For the reaction, A leftharpoons nB the concentration of A decreases from 0.06 to 0.03 mol L-1 and that of B rises from 0 to 0.06 mol L-1 at equilibrium. The values of n and the equilibrium constant for the reaction, respectively, are

Question

For the reaction, A leftharpoons nB the concentration of A decreases from 0.06 to 0.03 mol L-1 and that of B rises from 0 to 0.06 mol L-1 at equilibrium. The values of n and the equilibrium constant for the reaction, respectively, are (A) 2 and 0.12 (B) 2 and 1.2 (C) 3 and 0.12 (D) 3 and 1.2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/bf35cae1e6d921cd03f9502d3ffcbe9b-.png" /> Thus, the reaction becomes A leftharpoons 2B kc=([B]2/[A])=((0.06)2/0.03)=0.12

Q. For the reaction, $A ⇌ n B$ the concentration of $A$ decreases from $0.06$ to $0.03 m o l L^{- 1}$ and that of $B$ rises from $0$ to $0.06 m o l L^{- 1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction, respectively, are

2 and 0.12

2 and 1.2

3 and 0.12

3 and 1.2

Thus, the reaction becomes $A ⇌ 2 B$
$k_{c} = \frac{[ B ] ^{2}}{[ A ]} = \frac{( 0.06 ) ^{2}}{0.03} = 0.12$

Q. For the reaction, A⇌nB the concentration of A decreases from 0.06 to 0.03molL−1 and that of B rises from 0 to 0.06molL−1 at equilibrium. The values of n and the equilibrium constant for the reaction, respectively, are