Chemistry, 15.04.2020 02:57, christiannpettyy
The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → CH2CHF(g) + HF(g) At 460 °C, k = 5.8 × 10-6 s-1 and Ea = 265 kJ/mol. To what temperature (in K) would you have to raise the reaction to make it go four times as fast?
Answers: 2
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](/tpl/images/0600/9306/6d953.png)
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 265 kJ/mol = 265000 J/mol
= initial temperature = 
![\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]](/tpl/images/0600/9306/ddf14.png)
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The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydro...
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