Q. how to calculate activation energy

Activation energy is the minimum energy that reacting particles must have for a reaction to occur. The standard way to calculate activation energy uses the Arrhenius equation. I will derive the practical formulas step by step, explain units, and give a numerical example.

Step 1. Start from the Arrhenius equation. The rate constant \(k\) depends on temperature \(T\) as

\[
k = A \exp\!\left(-\frac{E_a}{R T}\right),
\]

where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, and \(R\) is the gas constant.

Step 2. Take the natural logarithm of both sides to linearize the expression. This gives

\[
\ln k = \ln A – \frac{E_a}{R T}.
\]

Step 3. If you have rate constants measured at two different temperatures, \(k_1\) at \(T_1\) and \(k_2\) at \(T_2\), subtract the two logarithmic expressions to eliminate \(\ln A\). This yields

\[
\ln k_2 – \ln k_1 = -\frac{E_a}{R}\!\left(\frac{1}{T_2} – \frac{1}{T_1}\right).
\]

Step 4. Solve this equation for the activation energy \(E_a\). Rearranging gives the two-point form commonly used to calculate \(E_a\):

\[
E_a = -R\,\frac{\ln\!\left(\dfrac{k_2}{k_1}\right)}{\dfrac{1}{T_2} – \dfrac{1}{T_1}}.
\]

It is often convenient to write the same expression with the sign changed inside the logarithm and the denominator, for clarity:

\[
E_a = R\,\frac{\ln\!\left(\dfrac{k_1}{k_2}\right)}{\dfrac{1}{T_2} – \dfrac{1}{T_1}}.
\]

Step 5. Pay attention to units. Use the gas constant \(R\) in consistent units with the desired unit for \(E_a\). Common choices are \(R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\) which yields \(E_a\) in joules per mole, or divide by 1000 to get kilojoules per mole.

Step 6. Numerical example. Suppose \(k_1 = 1.2\times 10^{-3}\ \mathrm{s^{-1}}\) at \(T_1 = 300\ \mathrm{K}\), and \(k_2 = 4.5\times 10^{-3}\ \mathrm{s^{-1}}\) at \(T_2 = 350\ \mathrm{K}\). Step through the calculation.

Compute the logarithm of the ratio:

\[
\ln\!\left(\frac{k_2}{k_1}\right) = \ln\!\left(\frac{4.5\times 10^{-3}}{1.2\times 10^{-3}}\right) = \ln(3.75) \approx 1.322.
\]

Compute the difference of inverse temperatures:

\[
\frac{1}{T_2} – \frac{1}{T_1} = \frac{1}{350} – \frac{1}{300} = -\frac{50}{105000} \approx -4.7619\times 10^{-4}\ \mathrm{K^{-1}}.
\]

Use \(R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\) and the two-point formula:

\[
E_a = -R\,\frac{\ln\!\left(\dfrac{k_2}{k_1}\right)}{\dfrac{1}{T_2} – \dfrac{1}{T_1}}
= -8.314\cdot\frac{1.322}{-4.7619\times 10^{-4}}\ \mathrm{J\,mol^{-1}}.
\]

Evaluate the arithmetic. The numerator is \( -8.314\times 1.322 \approx -10.9909\). Dividing by the denominator gives

\[
E_a \approx \frac{10.9909}{0.00047619}\ \mathrm{J\,mol^{-1}} \approx 2.308\times 10^{4}\ \mathrm{J\,mol^{-1}}.
\]

Convert to kilojoules per mole if desired:

\[
E_a \approx 23.08\ \mathrm{kJ\,mol^{-1}}.
\]

Step 7. Alternate method. If you have multiple \(k\) values at many temperatures, fit \(\ln k\) versus \(1/T\) to a straight line. The slope of that line is \(-E_a/R\). That is, perform a linear regression for the model

\[
\ln k = \ln A – \frac{E_a}{R}\cdot \frac{1}{T},
\]

and extract \(E_a = -(\text{slope})\cdot R\).

Summary. To calculate activation energy from two rate constants, use the two-point Arrhenius formula

\[
E_a = -R\,\frac{\ln\!\left(\dfrac{k_2}{k_1}\right)}{\dfrac{1}{T_2} – \dfrac{1}{T_1}},
\]

being careful to use consistent units for \(R\) and \(T\). Alternatively, use a linear fit of \(\ln k\) versus \(1/T\) and obtain \(E_a\) from the slope.