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Grade 12th passPhysical Chemistry

Rate of constant for given reaction get doubled for 20° rises in temperature fro 330k find the activation energy for this reaction

Profile image of Jyotika
6 Years agoGrade 12th pass
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1 Answer

Profile image of Arun
6 Years ago

 

Since \log \frac{K_{2}}{K_{1}} = \frac{Ea}{2.303\:R}\left ( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right )

K2  = 2K1 , T1 = 273 + 20 K = 293 K

T2 = 273 + 35 K = 308 K

R = 8.314 J mol-1 K-1

\therefore \log 2 = \frac{Ea}{2.303 \times 8.314}\left ( \frac{1}{293} - \frac{1}{308}\right )

\Rightarrow Ea = 34.7 \:KJ\:mol^{-1}


 
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