Consider the following reaction:
CO(g) + 2H2 (g) ⇌ CH3 OH (g)
In an experiment, 0.840 mol of CO and 0.840 mol of H2 were placed in a 2.00 L reaction vessel. When the reaction reached equilibrium there was 0.580 mol of CO remaining. Kc for the reaction is.
a) Kc = 0.357
b) Kc = 2.80
c) Kc = 0.0571
d) Kc = 0.608
e) Kc = 17.5
To find the equilibrium constant (Kc) for the reaction, we need to calculate the concentrations of CO, H2, and CH3OH at equilibrium.
Given:
Initial concentration of CO (COCO) = 0.840 mol
Initial concentration of H2 (CH2) = 0.840 mol
Final concentration of CO (COCOfinal) = 0.580 mol
Since the stoichiometric coefficient for CO is 1, the change in concentration (ΔCOCO) = COCO - COCOfinal = 0.840 mol - 0.580 mol = 0.260 mol.
Since the stoichiometric coefficient for H2 is 2, the change in concentration (ΔCH2) = 2 * ΔCOCO = 2 * 0.260 mol = 0.520 mol.
The concentration of CO at equilibrium (COCOequilibrium) = COCOfinal = 0.580 mol.
The concentration of H2 at equilibrium (CH2equilibrium) = CH2initial - ΔCH2 = 0.840 mol - 0.520 mol = 0.320 mol.
The concentration of CH3OH at equilibrium (CCH3OH) = 0.840 mol - ΔCOCO = 0.840 mol - 0.260 mol = 0.580 mol.
Now, we can substitute these values into the expression for Kc:
Kc = [CH3OH] / [CO] * [H2]^2
Kc = (0.580 mol) / (0.580 mol) * (0.320 mol)^2
Kc = 0.580 / 0.588
Kc ≈ 0.986
Therefore, the correct answer is not listed.