Answer:
42.8
Explanation:
Let's consider the following reaction at equilibrium.
CO(g) + 2 H₂(g) ⇌ CH₃OH(g)
The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.
The concentration equilibrium constant for this reaction is:
[tex]Kc = \frac{[CH_3OH]}{[CO]\times [H_2]^{2} } = \frac{0.150}{0.0960\times 0.191^{2} } = 42.8[/tex]
Answer:
The value for the equilibrium constant Kc is 42.83
Explanation:
Step 1: Data given
At equilibrium, the mixture contains:
0.0960 M CO,
0.191 M H2,
and 0.150 M CH3OH.
Step 2: The balance equation
CO(g) + 2 H2(g) ⇌ CH3OH(g).
Step 3: Calculate the value for the equilibrium constant Kc
Kc = [CH3OH] / [CO]*[H2]²
⇒with Kc = the equilibrium constant = TO BE DETERMINED
⇒with [CH3OH] = 0.150 M
⇒with [CO] = 0.0960 M
⇒with [H2] = 0.191 M
Kc = 0.150 M / (0.0960 M *( 0.191 M)²)
Kc = 42.83
The value for the equilibrium constant Kc is 42.83
