Factors influencing rate of reaction

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

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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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Factors influencing rate of reaction

Remember concepts with our Masterclasses.

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C?R=8.314Jmol−1K−1

logk2k1=Ea2.303R1T1−1T2 k2=2k1, T1=20+273=293K or T2=35+273=308K R=8.314 J mol−1K−1 log2=Ea2.303×8.3141293−1308 0.3010=Ea19.147×15293×308 Ea=34673Jmol−1 or 34.7kJmol−1

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