4 moles of A are mixed with 4 moles of B, then 2 moles of C are formed at equilibrium, according to the reaction,
A + B $⇌$ C + D.
The equilibrium constant is :

Q. For the reaction,

A + B ⇌ C + D

4

moles of

A

are mixed with

4

moles of

B

. At equilibrium,

2

moles of

C

and

D

are formed. The equilibrium constant for the reaction will be:

Q. 4 moles of A are mixed with 4 moles of B. At equilibrium for the reaction

A + B ⇌ C + D

, 2 moles of C and D are formed.

The equilibrium constant for the reaction will be-

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1.1 mole of A are mixed with 2.2 mole of B and the mixture is then kept in one litre flask till the equilibrium is attained A+2B⇌2C+D. At the equilibrium, 0.2 mole of C are formed. The equilibrium constant of the reaction is :

The correct option is A 0.001 Aaa−x+2Bbb−2x⇌2C02x+D0x Given, 2x=0.2, Also, a=1.1 x=0.1b=2.2∴Kc=4x3(a−x)(b−2x)2 =4×(0.1)3(1.1−0.1)(2.2−0.2)2=0.001

1.1 mole of $A$ are mixed with 2.2 mole of $B$ and the mixture is then kept in one litre flask till the equilibrium is attained $A + 2 B ⇌ 2 C + D$ . At the equilibrium, 0.2 mole of $C$ are formed. The equilibrium constant of the reaction is :

The correct option is A 0.001
$A a a - x + 2 B b b - 2 x ⇌ 2 C 0 2 x + D 0 x$

Given, $2 x = 0.2$ , Also, $a = 1.1$

$x = 0.1 b = 2.2$

$∴ K_{c} = \frac{4 x^{3}}{(a - x) {(b - 2 x)}^{2}}$

$= \frac{4 \times {(0.1)}^{3}}{(1.1 - 0.1) {(2.2 - 0.2)}^{2}} = 0.001$