Question

0.125 moles of oxygen gas is added to carbon in a 0.250L container. The mixture equilibrates at 500K. Calculate the equilibrium concentration of carbon monoxide knowing that K = 0.086 at 500K.

C(s) + O₂(g) <--> 2CO(g)

Answer 1

Given :

Moles of oxygen = 0.125

Volume of container = 0.250 L

Reaction :

C ( s) + O2 (g) -- > 2 CO(g)

We have to calculate equilibrium concentration of carbon monoxide.

K = 0086

Lets set up ICE

C ( s) +       O2 (g)                   -- >         2 CO(g)

I                          0.125 mol / 0.250 L                  0

C                      -x                                                +2x

E                        (0.125/0.250) L – x                      2x

Equilibrium expression

K = [ CO ] ² / [ O2 ]

Solid phase does not involve in the equilibrium.

Lets plug given values.

0.086 = (2x)² / ( 0.125/0.250) L – x)

0.086 = 4x²   / ( 0.5 – x )

x = 0.0935

Lets find equilibrium concentration of CO (g)

[CO2 (g) ] = 2 x = 2 x 0.0935

= 0.187 M