The decomposition of NH_3 on platinum surface is zero order...

The decomposition of

\mathrm{NH}_3

on platinum surface is zero order reaction. The rates of production of

\mathrm{N}_2 and \mathrm{H}_2

are respectively

\left(k=2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\right)

Options

A $7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
B $2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
C $5.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
D $2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 5.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

B. $2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} ; 7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$

For the reaction,

2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2+3 \mathrm{H}_2

From equation (i) and (ii) The decomposition of NH_3 on platinum surface is zero order reaction. The rates of production of N_2 and H_2 are

\therefore

Rate of production of

\mathrm{N}_2, \frac{d\left[\mathrm{~N}_2\right]}{d t}=k =2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

From equation (iii) Rate of production of

\mathrm{H}_2, \frac{d\left[\mathrm{H}_2\right]}{d t}=3 k

\begin{aligned} & =3 \times 2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =7.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\end{aligned}