To tackle the problem of calculating the degree of dissociation of PCl5, we first need to understand the dissociation reaction and how to apply the concepts of equilibrium and stoichiometry. The dissociation of phosphorus pentachloride (PCl5) can be represented as follows:
Dissociation Reaction
The reaction can be expressed as:
- PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
From this reaction, we can see that one mole of PCl5 dissociates into one mole of PCl3 and one mole of Cl2. Therefore, for every mole of PCl5 that dissociates, we produce one mole of PCl3 and one mole of Cl2.
Initial Conditions
Let's analyze the first scenario where 0.03 moles of PCl5 are brought to equilibrium in a volume of 2.09 liters at 500 K and 1 atm. We can set up an ICE (Initial, Change, Equilibrium) table to help us visualize the changes in concentration.
ICE Table
| Species |
Initial (mol) |
Change (mol) |
Equilibrium (mol) |
| PCl5 |
0.03 |
-α |
0.03 - α |
| PCl3 |
0 |
+α |
α |
| Cl2 |
0 |
+α |
α |
Here, α represents the degree of dissociation of PCl5. At equilibrium, the total number of moles will be:
- Total moles = (0.03 - α) + α + α = 0.03 + α
Calculating Concentrations
Next, we can calculate the equilibrium concentrations. The total volume is 2.09 L, so the equilibrium concentrations will be:
- [PCl5] = (0.03 - α) / 2.09
- [PCl3] = α / 2.09
- [Cl2] = α / 2.09
Using the ideal gas law and the equilibrium constant expression, we can derive the equilibrium constant (Kp) for the reaction:
- Kp = (PCl3)(PCl2) / (PCl5)
At equilibrium, we can express the partial pressures in terms of concentrations. Since we are at 1 atm, we can assume that the total pressure is equal to the sum of the partial pressures:
Using the ideal gas law, we can find the total moles at equilibrium and set up the equation to solve for α. However, since we know the total moles and the initial moles, we can directly calculate α.
Finding Degree of Dissociation
At equilibrium, the total moles can also be expressed as:
- 0.03 + α = (P_total * V) / (R * T)
Substituting the known values (R = 0.0821 L·atm/(K·mol), T = 500 K, V = 2.09 L, and P_total = 1 atm), we can solve for α. After calculating, we find that:
Second Scenario
Now, let's consider the second scenario where 0.2 moles of PCl5 are brought to equilibrium in a 3 L container at the same temperature. We can set up a similar ICE table:
ICE Table for Second Scenario
| Species |
Initial (mol) |
Change (mol) |
Equilibrium (mol) |
| PCl5 |
0.2 |
-β |
0.2 - β |
| PCl3 |
0 |
+β |
β |
| Cl2 |
0 |
+β |
β |
At equilibrium, the total moles will be:
- Total moles = (0.2 - β) + β + β = 0.2 + β
Using the same approach as before, we can find the equilibrium concentrations:
- [PCl5] = (0.2 - β) / 3
- [PCl3] = β / 3
- [Cl2] = β / 3
Now, we can set up the equilibrium expression and solve for β. After performing the calculations, we find that:
Final Results
Thus, the degree of dissociation of PCl5 in the first scenario is approximately 0.01, and in the second scenario, it is approximately 0.1. This illustrates how the degree of dissociation can vary based on the initial concentration and volume of the system.
