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In the reversible reaction A+B hArr C+D,...

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

A

`6.4`

B

`0.64`

C

`1.6`

D

16

Text Solution

Verified by Experts

The correct Answer is:
D
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.
`K_c = ([C] [D])/([A][B])=(0.8 xx0.8)/(0.2 xx0.2) = 16`
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