Question

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?

Answer 1

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 $T_2$ = $4\times K_1$

$Ea$ = 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