Chemistry: 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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The compound 1,1-difluoroethane decomposes at elevated

VelvetXace

23.01.2022

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

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

where,

K_1 = rate constant at 460^oC = $5.8\times 10^{-6}s^{-1}$

K_2 = rate constant at T_2 = $4\times K_1$

= activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 460^oC=273+460=733K

T_2 = final temperature = ?

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

$\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]$

T_2=791.1K

Therefore, the final temperature would be, 791.1 K

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