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The following data were obtained for a given reaction at $300 K$.
$\begin{array}{llc} & \text { Reaction } & \begin{array}{c}\text { Energy of activation } \\ \left(\mathrm{kJ} \mathrm{mol}^{-1}\right)\end{array} \\ \text { (i) } & \text { uncatalysed } & 76 \\ \text { (ii) } & \text { catalysed } & 57\end{array}$
The factor by which rate of catalysed reaction is increased, is
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$\begin{array}{llc} & \text { Reaction } & \begin{array}{c}\text { Energy of activation } \\ \left(\mathrm{kJ} \mathrm{mol}^{-1}\right)\end{array} \\ \text { (i) } & \text { uncatalysed } & 76 \\ \text { (ii) } & \text { catalysed } & 57\end{array}$
The factor by which rate of catalysed reaction is increased, is
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2293 Upvotes
Verified Answer
The correct answer is:
2000
Using Arrhenius equation,
$\begin{array}{l}
K=A \cdot e^{-\frac{E_{a}}{R t}}, \text { we get } \\
\log k=\log A-\frac{E_{a}}{2.303 R T} \quad \therefore
\end{array}$
$\log k_{1}=\log A-\frac{E_{a_{(1)}}}{2.303 R T}$
and $\log k_{2}=\log A-\frac{E_{a_{(2)}}}{2.303 R T}$
or $\log \frac{k_{2}}{k_{1}}=\frac{1}{2.303 R T}\left[E_{a(1)}-E_{a(2)}\right]$
(from (i) and (ii))
$=\frac{1}{2.303 \times 8.314 \times 300}(76000-57000)$
or $\log \frac{k_{2}}{k_{1}}=\frac{19000}{2.303 \times 8.314 \times 300}$ $=\frac{190}{6.9 \times 8.314}$
or $\frac{k_{2}}{k_{1}}=2000[$ taking antilog $]$
$\begin{array}{l}
K=A \cdot e^{-\frac{E_{a}}{R t}}, \text { we get } \\
\log k=\log A-\frac{E_{a}}{2.303 R T} \quad \therefore
\end{array}$
$\log k_{1}=\log A-\frac{E_{a_{(1)}}}{2.303 R T}$
and $\log k_{2}=\log A-\frac{E_{a_{(2)}}}{2.303 R T}$
or $\log \frac{k_{2}}{k_{1}}=\frac{1}{2.303 R T}\left[E_{a(1)}-E_{a(2)}\right]$
(from (i) and (ii))
$=\frac{1}{2.303 \times 8.314 \times 300}(76000-57000)$
or $\log \frac{k_{2}}{k_{1}}=\frac{19000}{2.303 \times 8.314 \times 300}$ $=\frac{190}{6.9 \times 8.314}$
or $\frac{k_{2}}{k_{1}}=2000[$ taking antilog $]$
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