Consider the following reversible chemical reactions: A _ 2 (g) + B _ 2 (g) k _ 1 ⇌ 2AB(g) .....(1) 6AB(g) k _ 2 ⇌ 3 A _ 2 (g) + 3 B _ 2 (g) .....(2) The relation between K _ 1 and K _ 2 is:

Correct answer: A. K _ 2 = K _ 1 ^ - 3 A _ 2 + B _ 2 ⇌ 2AB k _ 1 = [AB] ^ 2 [ A _ 2 ][ B _ 2 ] 6AB ⇌ 3 A _ 2 + 3 B _ 2 k _ 2 = [ A _ 2 ] ^ 3 [ B _ 2 ] ^ 3 [AB] ^ 6 So, 1 k _ 1 ^ 3 = k _ 2

JEE Main 20199 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2019 · 9 January · Evening Shift · Chemistry · Chemical Kinetics

Consider the following reversible chemical reactions: A _ 2 (g) + B...

Consider the following reversible chemical reactions:

{A}_{2}\left(g\right) + {B}_{2}\left(g\right)\overset{{k}_{1}}{⇌}2AB\left(g\right)

.....(1)

6AB\left(g\right)\overset{{k}_{2}}{⇌}3{A}_{2}\left(g\right) + 3{B}_{2}\left(g\right)

.....(2)

The relation between

{K}_{1}

and

{K}_{2}

is:

Options

A ${K}_{2} = {K}_{1}^{- 3}$
B ${K}_{1}{K}_{2} = \dfrac{1}{3}$
C ${K}_{2} = {K}_{1}^{3}$
D ${K}_{1}{K}_{2} = 3$

Correct answer

A. ${K}_{2} = {K}_{1}^{- 3}$

Step-by-step solution

{A}_{2} + {B}_{2} ⇌ 2AB

{k}_{1} = \dfrac{{\left[AB\right]}^{2}}{\left[{A}_{2}\right]\left[{B}_{2}\right]}

6AB ⇌ 3{A}_{2} + 3{B}_{2}

{k}_{2} = \dfrac{{\left[{A}_{2}\right]}^{3}{\left[{B}_{2}\right]}^{3}}{{\left[AB\right]}^{6}}

So,

\dfrac{1}{{k}_{1}^{3}} = {k}_{2}

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