The initial molar concentration of the reactants A and B were 0.1 M and 0.2 M respectively in the following reaction
`A+B hArr 2C`
When equilibrium was attained the concentration of A in the reaction mixture was found to be 0.06M . Calculate the equilibrium constant.

Answer

`K_(c) =0.67`

Explanation

the number of moles of A initially present =0.1 M
`The number of moles of A left =0.06 M
`:.` The number of moles of A reacted =0.1 -0.06 =0.04 M
According to the equation 0.04 M of A will react with 0.04 M of B to form 0.08 M of C. Therefore , the molar concentration of different species before the reaction and at the equilibrium point is :
`{:(,A ,+,B,hArr,2C),("Initial molar conc.",0.1 -0.04,,0.2-0.04,,2xx0.04),("Equilibrium molar conc.",0.06,,0.16,,0.08):}`
Applying Law of chemical equilibrium
`K_(C)=[[C]^(2)]/[[A][B]]=(0.08 xx 0.08)/(0.06xx0.16)=0.67`