First slide

Kinetics of chemical reactions

Question

For a reaction N₂ + 3H₂ $\to$ 2NH₃, the rate of formation of ammonia was found to be 2.0 × 10^-4 mol dm^-3 s^-1. The rate of consumption of H₂ will be

Moderate

A

$1.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$
B

$2.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$
C

$3.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$
D

$4.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$

Solution

$We will have - \frac{d [N_{2}]}{dt} = - \frac{1}{3} \frac{d [H_{2}]}{dt} = \frac{1}{2} \frac{d [{NH}_{3}]}{dt}$
$It is given that \frac{d [{NH}_{3}]}{dt} = 2 \times 10^{- 4} mol {dm}^{- 3} s^{- 1} . Hence, - \frac{d [H_{2}]}{dt} = \frac{3}{2} \frac{d [{NH}_{3}]}{dt} = 3 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App