

91,600 J mol^-1 = 91.6 kJ mol^-1

Therefore, the activation energy for the decomposition of N₂O₅ is 91.6 kJ mol^-1.

Using Arrhenius plots to determine activation energy and the Arrhenius constant

An Arrhenius plot graphs the logarithmic form of the Arrhenius equation, plotting ln k against $\frac{1}{\text{T}}$.

This yields a straight line from which we can determine:

1. Gradient - $\frac{-{\text{E}}_{\text{a}}}{\text{R}}$, allowing us to calculate E_a.
2. y-intercept - ln A, which lets us find A by taking the exponential (e^intercept).

Worked example 2 - Calculating the activation energy and pre-exponential factor

Calculate the activation energy (in kJ mol^-1) and the Arrhenius factor (in dm^3^ mol^-1^ s^-1^) from the Arrhenius plot above.

Give your answers to 2 significant figures.

The gas constant, R = 8.31 J K^-1^ mol^-1^.

Step 1: Calculate the gradient of the line

$\text{Gradient }=\frac{-36-(-6)}{(2.0\times {10}^{-3})-(0.5\times {10}^{-3})}$

$=\frac{-30}{(1.5\times {10}^{-3})}=-\mathrm{20,000}$

Step 2: Calculate activation energy (E_a)

${\text{E}}_{\text{a}}=-\text{Gradient }\times \text{ R }=-\mathrm{20,000}\times 8.31=\mathrm{166,200}{\text{ J mol}}^{-1}$

Step 3: Rearrange logarithmic form of the Arrhenius equation

$\ln (\text{A})=\frac{{\text{E}}_{\text{a}}}{\text{RT}}+\ln (\text{k})$

Step 4: Substitution and correct evaluation

Using the point (0.5 x 10^-3, $-$6) and the calculated gradient from step 1

$\ln (\text{A})=(\mathrm{20,000}\times (0.5\times {10}^{-3}))-6$

= $10-6=4$

Step 5: Calculate pre-exponential factor (A)

A = e⁴ = 55 mol dm^-3 s^-1.

Therefore, the activation energy is 170,000 J mol^-1 and the pre-exponential factor is 55 mol dm^-3 s^-1.