Question

For the reaction:

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

the value of Kc is 1.845 at a specific temperature. We place 0.500mol CO and 0.500mol H2O in a 1.00L container at this temperature and allow the reaction to reach equilibrium. Determine the equilibrium concentration of all species present in the container.

Answer 1

The equilibrium constant for CO now

= 0.212 M

For H₂O

= 0.212 M

For CO₂ = x = 0.2880 M

For H₂ = x = 0.2880 M

Step-by-step explanation:

The chemical equation for the reaction is:

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

The ICE Table for this reaction can be represented as follows:

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

Initial 0.5 0.5 - -

Change -x -x + x + x

Equilibrium 0.5 -x 0.5 - x

The equilibrium constant
`K_c = ([x][x])/([0.5-x][0.5-x])`

`K_c = ([x]^2)/([0.5-x]^2)`

where;
`K_c = 1.845`

`1.845 = \begin {pmatrix} (x)/(0.5-x) \end {pmatrix}^2`

`√(1.845)= \begin {pmatrix} (x)/(0.5-x) \end {pmatrix}`

`1.3583= \begin {pmatrix} (x)/(0.5-x) \end {pmatrix}`

1.3583 (0.5-x) = x

0.67915 - 1.3583x = x

0.67915 = x + 1.3583x

0.67915 = 2.3583x

x = 0.67915/2.3583

x = 0.2880

The equilibrium constant for CO now = 0.5 - x

= 0.5 - 0.2880

= 0.212 M

For H₂O = 0.5 - x

= 0.5 - 0.2880

= 0.212 M

For CO₂ = x = 0.2880 M

For H₂ = x = 0.2880 M