Chemistry, 13.03.2020 18:34 RoxanneDuartee
The rate constants of some reactions double with every 10-degree rise in temperature. Assume that a reaction takes place at 295 K and 305 K. What must the activation energy be for the rate constant to double as described?
Answers: 1
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](/tpl/images/0546/6834/6d953.png)
= rate constant at 295 K
= rate constant at 305 K = 
= initial temperature = 295 K
= final temperature = 305 K![\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}]](/tpl/images/0546/6834/d1fff.png)
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The rate constants of some reactions double with every 10-degree rise in temperature. Assume that a...
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