Question

At high temperatures, carbon reacts with O2 to produce CO as follows: C(s) O2(g) 2CO(g). When 0.350 mol of O2 and excess carbon were placed in a 5.00-L container and heated, the equilibrium concentration of CO was found to be 0.060 M. What is the equilibrium constant, Kc, for this reaction

Answer 1

Value of
`K_(c)` is 0.090.

Step-by-step explanation:

Initial molarity of
`O_(2)` =
`(0.350)/(5.00)M` = 0.0700 M

Construct an ICE table corresponding to the combustion reaction of carbon to determine
`K_(c)`

`C(s)+O_(2)(g)\rightarrow 2CO(g)`

I (M): - 0.0700 0

C (M): - -x +2x

E (M): - 0.0700-x 2x

So,
`K_(c)=([CO]^(2))/([O_(2)])` , where [CO] and
`[O_(2)]` represents equilibrium concentration of CO and
`O_(2)` respectively.

Here,
`[CO]=2x=0.060`

⇒x = 0.030

So,
`[O_(2)]` = 0.0700-x = (0.0700-0.030) = 0.040

Hence,
`K_(c)=((0.060)^(2))/(0.040)=0.090`