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  • The energy of activation for a reaction at  is  <img alt="100 \mathr

The energy of activation for a reaction at 300 \mathrm{~K} is  100 \mathrm{kJmol}^{-1}  For same concentration, presence of a catalyst lowers the energy of activation by 75 \% . Calculate the value of:
\mathrm{\log _{10}\left(\frac{r_2}{r_1}\right) }
\mathrm{r_2= }  rate in presence of catalyst
\mathrm{r_1= } rate in absence of catalyst.

Option: 1

13.06


Option: 2

10.02


Option: 3

7.65


Option: 4

3.73


Answers (1)

best_answer

The Arrhenius equation is,  \mathrm{k=A e^{-E_a / R T}}
In absence of catalyst,    \mathrm{k_1=A e^{-100 / R T}}
In presence of catalyst,    \mathrm{k_2=A e^{-25 / R T}}
So,

\mathrm{\frac{k_2}{k_1}=e^{75 / R T}}

or     \mathrm{2.303 \log \frac{k_2}{k_1}=\frac{75}{R T}}
or   \mathrm{2.303 \log \frac{k_2}{k_1}=\frac{75}{8.314 \times 10^{-3} \times 300}}
or
\mathrm{\log \frac{k_2}{k_1}=\frac{75}{8.314 \times 10^{-3} \times 300 \times 2.303} \approx 13.06}
Since
Rate\mathrm{ =k[\text { Conc }]^n}.

At a particular concentration,
\mathrm{Rate \propto k}
So,
\mathrm{ \frac{k_2}{k_1}=\frac{\text { rate in presence of catalyst }}{\text { rate in absence of catalyst }} }
\mathrm{ i.e., \frac{r_2}{r_1}=\frac{k_2}{k_1}}
Hence,
\mathrm{ \log _{10}\left(\frac{r_2}{r_1}\right)=13.06 }.

Posted by

sudhir kumar

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