Question

Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.

Answer 1

4.41

Step-by-step explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

`[CO]_i = (0.500mol)/(5.00L) = 0.100M`

`[H_2]_i = (1.500mol)/(5.00L) = 0.300M`

`[H_2O]_(eq) = (0.198mol)/(5.00L) = 0.0396M`

Step 3: Make an ICE chart

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I 0.100 0.300 0 0

C -x -3x +x +x

E 0.100-x 0.300-3x x x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

`Kc = ([CH_4] * [H_2O] )/([CO] * [H_2]^(3) ) = (0.0396 * 0.0396 )/(0.060 * 0.181^(3) ) = 4.41`