At a particular temperature, K = 2.0 10-6 mol/L for the following reaction.
2 CO2(g) 2 CO(g) + O2(g)
If 1.8 mol CO2 is initially placed into a 5.9-L vessel, calculate the equilibrium concentrations of all species.
You need to learn where the arrow keys are and USE them.
...........2CO2 ==> 2CO + O2
initial....1.8........0....0
change.....-2x......2x.....x
equi.....1.8-2x.....2x.....x
Substitute the equil concns from the ICE chart into the Kc expression and solve for x. Then evaluate each element/compound. PLEASE note that x = moles. You must divide each by 5.9L to obtain concn in molarity.
Also note that you can sart by 1.8mol/5.9L = ?M and use M through out the ICE chart in which case x will be in molarity.
To calculate the equilibrium concentrations of all species, we need to use the expression for the equilibrium constant (K) and set up an ICE (Initial-Change-Equilibrium) table.
Step 1: Write down the balanced equation and the expression for the equilibrium constant:
2 CO2(g) ⇌ 2 CO(g) + O2(g)
K = [CO]^2 • [O2] / [CO2]^2
Step 2: Set up the ICE table:
Let x represent the change in concentration of CO2, CO, and O2.
CO2(g) -> CO(g) + O2(g)
Initial (mol/L): 1.8 0 0
Change (mol/L): -2x +2x +x
Equilibrium (mol/L): 1.8-2x 2x x
Step 3: Substitute the equilibrium concentrations into the expression for K and solve for x:
K = (2x)^2 • x / (1.8-2x)^2
2.0 x 10^-6 = 4x^3 / (1.8-2x)^2
Since the value of K is very small, we can assume that x is also small. This allows us to solve the equation by neglecting the x term in the denominator:
2.0 x 10^-6 = 4x^3 / (1.8)^2
2.0 x 10^-6 = 4x^3 / 3.24
Step 4: Solve for x using algebra:
4x^3 = (2.0 x 10^-6) * 3.24
x^3 = (2.0 x 10^-6) * 3.24 / 4
x^3 = (2.0 x 10^-6) * 0.81 x 10^-6
x^3 = 1.62 x 10^-12
x ≈ (1.62 x 10^-12)^(1/3)
x ≈ 1.24 x 10^-4
Step 5: Calculate the equilibrium concentrations of CO2, CO, and O2:
CO2: 1.8 - 2x = 1.8 - 2(1.24 x 10^-4) = 1.799752 mol/L
CO: 2x = 2(1.24 x 10^-4) = 2.48 x 10^-4 mol/L
O2: x = 1.24 x 10^-4 mol/L
Therefore, the equilibrium concentrations of all species are approximately:
CO2(g): 1.799752 mol/L
CO(g): 2.48 x 10^-4 mol/L
O2(g): 1.24 x 10^-4 mol/L