the equilibrium contant kc=49 at 230^0C, for the reaction PCl3 + Cl2= PCL5. If 0.50mol of each of phosphorus trichloride and chlorine are added to a 5.0L reaction vessel, what is the equilibrium composition of the mixture at 230 degrees celcias?
To determine the equilibrium composition of the mixture at 230 degrees Celsius, we first need to calculate the initial moles and concentrations of each substance in the reaction vessel. Then, we can use the equilibrium constant and the stoichiometry of the reaction to calculate the concentrations at equilibrium.
Step 1: Calculate the initial moles of each substance.
Given:
- Initial moles of phosphorus trichloride (PCl3) = 0.50 mol
- Initial moles of chlorine (Cl2) = 0.50 mol
- Volume of the reaction vessel = 5.0 L
Since the volume is given, and assuming these gases behave ideally, we can directly use moles to determine the initial concentrations.
Initial concentration of PCl3 = Moles of PCl3 / Volume of the reaction vessel = 0.50 mol / 5.0 L = 0.10 M
Initial concentration of Cl2 = Moles of Cl2 / Volume of the reaction vessel = 0.50 mol / 5.0 L = 0.10 M
Step 2: Calculate the changes in concentration.
Since we know the stoichiometry of the reaction is 1:1 for PCl3 and Cl2, the changes in concentration will be equal.
Let the change in concentration of both PCl3 and Cl2 be 'x' M.
Change in concentration of PCl3 = -x
Change in concentration of Cl2 = -x
Change in concentration of PCl5 = +x
Step 3: Calculate the equilibrium concentrations.
The equilibrium concentrations can be calculated using the equilibrium constant (Kc) expression:
Kc = [PCl5] / ([PCl3] * [Cl2])
Given: Kc = 49
Set up the expression using the initial concentrations and changes in concentration:
49 = (0.10 - x) * (0.10 - x) / (x)
Step 4: Solve for 'x' and calculate the equilibrium concentrations.
Simplify the expression:
49 = (0.01 - 0.20x + x^2) / x
Rearrange the equation:
49x = 0.01 - 0.20x + x^2
Rearrange to make it a quadratic equation:
x^2 - 49x + 0.01 - 0.20x = 0
Solve the quadratic equation:
Using the quadratic formula, we find two solutions for 'x':
x ≈ 4.81 x 10^-3 M (rounded to four decimal places)
x ≈ 4.82 x 10^-5 M (rounded to four decimal places)
Since we're interested in the positive value of 'x', we use x ≈ 4.81 x 10^-3 M.
To find the equilibrium concentrations, substitute 'x' back into the expressions for PCl3, Cl2, and PCl5:
[PCl3] at equilibrium = [PCl3] initial - x = 0.10 M - 4.81 x 10^-3 M ≈ 0.0952 M (rounded to four decimal places)
[Cl2] at equilibrium = [Cl2] initial - x = 0.10 M - 4.81 x 10^-3 M ≈ 0.0952 M (rounded to four decimal places)
[PCl5] at equilibrium = [PCl5] initial + x = 0 + 4.81 x 10^-3 M ≈ 4.81 x 10^-3 M (rounded to four decimal places)
Therefore, the equilibrium composition of the mixture at 230 degrees Celsius is:
- PCl3 concentration: 0.0952 M
- Cl2 concentration: 0.0952 M
- PCl5 concentration: 4.81 x 10^-3 M