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Activation Energy from Two Rate Constants

A reaction's rate constant increases tenfold when temperature rises from 300 K300\ \text{K} to 350 K350\ \text{K}.

Take R=8.314 J/mol⋅KR = 8.314\ \text{J/mol·K}.

Find the activation energy.

Given
T1300 KLower temperature
T2350 KHigher temperature
k2/k110 -Rate constant ratio
Hint 1

The Arrhenius relation is linear in 1/T, not in T.

Hint 2

ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂).

Hint 3

Keep several significant figures on 1/300 and 1/350 — you are subtracting two close numbers.

Worked solution — try the problem first

Two-point Arrhenius form:

ln⁡ ⁣(k2k1)=−EaR(1T2−1T1)=EaR(1T1−1T2)\ln\!\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Reciprocal temperature difference:

1300−1350=3.3333×10−3−2.8571×10−3=4.762×10−4 K−1\frac{1}{300} - \frac{1}{350} = 3.3333\times10^{-3} - 2.8571\times10^{-3} = 4.762\times10^{-4}\ \text{K}^{-1}

Solve:

ln⁡(10)=2.3026=Ea8.314(4.762×10−4)\ln(10) = 2.3026 = \frac{E_a}{8.314}(4.762\times10^{-4})

Ea=2.3026×8.3144.762×10−4=19.144.762×10−4=40,200 J/molE_a = \frac{2.3026 \times 8.314}{4.762\times10^{-4}} = \frac{19.14}{4.762\times10^{-4}} = 40{,}200\ \text{J/mol}

Ea≈40.2 kJ/mol\boxed{E_a \approx 40.2\ \text{kJ/mol}}

Plausibility. Typical activation energies run 40-200 kJ/mol. A value near 40 kJ/mol indicates moderate temperature sensitivity — consistent with a 10× increase over a fairly wide 50 K span. A highly sensitive reaction (150 kJ/mol) would show that jump over roughly 15 K.

Concepts:Arrhenius equationActivation energyTemperature dependence of rate

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