The rate constant for a reaction at 300 K is 2.0 × 10⁻³ s⁻¹ and at 400 K is 2.0 × 10⁻¹ s⁻¹ — JEE Chemistry
The rate constant for a reaction at 300 K is $2.0 \times 10^{-3}$ s⁻¹ and at 400 K is $2.0 \times 10^{-1}$ s⁻¹. Calculate the activation energy. ($R = 8.314$ J/mol·K)
1 Answer
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· 17d ago
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$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
$$\ln\frac{2.0\times10^{-1}}{2.0\times10^{-3}} = \ln100 = 4.605$$
$$4.605 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{400}\right) = \frac{E_a}{8.314} \times \frac{1}{1200}$$
$$E_a = 4.605 \times 8.314 \times 1200 = 45,900 J/mol \approx 45.9 kJ/mol$$
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Discussion (3)
TS
Pro tip: memorise the standard result, it reappears in many problems.
R
For revision — the key formula used here comes up almost every year.
AB
Thanks a ton, I was stuck on this exact problem for an hour.