Discussion Forum

1)

The following data were  obtained for a given reaction at 300K

 Reaction                             Energy of activation(kJ mol^-1)

 (i) uncatalysed                             76

  (ii)  catalysed                              57

 the factor by which rate of catalysed reaction is increased , is 

Answer:

Option C

Explanation:

Using Arrhenius equation

 $K= A.e^{-\frac{E_{a}}{Rt}}$

 $\log k=\log A-\frac{E_{a}}{2.303 RT}$

$\log k_{1}=\log A-\frac{E_{a(1)}{}}{2.303 RT}$    ...............(i)

and $\log k_{2}=\log A-\frac{E_{a(2)}{}}{2.303 RT}$     ...........(ii)

 or     $\log \frac{k_{2}}{k_{1}}=\frac{1}{2.303 RT}[E_{a(1)}-E_{a(2)}]$

                                                             ( from (i) and (ii))

 $=\frac{1}{2.303 \times8.314\times300}[76000-57000]$

 or   $=\log\frac{k_{2}}{k_{1}}=\frac{19000}{2.303\times8.314\times300}$

 $=\frac{190}{6.9\times8.314}$

 or $=\frac{k_{2}}{k_{1}}=2000$    [taking antilog]