0.592 g of copper is deposited in 60 minutes by passing
0.5 A current through a solution of copper (II) sulphate. The electro chemical equivalent of copper (II) (in $\mathrm{gC}^{-1}$ ) is
( $F=96500 \mathrm{C} \mathrm{mol}^{-1}$ )
For the gaseous reaction, $\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$
the rate can be expressed as
$ \begin{array}{l} -\frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=K_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \\\\ +\frac{d\left[\mathrm{NO}_{2}\right]}{d t}=K_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \\\\ +\frac{d\left[\mathrm{O}_{2}\right]}{d t}=K_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \end{array} $
The correct relation between $K_{1}, K_{2}$ and $K_{3}$
Match the following
List I (Industrial process) | List II (Catalyst used) |
A Ostwald's process | I $\mathrm{CuCl}_2$ |
B Haber's process | II Zeolites |
C Deacon's process | III Pt gauge |
D Cracking of hydrocarbons | IV Fe |
The correct answer is