Consider the following reaction, 2 H _2( ~g )+2 NO ( g ) N _2( ~g )+2 H _2 O ( g ) which follows the mechanism given below: 2 NO ( g ) k_1 k_ -1 N _2 O _2( ~g ) (fast equlibrium) N _2 O _2( ~g )+ H _2( ~g ) k_2 ~N _2 O ( g )+ H _2 O ( g ) (slow reaction) N _2 O ( g )+ H _2( ~g ) k_3 ~N _2( ~g )+ H _

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Consider the following reaction, 2 H _2( ~g )+2 NO ( g ) N _2( ~g )+2 H _2 O ( g ) which follows the mechanism given below: 2 NO ( g ) k_1 k_ -1 N _2 O _2( ~g ) (fast equlibrium) N _2 O _2( ~g )+ H _2( ~g ) k_2 ~N _2 O ( g )+ H _2 O ( g ) (slow reaction) N _2 O ( g )+ H _2( ~g ) k_3 ~N _2( ~g )+ H _2 O ( g ) (fast reaction) The order of the reaction is_____

Quantrex Academy · Accepted Answer

Correct answer: A. A

Rate law = k _2 [ ~N _2 O _2 ] [ H _2 ] [ slowest step of reaction is RDS ] k _1 k _ -1 = [ N _2 O _2 ] [ NO ]^2 [ N _2 O _2 ]= k _1 k _ -1 [ NO ]^2 Rate = k _2 k _1 k _ -1 [ NO ]^2 [ H _2 ] Order of reaction is (3)

JEE Advanced Chemistry Question (2024) — Solution

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