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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.

Respuesta :

ajeigbeibraheem

Answer:

The equilibrium constant for CO now

= 0.212 M

For Hâ‚‚O

= 0.212 M

For COâ‚‚ = x = 0.2880 M

For Hâ‚‚ = 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 Hâ‚‚O = 0.5 - x

= 0.5 -  0.2880

= 0.212 M

For COâ‚‚ = x = 0.2880 M

For Hâ‚‚ = x = 0.2880 M

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