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?

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Alleei Alleei · Answer 1 · 2020-04-15T06:25:51+02:00

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]460^oC[/tex] = [tex]5.8\times 10^{-6}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]T_2[/tex] = [tex]4\times K_1[/tex]

[tex]Ea[/tex] = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]460^oC=273+460=733K[/tex]

[tex]T_2[/tex] = final temperature = ?

Now put all the given values in this formula, we get:

[tex]\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}][/tex]

[tex]T_2=791.1K[/tex]

Therefore, the final temperature would be, 791.1 K