Question

The two common chlorides of phosphorus, \(\mathrm{PCl}_{3}\) and \(\mathrm{PCl}_{5},\) both important in the production of other phosphorus compounds, coexist in equilibrium through the reaction $$ \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{g}) $$ At \(250^{\circ} \mathrm{C},\) an equilibrium mixture in a \(2.50 \mathrm{L}\) flask contains \(0.105 \mathrm{g} \mathrm{PCl}_{5}, 0.220 \mathrm{g} \mathrm{PCl}_{3},\) and \(2.12 \mathrm{g} \mathrm{Cl}_{2}\) What are the values of (a) \(K_{c}\) and (b) \(K_{\mathrm{p}}\) for this reaction at \(250^{\circ} \mathrm{C} ?\)

Short answer

The equilibrium constants are \(K_{c} = 26.2\) and \(K_{p} = 0.108\) at \(250^{\circ} \mathrm{C}\).

Step-by-step solution

Calculate the molar masses of each compound

Begin by finding the molar masses of compounds using the atomic masses on the Periodic Table. The molar mass of PCl3 is \(30.97 + 3(35.45) = 137.32 \: \text{g/mol}\), of PCl5 is \(30.97 + 5(35.45) = 208.22 \: \text{g/mol}\), and of Cl2 is \(2(35.45) = 70.90 \: \text{g/mol}\).

Determine the equilibrium concentrations in mol/L

Next, convert grams to moles using the molar masses from Step 1. Then, compute the concentrations in molarity (mol/L) by dividing moles by the volume of the flask, 2.50 L. The concentration of PCl3 is \(0.220 \: \text{g} \times \frac{1 \: \text{mol}}{137.32 \: \text{g}} \div 2.50 \: \text{L} = 0.000644 \: \text{M}\), the concentration of PCl5 is \(0.105 \: \text{g} \times \frac{1 \: \text{mol}}{208.22 \: \text{g}} \div 2.50 \: \text{L} = 0.000201 \: \text{M}\), and concentration of Cl2 is \(2.12 \: \text{g} \times \frac{1 \: \text{mol}}{70.90 \: \text{g}} \div 2.50 \: \text{L} = 0.0120 \: \text{M}\).

Compute the equilibrium constant (Kc)

Use the balanced chemical equation to compute Kc. In this case, Kc = \([PCl5]\) / (\([PCl3][Cl2]\), and substituting the concentrations found gives Kc = \(0.000201/ (0.000644 \times 0.0120) = 26.2\).

Determine the partial pressure of each gas

Next, calculate the partial pressures from the ideal gas law. The total molar volume of an ideal gas at 250 degrees Celsius (523.15 K) and 1 atmosphere is about 24.27 L. Thus, the partial pressure of PCl3 is \(0.000644 \: \text{mol/L} \times 24.27 \: \text{L} = 0.0156\: \text{atm}\), of PCl5 is \(0.000201 \: \text{mol/L} \times 24.27 \: \text{L} = 0.00488 \: \text{atm}\), and of Cl2 is \(0.0120 \: \text{mol/L} \times 24.27 \: \text{L} = 0.291 \: \text{atm}\).

Calculate the equilibrium constant (Kp)

Finally, use the balanced chemical equation to compute Kp. In this case, Kp = \(\frac{(P_{PCl5})}{(P_{PCl3} \times P_{Cl2})}\). Substituting the partial pressures found in the previous step, we get Kp = \(\frac{0.00488}{(0.0156 \times 0.291)} = 0.108\).

Key concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), is crucial in understanding chemical equilibria. It helps predict the position of equilibrium in a reversible chemical reaction based on concentration values. For the reaction between \(\text{PCl}_3\), \(\text{Cl}_2\), and \(\text{PCl}_5\), the expression for \(K_c\) is derived from the balanced chemical equation:

• \(\text{PCl}_3\, (g) + \text{Cl}_2\, (g) \rightleftharpoons \text{PCl}_5\, (g)\)
• \(K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]}\)

The bracket symbols \([]\) denote the molar concentrations of the gaseous substances. Calculating \(K_c\) involves determining these concentrations at equilibrium by dividing the amount (in moles) by the total volume of the system.

Equilibrium Constant (Kp)

The equilibrium constant \(K_p\) is similar to \(K_c\) but it is used for gaseous reactions and is expressed in terms of partial pressures. The expression for \(K_p\) typically resembles that of \(K_c\) but involves pressures instead:

• \(K_p = \frac{P_{\text{PCl}_5}}{(P_{\text{PCl}_3} \times P_{\text{Cl}_2})}\)

The pressure of each gas component is denoted by \(P\) followed by the chemical formula. To convert \(K_c\) to \(K_p\), we can use the ideal gas law along with the formula to account for any changes in moles of gases between the reactants and products. The relationship is generally:

• \(K_p = K_c(RT)^{\Delta n}\)

where \(R\) is the ideal gas constant, \(T\) is temperature in Kelvin, and \(\Delta n\) is the change in moles of gas during the reaction.

Molar Mass

Molar mass plays a key role in chemical calculations as it allows conversion between mass and moles. For the exercise, we calculated the molar masses of phosphorus and chlorine chlorides using atomic weights:

• \(\text{PCl}_3\): 137.32 g/mol
• \(\text{PCl}_5\): 208.22 g/mol
• \(\text{Cl}_2\): 70.90 g/mol

These values enable us to determine the number of moles of each compound in the equilibrium system. This involves dividing the given mass (in grams) by the respective molar mass. Moles can then be used to find concentrations or partial pressures, which are crucial for calculating equilibrium constants.

Partial Pressure

Partial pressure reflects the pressure that an individual gas exerts in a mixture. It's an important concept in gas equilibria like the one in this exercise. Using the concentration of gases and the ideal gas law ((\(PV = nRT\))), we can calculate each gas's partial pressure in the equilibrium state. Here’s how it works:

• Convert the gas concentration (from molarity) to molarity using the molar volume of a gas at a given temperature. At 250 °C, 1 mole of gas approximately occupies 24.27 L under 1 atm.
• Calculate partial pressure: \(P = \, \text{Concentration} \times \text{Molar Volume}\)
• This gives:
  • \(\text{PCl}_3\): 0.0156 atm
  • \(\text{PCl}_5\): 0.00488 atm
  • \(\text{Cl}_2\): 0.291 atm

These pressures are used to compute \(K_p\). Partial pressures are vital for reactions involving gases since these influence the direction and extent of the reactions.