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Factors influencing rate of reaction

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Question

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C?
$(R = 8 .314 {Jmol}^{- 1} K^{- 1})$

Moderate

Solution

$\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2 .303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) k_{2} = 2 k_{1}, T_{1} = 20 + 273 = 293 K or T_{2} = 35 + 273 = 308 K R = 8 .314 J {mol}^{- 1} K^{- 1} \log 2 = \frac{E_{a}}{2 .303 \times 8 .314} (\frac{1}{293} - \frac{1}{308}) 0 .3010 = \frac{E_{a}}{19 .147} \times \frac{15}{293 \times 308} E_{a} = 34673 {Jmol}^{- 1} or 34 .7 {kJmol}^{- 1}$

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