Answer:
(A) 0.23 bar
Step-by-step explanation:
You want to know the final pressure of Cl₂ in a sealed flask in which equal numbers of moles of SO₂Cl₂, SO₂, and Ar are introduced to give a total pressure of 1.80 bar. Temperature remains constant. The Kp of the reaction is 0.50 for ...
SO₂Cl₂(g) ⇔ SO₂(g) +Cl₂(g)
Partial pressures
When the SO₂Cl₂ dissociates, the total number of moles of gas in the flask is increased by the number of moles of Cl₂. Without loss of generality, we can assume that the number of moles of each gas is initially 1. If we let x represent the number of moles of Cl₂, then the partial pressures of the flask contents are ...
- SO₂Cl₂ — (1.80 bar)·(1 -x)/3
- SO₂ — (1.80 bar)·(1 +x)/3
- Cl₂ — (1.80 bar)·(x)/3
- Ar — (1.80 bar)/3
Chlorine
The number of moles of chlorine gas at equilibrium will be given by the equation for the reaction constant.
[SO₂]·[Cl₂]/[SO₂Cl₂] = 0.5 = [0.6(1 +x)][0.6(x)]/[0.6(1 -x)]
Writing this as a quadratic, we have ...
6x(1 +x) = 5(1 -x)
6x² +11x -5 = 0
x = (√241 -11)/12 ≈ 0.377015 . . . . . moles
Then the final pressure of the chlorine gas is ...
(1.80 bar)·(0.377015)/3 ≈ 0.23 bar
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Additional comment
You may notice that the units of the reaction constant equation as used here are bar. Ordinarily, the reaction constant Kp is unitless. In the development of the equation for Kp, we find there is an extra factor of (RT)^(∆n) where ∆n is the difference in the numbers of moles between left-side and right-side reactants in the equation.
In order for this solution to work, we have to assume that RT=1, since there is not enough information otherwise.
The solution to the quadratic can be found any number of ways. We used the formula, a graph, and an equation solver to check this solution. Only the positive solution is of interest.
In the absence of any extra SO₂, half the initial quantity of SO₂Cl₂ would dissociate. The effect of the extra SO₂ is to reduce the quantity of Cl₂ that would otherwise be produced.