Question

The activation energy for the gas phase decomposition of t- butyl propionate is 164 KJ .C2H5COOC(CH3)3 changes to (CH3)2C=CH2 + C2HCOOH.the rate constant at 528 k is 3.80 *10^-4/s.what will be the rate constant be at 569k ?

Answer 1

The rate constant be at 569 K is $5.61\times 10^{-3}s^{-1}$

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 528 K= $3.80\times 10^{-4}s^{-1}$

$K_2$ = rate constant at 569 K = ?

$Ea$ = activation energy for the reaction = 164 kJ = 164000 J  (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 528 K

$T_2$ = final temperature = 569 K

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

$\log (\frac{K_2}{3.80\times 10^{-4}})=\frac{164000}{2.303\times 8.314J/mole.K}[\frac{1}{528K}-\frac{1}{569K}]$

$K_2=5.61\times 10^{-3}s^{-1}$

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