For the two gaseous reactions, following data are given $A \rightarrow B;k_{1} = 10^{10}e^{- 20,000/T}$ $C \rightarrow D;k_{2} = 10^{12}\ e^{- 24,606/T}$ The temperature at which $k_{1}$ becomes equal to $k_{2}$ is

Question

Dr. Abubacker · Accepted Answer

(b)
Given, $k_{1}10^{10}e^{- 20,000/T}$
$k_{2} = 10^{12}e^{- 24,606/T}$
$k_{1} = k_{2}$
$10^{10}e^{- 20,000/T} = 10^{12}e^{- 24,606/T}$
$e^{\frac{- 20,000}{T} + \frac{24,606}{T}} = 10^{2}$
$e^{\frac{4,606}{T}} = 10^{2}$
On taking log both sides,
$\frac{4606}{2.303T} = \log 10^{2}$
$2\log{10 	imes T = \frac{4606}{2.303}}$
$T = \frac{4606}{2.303 	imes 2}$
$= \frac{4606}{4.606} = 1000\ K$

Q. For the two gaseous reactions, following data are given $A \rightarrow B;k_{1} = 10^{10}e^{- 20,000/T}$ $C \rightarrow D;k_{2} = 10^{12}\ e^{- 24,606/T}$ The temperature at which $k_{1}$ becomes equal to $k_{2}$ is

Explanation

Reviewed by Dr. Abubacker

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