At 100 K the partial pressure of `CO_(2)` (g ) and CO (g) for the reaction
`CO_(2)(g) + C(s) hArr 2CO (g )` in a closed vessel at equilibrium are 0.15 and 0.60 bar respectively . The `K_(c )` for this reaction at the same temperature is approximately

Options

A. `2.0 xx 10^(-4)`
B. `2.89 xx 10^(10^(-2))`
C. `2.89 xx 10^(-3)`
D. `5.78 xx 10^(-3)`

Answer

B

Explanation

For the reaction
`CO_(2)(g) + c(s) hArr 2CO (g)`
`K_(sp)= ([CO]^(2) (g) )/([CO_(2)](g)) = (p[CO]^(2) (g))/(p[CO_(2)](g))`
Given `p[CO]-0.6 bar`
`K_(p) = [[0.6]^(2)])/[[0.15]] =24`
Also `K_(p) = K_(c ) (RT)^(Delta n)`
Where `Delta n` = moles of gaseous products = moles of gaseous reactants =2- l =1
`K_(p) =k_(r ) xx R xx T`
Where `K_(e )= (k_(p))/(RT)`
`K_(p) - 2.4`
`R= 0.082 L atn k^(-1) mol^(-1)`
`T= 1000 K rARr Delta n = + 1`
`K_(c ) = (K_(p))/(RT) = (2.4)/(0.0821 xx 1000) = 0.0289`
`K_(c )= 2.89 xx 10^(-2)`
Hence option ( b) is the correct answer