The Arrhenius Equation and Finding Activation Energy
The rate constant depends on temperature through the activation energy. Plotting the logarithm of the rate constant against one over temperature turns that dependence into a straight line whose gradient gives the barrier.
What a learner can do afterwards
- Explains why a small temperature rise changes the rate so much, using the fraction of particles above the barrier
- Rearranges the Arrhenius equation into straight-line form and says what to plot against what
- Finds an activation energy from the gradient of that plot
- Says what adding a catalyst does to the gradient and what it leaves alone
1 · Read
Only particles with enough energy to clear the barrier can react, and they sit in the thin fast tail of the spread. Warm the mixture a little and that tail fattens a lot, which is why a small rise in temperature lifts the rate so much.
The Arrhenius equation links the rate constant k to temperature through the activation energy. Its straight-line form is ln k = -Ea/R times 1/T plus ln A, so you plot ln k against 1/T, with T always in kelvin.
The line has gradient -Ea/R, so multiply the gradient by -R to get Ea, where R is the gas constant, 8.31 J per mol per kelvin. A gradient of -10000 kelvin gives Ea of 10000 times 8.31, which is 83100 J per mole, or 83.1 kJ per mole.
A catalyst lowers the activation energy, so the plotted line gets less steep, while the intercept ln A stays where it was. If you used Celsius by mistake, 1/T shifts and both the gradient and the answer break.
Plot ln k against 1/T in kelvin, read Ea from the gradient, and expect a catalyst to flatten the slope.
2 · Watch
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.