`1.1` mole of A mixed with `2.2` mole of B and the mixture is kept in a 1 litre at the equilibrium `0.2` mole of C is formed, then the value of `K_(c)` will be:

Options

A. `0.005`
B. `0.001`
C. `0.01`
D. `0.0001`

Answer

B

Explanation

`{:(,A,+,2B,hArr,2C,+,D),("Initial mole",1.1,,2.2,,0,,0),("At Eq.",1.1-x,,2.2-2x,,2x,,x),(,1.1-0.1,,2.2-0.2,,0.2,,0.1),(,1,,2,,0.2,,0.1),("Active mass",1/1,,2/1,,0.2,,0.1):}`
`K_(c) = ([C]^(2)[D])/([A][B]^(2)) = ((2)/(10)xx(2)/(10)xx(1)/(10))/(1xx2xx2)`
`= (1)/(1000) =0.001`