Question

For the reaction:

Answer 1

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$K_c = \dfrac{[x][x]}{[0.5-x][0.5-x]}$

$K_c = \dfrac{[x]^2}{[0.5-x]^2}$

where; $K_c = 1.845$

$1.845 = \begin {pmatrix} \dfrac{x}{0.5-x} \end {pmatrix}^2$

$\sqrt{1.845}= \begin {pmatrix} \dfrac{x}{0.5-x} \end {pmatrix}$

$1.3583= \begin {pmatrix} \dfrac{x}{0.5-x} \end {pmatrix}$

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