Answers: 2
![\ln(\frac{K_{65^oC}}{K_{40^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](/tpl/images/0525/7564/05a35.png)
= equilibrium constant at 65°C = ?
= equilibrium constant at 40°C = 
= Activation energy of the reaction = 65.5 kJ/mol = 65500 J/mol (Conversion factor: 1 kJ = 1000 J)
= initial temperature = ![40^oC=[40+273]K=313K](/tpl/images/0525/7564/84088.png)
= final temperature = ![65^oC=[65+273]K=338K](/tpl/images/0525/7564/eb1a6.png)
![\ln(\frac{K_{65^oC}}{5.45\times 10^{-2}})=\frac{65500J/mol}{8.314J/mol.K}[\frac{1}{313}-\frac{1}{338}]\\\\K_{65^oC}=0.350s^{-1}](/tpl/images/0525/7564/e250a.png)
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The rate constant of a particular first order reaction is 5.45 x 10^-2 sec^-1 at 40.0 oC. What is th...
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