Question

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.

Answer 1

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

$CO_{(g)}$    +      $2H_2O_{(g)}$   ⇄  $CH_3OH_{(g)}$

The ICE table can be represented as:

                                  $CO_{(g)}$    +      $2H_2_{(g)}$   ⇄  $CH_3OH_{(g)}$

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  $CH_3OH_{(g)}$ at equilibrium is x;

Let's not forget that at equilibrium  $CH_3OH_{(g)}$ = 0.11 M

So:

x =  [$CH_3OH_{(g)}$] = 0.11 M

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

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

[$CO_{(g)}$] = 0.16 M

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

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

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

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

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

$K_C = \frac{(0.11)}{(0.16)[(0.27)^2}$

$K_C = 9.4307$

$K_C$ = 9.4

∴ The equilibrium constant at that temperature = 9.4