Arrhenius equationAQA A-Level Chemistry: Revision notes

The rate constant k increases with temperature. At any temperature, molecules have a range of energies, shown by the Maxwell–Boltzmann distribution. Only collisions between molecules with energy equal to or greater than the activation energy (Ea) lead to reaction.

Raising the temperature moves the peak of the curve to a higher energy and flattens it, but the total area stays the same. Because the high-energy tail is small, even a small rise in temperature gives a large increase in the fraction of molecules with E ≥ Ea. Collisions are also slightly more frequent, but this is a small effect. This is why k, and so the rate, often roughly doubles for a rise of about 10 K.

A catalyst does not change the temperature; it provides a route with a lower Ea.

The variation of k with temperature is given by the Arrhenius equation:

$k = Ae^{-E_a/RT}$

• k is the rate constant (units depend on the order)
• A is the Arrhenius constant, which has the same units as k and is related to the frequency of collisions and their orientation
• Ea is the activation energy in J mol⁻¹
• R is the gas constant, 8.31 J K⁻¹ mol⁻¹ (given in the exam)
• T is the temperature in kelvin

The term e^(−Ea/RT) is the fraction of molecules with energy equal to or greater than Ea. As T rises, −Ea/RT becomes less negative and the term increases, so k increases. A larger Ea gives a smaller fraction, so a smaller k at a given temperature.

Worked example (calculating k): A = 2.0 × 10¹⁰ s⁻¹, Ea = 75.0 kJ mol⁻¹, T = 300 K. Ea/RT = 75 000 ÷ (8.31 × 300) = 30.08, so e^(−30.08) = 8.6 × 10⁻¹⁴, and k = 2.0 × 10¹⁰ × 8.6 × 10⁻¹⁴ = 1.7 × 10⁻³ s⁻¹.

Comparing two temperatures: dividing the equation at T₂ by the equation at T₁ cancels A:

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

This can be used to find Ea from two values of k, or to find k at a second temperature.

To find T for a given k, rearrange: T = Ea ÷ (R(ln A − ln k)). Keep several significant figures in intermediate steps because e raised to a large power magnifies rounding errors.

Taking natural logarithms of the Arrhenius equation gives

$\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A$

This has the form y = mx + c. A graph of ln k (y) against 1/T (x) is a straight line with:

• gradient = −Ea/R, so Ea = −gradient × R
• intercept on the ln k axis = ln A (at 1/T = 0)

Worked example: a line has gradient −7.5 × 10³ K. Ea = 7.5 × 10³ × 8.31 = 6.2 × 10⁴ J mol⁻¹ = 62 kJ mol⁻¹.

A steeper gradient means a larger Ea, so k changes more with temperature. The gradient is negative, but Ea is positive.