4 moles of A are mixed with 4 moles of B, when 2 moles of C and D are formed at equilibrium according to the reaction, A+B leftharpoons C+D the value of equilibrium constant is

Question

4 moles of A are mixed with 4 moles of B, when 2 moles of C and D are formed at equilibrium according to the reaction, A+B leftharpoons C+D the value of equilibrium constant is (A) 4 (B)  1 (C) 1 / 2 (D) 1 / 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/136cb85385f1dbbf03f2ab577f6737a4-.png" /> K=([C][D]/[A][B])=(2 × 2/2 × 2)=1

Q. $4$ moles of $A$ are mixed with $4$ moles of $B$ , when $2$ moles of $C$ and $D$ are formed at equilibrium according to the reaction,
$A + B ⇌ C + D$
the value of equilibrium constant is

$4$

$1$

$1/2$

$1/4$

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

Q. 4 moles of A are mixed with 4 moles of B, when 2 moles of C and D are formed at equilibrium according to the reaction, A+B⇌C+D the value of equilibrium constant is

4

1

1/2

1/4

K=[A][B][C][D]​=2×22×2​=1

Q. $4$ moles of $A$ are mixed with $4$ moles of $B$ , when $2$ moles of $C$ and $D$ are formed at equilibrium according to the reaction,
$A + B ⇌ C + D$
the value of equilibrium constant is

$4$

$1$

$1/2$

$1/4$

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