The value of `K_(C)` for the equilibrium reaction `CO_(2)(g) +C(s)hArr 2 CO(g)`
at T (K) is 0 . 0 36 . If the equilibrium concentration of `CO_(2) (g) ` is 0 . 004 M , the concentration of CO (g) in mol L^(-1)` is

Options

A. `3 . 6 xx 10^(-2)`
B. `2 . 0 xx 10^(-2)`
C. `1 . 2 xx 10^(-2)`
D. `1 . 2 xx 10 ^(-3)`

Answer

C

Explanation

For the reaction,
`CO_(2) (g) +C(s) hArr 2 CO_(g)`
Given,
`K_(c) = 0. 036`
Equilibrium conc, of `[CO_(2)(g) ] = 0 . 004 M`
As we know that
` K_(c)= ([CO(g)]^(2))/([CO_(2) (g)])" " [ C (s) = 1 ] . . . (i)`
On substituting the given values in eq. (i). we get
` 0 . 036 = ([CO(g)]^(2))/(0.004)`
`[ CO(g)] = sqrt(0 . 0 36 xx 0 . 00 4 )`
` = sqrt( 1.44 xx 10^(-4))`
` = 1 . 2 xx 10^(-2) mol //L`