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Question: In the reversible reaction *A* + *B*⇌*C* + *D*, the concentration of each *C* and *D* at equilibrium...

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 KcK_{c} will be

A

6.4

B

0.64

C

1.6

D

16.0

Answer

16.0

Explanation

Solution

Suppose 1 mole of A and B 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 each.

Kc=[C][D][A][B]=0.8×0.80.2×0.2=16.0K_{c} = \frac{\lbrack C\rbrack\lbrack D\rbrack}{\lbrack A\rbrack\lbrack B\rbrack} = \frac{0.8 \times 0.8}{0.2 \times 0.2} = 16.0