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?

A. 0.001
B. 0.072
C. 0.090
D. 1.2

Answer 1

Answer : The correct option is, (C) 0.090

Solution : Given,

Initial moles of
`O_2` = 0.350 mole

volume of solution = 5.00 L

First we have to calculate the concentration
`O_2`.

`\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}`

`\text{Concentration of }O_2=(0.350moles)/(5.00L)=0.07M`

The given equilibrium reaction is,

`C(s)+O_2(g)\rightleftharpoons 2CO(g)`

Initially 0.07 0

At equilibrium (0.07-x) 2x

The expression of
`K_c` will be,

`K_c=([CO]^2)/([O_2])`

As we are given the concentration of
`CO` at equilibrium is, 0.060 M

That means,

2x = 0.060 M

x = 0.030 M

The concentration of
`O_2` at equilibrium = 0.07 - x = 0.07 - 0.03 = 0.04 M

Now put all the given values in the above expression, we get:

`K_c=([CO]^2)/([O_2])`

`K_c=((0.060)^2)/((0.04))`

`K_c=0.090`

Therefore, the value of equilibrium constant for this reaction is, 0.090