For the reaction `A+B hArr C+D` taking place in a 1 L vessel, equilibrium concentration of `[C]=[D]=0.5M` if we start with 1 mole each A and B. Percentage of A converted into C if we start with 2 moles of A and 1 mole of B, is

Options

A. `25%`
B. `40%`
C. `66.66%`
D. `33.33%`

Answer

D

Explanation

`{:("Initial conc. ",A,+,B,hArr,C,+,D),("Eqm. conc.",(1-0.5)M,,(1-0.5)M,,0.5M,,0.5M):}`
`[A]=[B]=[C]=[D]=0.5`
`:. K=([C][D])/([A][B])`
In second case,
`{:(,A,+,B,hArr,C,+,D),("Initial eonc. ",2M,,1M,,,,),("Eqm. conc.",(2-x)M,,(1-x)M,,xM,,xM):}`
`K=([C][D])/([A][B])`
`:. (x^(2))/((2-x)(1-x))=1`
`x^(2)=(2-x)(1-x)`
`x^(2)=-2x-x+2+x^(2)`
`:. x=(2)/(3)`
`:. %` of A converted to `C=(x)/(2)xx100`
`=(2xx100)/(3xx2)=33.33%`