Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g)

The reaction between CO and H2 is carried out at a specific temperature with initial concentrations of CO = 0.32 M and H2 = 0.53 M. At equilibrium, the concentration of CH3OH is 0.16 M. Find the equilibrium constant at this temperature.

Kc = 4.76

Explanation:

To find the concentrations of CO and H₂ at equilibrium, you have to set up an ICE (Initial, Change, Equilibrium) table.

    CO (g) + 2H₂(g) ⇌ CH₃OH (g)

I   0.32 M   0.53 M          0

C    -x            -2x            +x

E  0.32-x    0.53-2x        0.16 M

Since you know the concentration of CH₃OH at equilibrium, it would be equal to x since 0 + x = 0.16. So,

[CH₃OH] = 0.16 M

[CO] = 0.32 - 0. 16 = 0.16 M

[H₂] = 0.53 - 2(0.16) = 0.21 M

Now that you have all the concentrations at equilibrium, you can calculate the equilibrium constant.

Kc  = products ÷ reactants

     = [CH₃OH] ÷ [CO][H₂]²

     =  0.16 ÷ (0.16)(0.21)

Kc =   4.76

The equilibrium constant at this temperature is 4.76.

Hope this helps.

a is the answer to this question

explanation:

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2