Arrhenius equation — calculate activation energy from rate constant data

Question

The rate constant of a reaction is $1.5 \times 10^{-3}$ s⁻¹ at 300 K and $4.5 \times 10^{-3}$ s⁻¹ at 320 K. Calculate the activation energy of the reaction. ( $R = 8.314$ J mol⁻¹ K⁻¹)

(NEET 2023, similar pattern)

Solution — Step by Step

k = A \cdot e^{-E_a/(RT)}

where $k$ = rate constant, $A$ = pre-exponential factor, $E_a$ = activation energy, $R$ = gas constant, $T$ = temperature in Kelvin.

For two temperatures $T_1$ and $T_2$ with rate constants $k_1$ and $k_2$ :

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

Or equivalently using log base 10:

\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

$k_1 = 1.5 \times 10^{-3}$ (at $T_1 = 300$ K), $k_2 = 4.5 \times 10^{-3}$ (at $T_2 = 320$ K)

\ln\frac{k_2}{k_1} = \ln\frac{4.5 \times 10^{-3}}{1.5 \times 10^{-3}} = \ln 3 = 1.0986

\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = 2.083 \times 10^{-4} \text{ K}^{-1}

1.0986 = \frac{E_a}{8.314} \times 2.083 \times 10^{-4}

E_a = \frac{1.0986 \times 8.314}{2.083 \times 10^{-4}}

E_a = \frac{9.133}{2.083 \times 10^{-4}}

\boxed{E_a \approx 43,850 \text{ J/mol} \approx 43.85 \text{ kJ/mol}}

Why This Works

The Arrhenius equation tells us that only a fraction of molecules have enough energy to cross the activation energy barrier. Higher temperature means more molecules have sufficient energy, so the rate constant increases.

The two-temperature form eliminates the pre-exponential factor $A$ (which is hard to measure directly). By taking the ratio $k_2/k_1$ , the $A$ cancels out, leaving only $E_a$ as the unknown.

A useful check: for most reactions, activation energy falls in the range 40-200 kJ/mol. If your answer is 4 kJ or 4000 kJ, something went wrong with the units.

Alternative Method — Using log₁₀

\log\frac{4.5 \times 10^{-3}}{1.5 \times 10^{-3}} = \frac{E_a}{2.303 \times 8.314} \times 2.083 \times 10^{-4}

\log 3 = 0.4771

0.4771 = \frac{E_a}{19.147} \times 2.083 \times 10^{-4}

E_a = \frac{0.4771 \times 19.147}{2.083 \times 10^{-4}} = 43,850 \text{ J/mol}

For NEET, keep these values ready: $\ln 2 = 0.693$ , $\ln 3 = 1.099$ , $\log 2 = 0.301$ , $\log 3 = 0.477$ . Most Arrhenius problems give rate constant ratios of 2 or 3 — knowing these logarithm values saves calculation time.

Common Mistake

Two frequent errors: (1) Getting $T_1$ and $T_2$ mixed up in the formula — always put the lower temperature as $T_1$ and higher as $T_2$ . If you swap them, you get a negative $E_a$ , which is physically meaningless. (2) Forgetting to convert the final answer from J/mol to kJ/mol when the options are in kJ. Check your answer against the typical range (40-200 kJ/mol) as a sanity check.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using log₁₀

Common Mistake

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