a 1.00-mol sample of phosphorus pentachloride placed in 10.0-L reaction flask and allowed to come to equilibrium at 250 degrees celsius; PCL5(g) = PCL3(g) Cl2(g) if the amount of chlorine in the equilibrium mixture is0.470 mol, calculate (a) the equilibrium concentration of each gas, (b) Kc and (c) the fraction of PCL5 dissociated
.............PCl5 ==> PCl3 + Cl2
initial......1.0 mol....0.....0
change.......-x.........x.....x
equil......................0.470
Therefore, at equilibrium PCl3 must be the same or 0.470 mol and PCl5 must be 1.0-0.470.
Cl2 = 0.470/10L = ?
PCl3 = 0.470/10L = ?
PCl5 = (1-0.470)/10L = ?
Kc = (PCl3)(Cl2)/(PCl5).
Substitute and solve for Kc.
fraction = moles Cl2/initial moles PCl5
To solve this problem, we need to start by writing the balanced chemical equation for the reaction:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
(a) To calculate the equilibrium concentration of each gas, we need to know the initial moles of PCl₅ and the change in moles for each species at equilibrium.
Given:
Initial moles of PCl₅ = 1.00 mol
Moles of Cl₂ at equilibrium = 0.470 mol
Since 1 mole of PCl₅ produces 1 mole of Cl₂ and 1 mole of PCl₃, the moles of PCl₃ at equilibrium will also be 0.470 mol.
To calculate the equilibrium concentration (in mol/L), we divide the moles by the volume (10.0 L):
Concentration of PCl₅ = (1.00 mol) / (10.0 L) = 0.100 M
Concentration of PCl₃ = (0.470 mol) / (10.0 L) = 0.047 M
Concentration of Cl₂ = (0.470 mol) / (10.0 L) = 0.047 M
(b) To find the equilibrium constant, Kc, we need to use the concentrations of the reactants and products at equilibrium:
Kc = [PCl₃][Cl₂] / [PCl₅]
Kc = (0.047 M)(0.047 M) / (0.100 M) = 0.022 M²
(c) The fraction of PCl₅ dissociated can be calculated by comparing the change in moles of PCl₅ to the initial moles:
Change in moles of PCl₅ = initial moles - moles at equilibrium
= 1.00 mol - 0.100 mol = 0.900 mol
Fraction of PCl₅ dissociated = change in moles / initial moles
= 0.900 mol / 1.00 mol
= 0.90 or 90% (as a decimal)
Therefore, (a) the equilibrium concentration of each gas is PCl₅ = 0.100 M, PCl₃ = 0.047 M, and Cl₂ = 0.047 M. (b) The equilibrium constant, Kc, is 0.022 M². (c) The fraction of PCl₅ dissociated is 90%.
To solve this problem, we need to use the principles of chemical equilibrium and the equation Kc = ([PCL3] * [Cl2]) / [PCL5]. Let's break down each part of the problem:
(a) The equilibrium concentration of each gas:
We are given the amount of chlorine in the equilibrium mixture (0.470 mol). Since PCl5 dissociates into PCl3 and Cl2 in a 1:1 ratio, the equilibrium concentration of each gas will also be 0.470 mol. Therefore, [PCL3] = 0.470 mol/L, [Cl2] = 0.470 mol/L, and [PCL5] = 1.00 mol/L.
(b) Kc Calculation:
Using the formula Kc = ([PCL3] * [Cl2]) / [PCL5], we can substitute the given values to calculate Kc. Kc = (0.470 mol/L * 0.470 mol/L) / 1.00 mol/L = 0.22075.
Therefore, the equilibrium constant Kc is approximately 0.22075.
(c) Fraction of PCL5 dissociated:
To find the fraction of PCL5 dissociated, we need to compare the initial and equilibrium concentrations. The initial concentration of PCL5 is 1.00 mol/L, and the equilibrium concentration is determined by (1 - x), where x is the fraction of PCL5 dissociated.
Since we know the concentration of chlorine at equilibrium is 0.470 mol/L, and PCL5 dissociates into PCL3 and Cl2 in a 1:1 ratio, we can assume that 0.470 mol/L from the original 1.00 mol/L comes from the dissociated PCL5.
Setting up the equation:
(1 - x) = 0.47
1 - x = 0.47
x = 0.53
Therefore, the fraction of PCL5 dissociated is approximately 0.53.
In summary:
(a) The equilibrium concentration of each gas: [PCL3] = 0.470 mol/L, [Cl2] = 0.470 mol/L, [PCL5] = 1.00 mol/L.
(b) The equilibrium constant Kc: Kc = 0.22075.
(c) The fraction of PCL5 dissociated: 0.53.