A+B leftharpoons C+D. If initially the concentration of A and B are both equal but at equilibrium, concentration of D will be twice of that of A, then what will be the equilibrium constant of reaction?

Question

A+B leftharpoons C+D. If initially the concentration of A and B are both equal but at equilibrium, concentration of D will be twice of that of A, then what will be the equilibrium constant of reaction? (A) 4/9 (B) 9/4 (C) 1/9 (D) 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/0fe07aa1ae4907cc590abf13f76e6179-.png" /> Kc=([C][D]/[A][B])=(2 x × 2 x/x × x)=4

Q. $A + B ⇌ C + D .$ If initially the concentration of $A$ and $B$ are both equal but at equilibrium, concentration of $D$ will be twice of that of $A$ , then what will be the equilibrium constant of reaction?

4/9

9/4

1/9

4

$K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{2 x \times 2 x}{x \times x} = 4$

Q. A+B⇌C+D. If initially the concentration of A and B are both equal but at equilibrium, concentration of D will be twice of that of A, then what will be the equilibrium constant of reaction?

4/9

9/4

1/9

4

Kc​=[A][B][C][D]​=x×x2x×2x​=4

Q. $A + B ⇌ C + D .$ If initially the concentration of $A$ and $B$ are both equal but at equilibrium, concentration of $D$ will be twice of that of $A$ , then what will be the equilibrium constant of reaction?

$K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{2 x \times 2 x}{x \times x} = 4$