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← JEE Advanced 2024 · Paper 1
ChemistryInteger+4Question 43 of 51

JEE Advanced 2024 · Paper 1, Chemistry question 9

Consider the following reaction,

2 HX2(g)+2 NO(g)→NX2(g)+2 HX2O(g)\ce{2H2(g) + 2NO(g) -> N2(g) + 2H2O(g)}

which follows the mechanism given below:

2 NO(g)⇌k1k−1NX2OX2(g)(fast equlibrium)\ce{2NO(g)} \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \ce{N2O2(g)} \qquad \text{(fast equlibrium)}
NX2OX2(g)+HX2(g)→k2NX2O(g)+HX2O(g)(slow reaction)\ce{N2O2(g) + H2(g)} \xrightarrow{k_2} \ce{N2O(g) + H2O(g)} \qquad \text{(slow reaction)}
NX2O(g)+HX2(g)→k3NX2(g)+HX2O(g)(fast reaction)\ce{N2O(g) + H2(g)} \xrightarrow{k_3} \ce{N2(g) + H2O(g)} \qquad \text{(fast reaction)}

The order of the reaction is _____.

Answer and worked solution

Official answer: 3

The slow step sets the rate: r=k2[NX2OX2][HX2]r = k_2[\ce{N2O2}][\ce{H2}].

The fast pre-equilibrium gives [NX2OX2]=K1[NO]2[\ce{N2O2}] = K_1[\ce{NO}]^2 with K1=k1k−1K_1 = \frac{k_1}{k_{-1}}, so
r=k2K1[NO]2[HX2].r = k_2K_1[\ce{NO}]^2[\ce{H2}].
Order =2+1=3= 2 + 1 = 3.