Question

At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initially placed into a 4.9-L vessel, calculate the equilibrium concentrations of all species.

Answer 1

concentration of
`[O_2]` = 0.0124 = 12.4 ×10⁻³ M

concentration of
`[CO]` = 0.0248 = 2.48 ×10⁻² M

concentration of
`[CO_2]` = 0.4442 M

Step-by-step explanation:

Equation for the reaction:

`2CO_2_{(g)` ⇄
`2CO_{(g)` +
`O_2_{(g)`

Concentration of
`CO_2_{(g)` =
`(2.3)/(4.9)` = 0.469

For our ICE Table; we have:

`2CO_2_{(g)` ⇄
`2CO_{(g)` +
`O_2_{(g)`

Initial 0.469 0 0

Change - 2x +2x +x

Equilibrium (0.469-2x) 2x x

K =
`([CO]^2[O])/([CO_2]^2)`

K =
`([2x]^2[x])/([0.469-2x]^2)`

`4.1*10^(-6)=(2x^3)/((0.469-2x)^2)`

Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²

`4.1*10^(-6) = (2x^3)/((0.938))`

`2x^3 =3.8458*10^{-6`

`x^3 =(3.8458*10^(-6))/(2)`

`x^3=1.9229*10^{-6`

`x=\sqrt[3]{1.9929*10^(-6)}`

x = 0.0124

∴ at equilibrium; concentration of
`[O_2]` = 0.0124 = 12.4 ×10⁻³ M

concentration of
`[CO]` = 2x = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of
`[CO_2]` = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M