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CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations of [CO]=0.27M[CO]=0.27M and [H2]=0.49M[H2]=0.49M. At equilibrium, the concentration of CH3OHCH3OH is 0.11 MM. Find the equilibrium constant at this temperature.

Respuesta :

ajeigbeibraheem

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

[tex]CO_{(g)}[/tex]    +      [tex]2H_2O_{(g)}[/tex]   ⇄  [tex]CH_3OH_{(g)}[/tex]

The ICE table can be represented as:

                                  [tex]CO_{(g)}[/tex]    +      [tex]2H_2_{(g)}[/tex]   ⇄  [tex]CH_3OH_{(g)}[/tex]

Initial                          0.27             0.49              0.0

Change                      -x                  -2x                 x

Equilibrium               0.27 - x         0.49 -2x          x

We can now say that the concentration of  [tex]CH_3OH_{(g)}[/tex] at equilibrium is x;

Let's not forget that at equilibrium  [tex]CH_3OH_{(g)}[/tex] = 0.11 M

So:

x =  [[tex]CH_3OH_{(g)}[/tex]] = 0.11 M

[[tex]CO_{(g)}[/tex]] = 0.27 - x

[[tex]CO_{(g)}[/tex]] = 0.27 - 0.11

[[tex]CO_{(g)}[/tex]] = 0.16 M

[[tex]2H_2_{(g)}[/tex]] = (0.49 - 2x)

[[tex]2H_2_{(g)}[/tex]] = (0.49 - 2(0.11))

[[tex]2H_2_{(g)}[/tex]] = 0.49 - 0.22

[[tex]2H_2_{(g)}[/tex]] = 0.27 M

[tex]K_C = \frac{[CH_3OH]}{[CO][H_2]^2}[/tex]

[tex]K_C = \frac{(0.11)}{(0.16)[(0.27)^2}[/tex]

[tex]K_C = 9.4307[/tex]

[tex]K_C[/tex] = 9.4

∴ The equilibrium constant at that temperature = 9.4