Answer:
4.41
Explanation:
Step 1: Write the balanced equation
CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)
Step 2: Calculate the respective concentrations
[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]
[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]
[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]
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)
[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]
