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:

\frac{1}{4}

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\frac{1}{2}

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1

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Solution

The correct option is A $1$
$A + B ⇌ C + D$
Initial moles $4$ $4$ $0$ $0$

At eq. $4 - x$ $4 - x$ $x$ $x$

Given,

At equilibrium, $[C] = 2$
$∴ x = 2$

$K_{c} = \frac{[C] [D]}{[A] [B]}$

Hence, $K_{c} = \frac{2 \times 2}{2 \times 2} = 1$

The correct answer is $K = 1$

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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-

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

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 :

4

moles of

A

are mixed with

4

moles of

B

when

2

moles of

C

are formed at equilibrium according to the reaction

A + B ⇌ C + D

The value of the equilibrium constant is :

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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:

The correct option is A 1 A+B⇌C+DInitial moles 4 4 0 0 At eq. 4−x 4−x x xGiven, At equilibrium, [C]=2∴x=2 Kc=[C][D][A][B]Hence, Kc=2×22×2=1The correct answer is K=1

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:

The correct option is A $1$
$A + B ⇌ C + D$
Initial moles $4$ $4$ $0$ $0$

At eq. $4 - x$ $4 - x$ $x$ $x$

Given,

At equilibrium, $[C] = 2$
$∴ x = 2$

$K_{c} = \frac{[C] [D]}{[A] [B]}$

Hence, $K_{c} = \frac{2 \times 2}{2 \times 2} = 1$

The correct answer is $K = 1$