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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 be
Uttarkhand PMTUttarkhand PMT 2009
Solution:
Arrhenius equation is $k=A e^{-E_{a} / R T}$
In the presence of catalyst, rate constant,
$k=A e^{-E_{a} / R T} Eq$ (ii) / Eq (i), we get
$\frac{k}{k}=e^{\left(E_{a}-E_{a}\right) / R T} $
$\Rightarrow \frac{k}{k}=e^{\left(\frac{2000}{2 \times 300}\right)}=e^{3.33} $
$\left[\because E_{a}-E_{a}=2000\right. \text { (given)] }$
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{R}{R}=\frac{k}{k}=28 $ Or $ R=28\, R$