The equilibrium constant for a reaction
`A+2B hArr 2C` is `40`. The equilibrium constant for reaction `C hArr B+1//2 A` is

Options

A. (a)`1//40`
B. (b)`(1//40)^(1//2)`
C. (c )`(1//40)^(2)`
D. (d)`40`

Answer

B

Explanation

`A +2B hArr 2C, " "…(i)" " K=40`
And for `C hArr B +(1)/(2)A," "…(ii)" "K_(1) =?`
`K =(1)/(sqrt(k))=(1)/((40)^(1//2))`
Since equation (ii) is obtained by reversing and dividing equation (i) by 2.