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  4. In the presence of a catalyst, activation energy of a reaction is lowered by 2 kcal at 27° C. Hence, rate will

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Q. In the presence of a catalyst, activation energy of a reaction is lowered by $2$ kcal at $ 27^\circ C$. Hence, rate will

Delhi UMET/DPMTDelhi UMET/DPMT 2010Chemical Kinetics

Solution:

Arrhenius equation is
$ k = Ae^{ - E_a / RT } $
In the presence of catalyst rate constant $k' = Ae^{ - E_a / RT } $
Eq. (ii ) /Eq. (i), we get
$ \frac{ k' }{ k } = e^{( E_a / E_a ' )} / RT $
$ \frac{ k' }{ k } = e^{ \bigg( \frac{ 2000}{ 2 \times 300} \bigg) } = e^{ 3.33} $
$ ( \because E_a - E_a ' = 2000 cal$ (given)]
$ \because $ On taking log both sides,
$\log \frac{ k ' }{ k } = \frac{ 3. 33}{ 2.303 } $
$ \frac{ k ' }{ k } = $ antilog $1.446 = 27.92 = 28$
We know that, rate $(R) \propto k $
$ \therefore \frac{ kR' }{ R} = \frac{ k ' }{ k } = 28 $
or $R' =28 R$