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. 4 mole of

A

are mixed with 4 mole of

B

so, 2 moles of

C

are formed at equilibrium according to the following reaction,

A + B

⇌

C + D

The value of the equilibrium constant is:

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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

The correct option is C 1 A + B ⇌ C + D Initially, [A] = 4, [B] = 4, [C] = 0, [D] = 0 After time T, [A] = (4 - 2), [B] = (4 – 2), [C] = 2, [D] = 2 Equilibrium constant = [C][D][A][B] = 2 × 22 × 2 = 1

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

The correct option is C
1

A + B $⇌$ C + D

Initially, [A] = 4, [B] = 4, [C] = 0, [D] = 0

After time $T$ , [A] = (4 - 2), [B] = (4 – 2), [C] = 2, [D] = 2

Equilibrium constant = $\frac{[C] [D]}{[A] [B]} = \frac{2 \times 2}{2 \times 2}$ = 1