For the reversible reaction net rate is : $2 NO (g) + O_{2} (g) ⇌ 2 NO_{2} (g)$
$(\frac{d x}{d t})_{n e t} = 2.6 \times 1 0^{3} [NO]^{2} [O_{2}] - 4.1 [N O_{2}]^{2}$
If a reaction mixture contains 0.01 mol each of NO and O₂and 0.1 mol of NO₂ in 1 L closed flask, then above reaction is :

Question

For the reversible reaction net rate is : $2 NO (g) + O_{2} (g) ⇌ 2 NO_{2} (g)$

$(\frac{d x}{d t})_{n e t} = 2.6 \times 1 0^{3} [NO]^{2} [O_{2}] - 4.1 [N O_{2}]^{2}$

If a reaction mixture contains 0.01 mol each of NO and O₂and 0.1 mol of NO₂ in 1 L closed flask, then above reaction is :

TestHub · Accepted Answer

2 NO ( g ) + O 2 ​ ( g ) ⇌ 2 N O 2 ​ ( g ) Net rate: R a t e = 2.6 × 1 0 3 [ NO ] 2 [ O 2 ​ ] − 4.1 [ N O 2 ​ ] 2 In a 1 L flask: * [ NO ] = 0.01 * [ O 2 ​ ] = 0.01 * [ N O 2 ​ ] = 0.1 Forward rate R a t e f ​ = 2.6 × 1 0 3 ( 0.01 ) 2 ( 0.01 ) = 2600 × 1 0 − 4 × 1 0 − 2 = 2.6 × 1 0 − 3 Backward rate R a t e b ​ = 4.1 ( 0.1 ) 2 = 4.1 × 0.01 = 0.041 Compare rates R a t e f ​ = 0.0026 R a t e b ​ = 0.041 Backward rate > Forward rate

Chemistry - Chemical Kinetics Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Main 2028 (Class 11)

JEE Advanced 2027 (Class 12)