1. Tardigrade
  2. Question
  3. Chemistry
  4. 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

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

Equilibrium

Solution:

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.
$K_{C} =\frac {[C][D]}{[A][B]} =\frac {0.8 \times 0.8 }{0.2 \times 0.2} =16$