3.0 moles each of carbon monoxide, hydrogen, and carbon are placed in a 2.0 liter vessel and allowed to come to equilibrium according to the equation: CO(g)+H2(g)-->C(s)+H20(g)
if the equilibrium constant at the temperature of the experiment is 4.0, what is the equilibrium concentration of water vapor?
[H2O]=1M
3.0 mols/2.0 L = 1.5M
..........CO + H2 ==> C + H2O
initial..1.5..1.5....1.5...0
change....-x...-x.....x....x
equil...1.5-x..1.5-x..x....x
Kc = 4.0 = (H2O)/(C0)(H2O)
Substitute from the ICE chart into the Kc expression and solve for x.
[H2O]=.81
To find the equilibrium concentration of water vapor, we first need to determine the initial moles of carbon monoxide (CO) and hydrogen (H2) in the 2.0 liter vessel.
Given that there are 3.0 moles each of CO and H2, and the total volume is 2.0 liters, we can determine the initial concentration (C0) of both gases by dividing the number of moles by the volume:
C0(CO) = 3.0 moles / 2.0 liters = 1.5 M
C0(H2) = 3.0 moles / 2.0 liters = 1.5 M
Next, we need to consider the stoichiometry of the balanced equation. From the equation:
CO(g) + H2(g) → C(s) + H2O(g)
We see that for every 1 mole of CO that reacts, 1 mole of H2O is produced. Therefore, the reaction will consume the same number of moles of CO as the number of moles of H2O produced.
Since there are an equal number of moles for CO and H2 (3.0 moles each), it means that all the initial CO will be consumed at equilibrium, leaving no CO remaining. Thus, we can assume the equilibrium concentration of CO (CCO) is zero.
Using the information above, we can construct an ICE (Initial, Change, Equilibrium) table:
CO(g) + H2(g) ⇌ C(s) + H2O(g)
-------------------------------------------
Initial 1.5 M 1.5 M 0 M 0 M
Change -1.5 M -1.5 M +1.5 M +1.5 M
Equilibrium 0 M 0 M 1.5 M 1.5 M
Now, we can use the equilibrium constant expression (Keq) to find the equilibrium concentration of water vapor (C(H2O)).
Keq = [C(H2O)] / [C(CO) x C(H2)]
Given that Keq = 4.0, C(CO) = 0 M, and C(H2) = 1.5 M, we plug these values into the equation and solve for C(H2O):
4.0 = [C(H2O)] / (0 M x 1.5 M)
Therefore, [C(H2O)] = 4.0 x 0 M x 1.5 M
= 0 M
Hence, the equilibrium concentration of water vapor is 0 M.