Answer:
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
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[tex]K_c = \dfrac{[x][x]}{[0.5-x][0.5-x]}[/tex]
[tex]K_c = \dfrac{[x]^2}{[0.5-x]^2}[/tex]
where; [tex]K_c = 1.845[/tex]
[tex]1.845 = \begin {pmatrix} \dfrac{x}{0.5-x} \end {pmatrix}^2[/tex]
[tex]\sqrt{1.845}= \begin {pmatrix} \dfrac{x}{0.5-x} \end {pmatrix}[/tex]
[tex]1.3583= \begin {pmatrix} \dfrac{x}{0.5-x} \end {pmatrix}[/tex]
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
