In the reversible reaction A + B leftharpoons C + D, the concentration of each C and D at equilibrium was 0.8 mole/litre, then the equilibrium constant Kc will be

Question

In the reversible reaction A + B leftharpoons C + D, the concentration of each C and D at equilibrium was 0.8 mole/litre, then the equilibrium constant Kc will be (A) 6.4 (B) 0.64 (C) 1 .6 (D) 16.0

Tardigrade Admin · Accepted Answer

Suppose 1 mole of A and B is each taken, then 0.8 mole/litre of C and D each formed remaining concentration of A and B will be (1 - 0 .8) = 0.2 mole/litre. KC = ([C][D]/[A][B]) = (0.8 × 0.8 /0.2 × 0.2) =16

Q. In the reversible reaction $A + B ⇌ C + D$ , the concentration of each $C$ and $D$ at equilibrium was $0.8$ mole/litre, then the equilibrium constant $K_{c}$ will be

$6.4$

$0.64$

$1.6$

$16.0$

Suppose $1$ mole of $A$ and $B$ is each taken, then $0.8$ mole/litre of $C$ and $D$ each formed remaining concentration of $A$ and $B$ will be $(1 - 0.8) = 0.2$ mole/litre.
$K_{C} = \frac{[ C ] [ D ]}{[ A ] [ B ]} = \frac{0.8 \times 0.8}{0.2 \times 0.2} = 16$

Q. In the reversible reaction A+B⇌C+D, the concentration of each C and D at equilibrium was 0.8 mole/litre, then the equilibrium constant Kc​ will be

6.4

0.64

1.6

16.0