Question

The rate constant for a reaction at 25ºC is half of the rate constant at 35ºC. Calculate the energy of activation of the reaction. (R = 1.987 cal K -1 mol -1 )
give the answer?​

Answer 1

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Answer 2

Given:

Temperatures,

$\bold{T_1=25^oC}$ , $\bold{T_2=35^oC}$

Rate constant for $\bold{T_1}$ is half of rate constant at $\bold{T_2}$.

And $\bold{R = 1.987 cal K ^{-1} mol^ {-1}}$

To Find:

The energy of activation of the reaction.

Solution:

Let, rate constant for $\bold{T_1}$ is $\bold{k_1}$ and rate constant for $\bold{T_2}$ is $\bold{k_2}$

∴ From the question we get,

$\bold{k_1=\frac{k_2}{2}=>2k_1=k_2 }$

Now, $\bold{T_1=25^oC=298.15 K}$ and $\bold{T_2=35^oC=308.15K}$

From Arrhenius equation, we get

$\bold{ln\frac{k_2}{k_1} =\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2} ) }$ ......... (1)

∴Putting all the values in respective positions we get,

$\bold{ln\frac{2k_1}{k_1} =\frac{E_a}{1.987}(\frac{1}{298.15}-\frac{1}{308.15} ) }$

$=>\bold{ln2 =\frac{E_a}{1.987}(0.0033-0.0032 ) }$

$=>\bold{1.987 \times ln2 =E_a \times 10^{-4} }$

$=>\bold{E_a \approx 1.4\times 10^4Jmol^{-1}}$

∴Activation energy for the reaction is $\bold{ 1.4\times 10^4\hspace{1mm}Jmol^{-1}}$