In reaction A + 2B leftharpoons 2C + D, initial concentration of B was 1.5 times of [A], but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is :

Question

In reaction A + 2B leftharpoons 2C + D, initial concentration of B was 1.5 times of [A], but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is : (A) 8 (B) 4 (C) 12 (D) 6

Tardigrade Admin · Accepted Answer

beginmatrixA +&2B&leftharpoons&2C +&D a&1.5a&&0&0 (a-x)&(1.5a-2x)&&2x&x endmatrix Hence Kc = ((2x)2× x/(a-x)(1.5a-2x)2 ) Given, at equilibrium ∴ (a-x)(1.5a-2x) ∴ a = 2x On solving Kc = 4

Q. In reaction $A + 2 B ⇌ 2 C + D$ , initial concentration of $B$ was $1.5$ times of $[A]$ , but at equilibrium the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is :

8

4

12

6

$A + a (a - x) 2 B 1.5 a (1.5 a - 2 x) ⇌ 2 C + 0 2 x D 0 x$
Hence $K_{c} = \frac{( 2 x ) ^{2} \times x}{( a - x ) ( 1.5 a - 2 x ) ^{2}}$
Given, at equilibrium
$∴ (a - x) (1.5 a - 2 x)$
$∴ a = 2 x$
On solving $K_{c} = 4$

Q. In reaction A+2B⇌2C+D, initial concentration of B was 1.5 times of [A], but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is :

8

4

12

6

A+a(a−x)​2B1.5a(1.5a−2x)​⇌​2C+02x​D0x​ Hence Kc​=(a−x)(1.5a−2x)2(2x)2×x​ Given, at equilibrium ∴(a−x)(1.5a−2x) ∴a=2x On solving Kc​=4

Q. In reaction $A + 2 B ⇌ 2 C + D$ , initial concentration of $B$ was $1.5$ times of $[A]$ , but at equilibrium the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is :

$A + a (a - x) 2 B 1.5 a (1.5 a - 2 x) ⇌ 2 C + 0 2 x D 0 x$
Hence $K_{c} = \frac{( 2 x ) ^{2} \times x}{( a - x ) ( 1.5 a - 2 x ) ^{2}}$
Given, at equilibrium
$∴ (a - x) (1.5 a - 2 x)$
$∴ a = 2 x$
On solving $K_{c} = 4$