The equilibirum constant `K_(c)` of the reaction, `2AhArrB+C is 0.5 at 25^(@)C and 1` atm. The reaction will proceed in the backward direction when concentrations [A], [B] and [C] are respectively-

Options

A. `10^(-3),10^(-2) and 10^(-2) M`
B. `10^(-1),10^(-2) and 10^(-2) M`
C. `10^(-2),10^(-2) and 10^(-3) M`
D. `10^(-2),10^(-3) and 10^(-3) M`

Answer

A

Explanation

`Q=([B][C])/([A^(2)])&K_(C)=0.5`
For option (A)
`Q=((10^(-2))xx(10^(-2)))/([10^(-3)]^(2))=100`
`& Qgt K_(C)` i.e., reaction proceed in backward direction