Question

The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K, 600.0 K, the rate constant, k , k, is 6.1 × 10 − 8 s − 1 . 6.1×10−8 s−1. What is the value of the rate constant at 775.0 K?

Answer 1

Answer: $4.3\times 10^{-13}s^{-1}$

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 $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $775.0$ = $?$

$Ea$ = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $775.0K$

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

$\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]$

$\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150$

$(\frac{6.1\times 10^{-8}}{K_2})=141253.8$

Therefore, the value of the rate constant at 775.0 K is $4.3\times 10^{-13}s^{-1}$