Question

The decomposition of $NH_{3}$ on platinum surface is zero order reaction. What are the rates of production of $N_{2}$ and $H_{2}$ if $k=2.5\times 10^{- 4}molL^{- 1}sec^{- 1}$ ?
$2 NH _{3}( g ) \rightarrow N _{2}( g )+3 H _{2}( g )$

• A. $1.25\times 10^{- 4} \, Ms^{- 1},3.75\times 10^{- 4} \, Ms^{- 1}$
• B. $3.00\times 10^{- 4} \, Ms^{- 1},7.50\times 10^{- 4} \, Ms^{- 1}$
• C. $2.50\times 10^{- 4} \, Ms^{- 1}, \, 1.25\times 10^{- 4} \, Ms^{- 1}$
• D. $3.75\times 10^{- 4} \, Ms^{- 1},2.50\times 10^{- 4} \, Ms^{- 1}$

Answer 1

Answer: (A)

$2 NH _{3}( g ) \rightarrow N _{2}( g )+3 H _{2}( g )$
Rate = k[NH₃]^o = k = 2.5 × 10^-4 M s^-1.
$\frac{1}{2} \frac{- \text{d}}{\text{dt}} \left[\text{NH}_{3}\right] = \frac{1}{1} \left\{\frac{\text{d}}{\text{dt}} \left[\text{N}_{\text{2}}\right]\right\} = \frac{1}{3} \left\{\frac{\text{d}}{\text{dt}} \left[\text{H}_{\text{2}}\right]\right\}$
$\frac{ d }{ d ⁡ t} \left[\text{N}_{2}\right] = \frac{2 \cdot 5 \times 1 0^{- 4}}{2} = 1 \cdot 2 5 \times 1 0^{- 4} \text{M s}^{- 1}$
$\frac{ d }{ d ⁡ t} \left[ H ⁡_{2}\right] = \frac{3}{2} \times 2 \cdot 5 \times 1 0^{- 4} = 3 \cdot 7 5 \times 1 0^{- 4} \text{M s}^{- 1}$