To tackle this problem, we need to analyze the equilibrium reaction between phosphorus pentachloride (PCl5) and its products, phosphorus trichloride (PCl3) and chlorine gas (Cl2). The first step is to set up the reaction and establish the changes that occur at equilibrium. Let's break this down systematically.
Understanding the Reaction
The reaction we are considering is:
- PCl5(g) ↔ PCl3(g) + Cl2(g)
From the stoichiometry of the reaction, we can see that 1 mole of PCl5 dissociates into 1 mole of PCl3 and 1 mole of Cl2. Thus, for every mole of PCl5 that dissociates, we produce 1 mole of PCl3 and 1 mole of Cl2.
Initial Moles and Changes at Equilibrium
We start with 3 moles of PCl5 and no products. If we let x be the degree of dissociation of PCl5, then at equilibrium, we can express the moles as follows:
- Moles of PCl5 = 3 - x
- Moles of PCl3 = x
- Moles of Cl2 = x
Total Moles at Equilibrium
The total number of moles at equilibrium would then be:
Total moles = (3 - x) + x + x = 3 + x
Applying Ideal Gas Law
Using the ideal gas law, we can relate the total pressure to the total number of moles. The total pressure (P) is given as 2.05 atm in a 100-litre vessel, at a temperature of 227°C (which we convert to Kelvin as follows: 227 + 273 = 500 K).
Using the ideal gas law in the form of P = (nRT)/V, we can rearrange it to find the number of moles:
Substituting in the values: P = 2.05 atm, V = 100 L, R = 0.0821 L·atm/(K·mol), and T = 500 K:
n = (2.05 atm × 100 L) / (0.0821 L·atm/(K·mol) × 500 K) = 5.01 moles
Relating Moles to Degree of Dissociation
Setting the total moles at equilibrium equal to our calculated value, we have:
3 + x = 5.01
Solving for x gives:
x = 5.01 - 3 = 2.01 moles
Calculating Degree of Dissociation
The degree of dissociation (α) is defined as the fraction of the original PCl5 that has dissociated:
α = x / initial moles of PCl5 = 2.01 / 3 = 0.670 or 67.0%
Finding Kp
To calculate the equilibrium constant Kp, we use the expression:
Kp = (P_PCl3 * P_Cl2) / P_PCl5
First, we need to find the partial pressures of each component at equilibrium:
- Partial pressure of PCl5 = (3 - x) / total moles × P_total
- Partial pressure of PCl3 = x / total moles × P_total
- Partial pressure of Cl2 = x / total moles × P_total
Substituting the values:
- P_PCl5 = (3 - 2.01) / 5.01 × 2.05 atm = 0.197 atm
- P_PCl3 = 2.01 / 5.01 × 2.05 atm = 0.797 atm
- P_Cl2 = 2.01 / 5.01 × 2.05 atm = 0.797 atm
Now, substituting these values into the Kp expression:
Kp = (0.797 atm * 0.797 atm) / 0.197 atm = 3.21 atm
Summary of Results
Thus, we have:
- Degree of dissociation of PCl5 (α) = 67.0%
- Equilibrium constant (Kp) = 3.21 atm
This analysis shows how we can approach equilibrium problems using the ideal gas law, stoichiometry, and the concept of partial pressures. Each step builds on the previous ones, leading us to a comprehensive understanding of the equilibrium state of this reaction.
