To calculate the number of moles of Cl2 produced at equilibrium, we'll use the given equilibrium constant (Kc) and the stoichiometry of the balanced equation.
The balanced equation for the reaction is: PCl5 ⇌ PCl3 + Cl2
The equilibrium expression for this reaction is: Kc = [PCl3] * [Cl2] / [PCl5]
Given: Kc = 0.041 Initial moles of PCl5 = 1 mole Volume of the vessel = 10 dm^3
To find the number of moles of Cl2 at equilibrium, we'll use an ICE (Initial-Change-Equilibrium) table and apply the stoichiometry of the reaction.
Let x be the number of moles of Cl2 produced at equilibrium.
The initial moles of PCl5 are 1 mole, and no PCl3 or Cl2 is present initially. Therefore, the initial concentrations of PCl5, PCl3, and Cl2 are:
[PCl5] = 1 mol / 10 dm^3 = 0.1 mol/dm^3 [PCl3] = 0 mol/dm^3 [Cl2] = 0 mol/dm^3
At equilibrium, the change in moles of PCl5 is (-x), the change in moles of PCl3 is (+x), and the change in moles of Cl2 is (+x). Therefore, the equilibrium concentrations are:
[PCl5] = (1 - x) mol / 10 dm^3 [PCl3] = x mol / 10 dm^3 [Cl2] = x mol / 10 dm^3
Substituting these values into the equilibrium expression, we have:
Kc = ([PCl3] * [Cl2]) / [PCl5] 0.041 = (x/10) * (x/10) / ((1 - x)/10) 0.041 = (x^2) / (10 * (1 - x))
Simplifying the equation, we get:
0.041 * 10 * (1 - x) = x^2 0.41 - 0.041x = x^2 0.041x^2 + 0.041x - 0.41 = 0
Now, we can solve this quadratic equation to find the value of x, which represents the number of moles of Cl2 at equilibrium.
Using a quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / (2a)
a = 0.041 b = 0.041 c = -0.41
Plugging in these values, we get:
x = (-0.041 ± √(0.041^2 - 4 * 0.041 * -0.41)) / (2 * 0.041)
Calculating this expression, we find:
x ≈ 0.532 or x ≈ -9.919
Since the number of moles cannot be negative, we discard the negative value.
Therefore, at equilibrium, approximately 0.532 moles of Cl2 will be produced.