The rate for the decomposition of $N H_{3}$ on platinum surface is zero order. What are the rate of production of $N_{2}$ and $H_{2}$ if $K = 2.5 \times 10^{- 4} m o l l i t r e^{- 1} s^{- 1}$ .

Open in App

Solution

$2 N H_{3} \to N_{2} + 3 H_{2}$
Rate of reaction $= - \frac{1}{2} \frac{d}{d t} [N H_{3}] = \frac{d}{d t} [N_{2}] = \frac{1}{3} \frac{d}{d t} [H_{2}] = k$

For a zero order reaction,
Rate $= k = 2.5 \times 10^{- 4} M / s$

Rate of production of $N_{2} = \frac{d}{d t} [N_{2}] = 2.5 \times 10^{- 4} M / s$
Rate of production of $H_{2} = \frac{d}{d t} [H_{2}] = 3 \times 2.5 \times 10^{- 4} M / s = 7.5 \times 10^{- 4} M / s$

Suggest Corrections

Similar questions

The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10⁻⁴ mol⁻¹ L s⁻¹?

N H_{3}

N_{2}

H_{2}

(k = 2.5 \times 10^{- 4} m o l L^{- 1} s^{- 1})

{N H}_{3}

N_{2}

H_{2}

{m o l e . l i t}^{- 1} . {s e c}^{- 1}

K = 2.5 \times 10^{- 4} m o l e . l i t^{- 1} . {s e c}^{- 1}

N H_{3}

K

5 \times 10^{- 4} M

s^{- 1}

N_{2}

N H_{3}

2.5 \times 10^{- 4}

m o l . d m^{- 3}

s^{- 1}

350

Join BYJU'S Learning Program