Chemical Equilibrium: Calculating Kc, Kp, and Equilibrium Concentrations/Pressures

Q1. At a temperature of 1000 K, a sealed 5.0 L vessel is filled with 1.00 mol SO2 and 1.00 mol O2. At equilibrium, the amount of SO3 is found to be 0.85 mol. Determine Kp and Kc for this reaction at 1000 K. 2 SO2(g) + O2(g) ↔ 2 SO3(g)

Background

Topic: Chemical Equilibrium, Equilibrium Constants (Kc and Kp)

This question tests your ability to use initial and equilibrium amounts to calculate equilibrium concentrations and then determine the equilibrium constants Kc (using concentrations) and Kp (using partial pressures) for a gaseous reaction.

Key Terms and Formulas

• Kc: Equilibrium constant in terms of concentration (mol/L)

• Kp: Equilibrium constant in terms of partial pressure (atm)

• Relationship: $K_p = K_c (RT)^{\Delta n}$

• R = 0.0821 L·atm/(mol·K)

• \( \Delta n = \) change in moles of gas (products - reactants)

Step-by-Step Guidance

1. Write the balanced equation and set up an ICE (Initial, Change, Equilibrium) table for the reaction, using moles and then converting to concentrations (mol/L).

2. Calculate the equilibrium concentrations of SO2, O2, and SO3 using the initial moles, the change (based on stoichiometry), and the final amount of SO3 given.

3. Use the equilibrium concentrations to write the expression for Kc: $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$

4. Calculate \( \Delta n \) for the reaction (moles of gaseous products minus moles of gaseous reactants).

5. Set up the formula to convert Kc to Kp: $K_p = K_c (RT)^{\Delta n}$

Try solving on your own before revealing the answer!

Final Answer:

Kc = 0.122 Kp = 0.0041

We calculated equilibrium concentrations using the ICE table and then applied the formulas for Kc and Kp with the correct value of R and Δn.

Q2. A mass of 7.24 grams of IBr is placed in a 225 mL container and heated to 500 K. Some IBr decomposes to form I2(g) and Br2(g). At equilibrium the partial pressure of Br2 is found to be 3.01 atm. Determine Kp and Kc for this reaction at 500 K. 2 IBr(g) ↔ I2(g) + Br2(g)

Background

Topic: Chemical Equilibrium, Equilibrium Constants (Kc and Kp)

This question tests your ability to use mass, volume, and equilibrium partial pressure to determine equilibrium concentrations and calculate Kc and Kp for a decomposition reaction.

Key Terms and Formulas

• Molar mass of IBr (needed to convert grams to moles)

• PV = nRT (to relate pressure, volume, and moles)

• Kc and Kp expressions for the reaction:

• $K_c = \frac{[I_2][Br_2]}{[IBr]^2}$

• $K_p = \frac{P_{I_2} P_{Br_2}}{(P_{IBr})^2}$

• R = 0.0821 L·atm/(mol·K)

• Relationship: $K_p = K_c (RT)^{\Delta n}$

Step-by-Step Guidance

1. Calculate the initial moles of IBr using its molar mass and the given mass.

2. Set up an ICE table for the reaction, using moles and then converting to concentrations (mol/L) and partial pressures (atm).

3. Use the equilibrium partial pressure of Br2 to determine the change in moles and the equilibrium amounts of all species.

4. Write the expressions for Kc and Kp using the equilibrium values.

5. Set up the formula to convert between Kc and Kp if needed.

Try solving on your own before revealing the answer!

Final Answer:

Kc = 0.053 Kp = 0.041

We used the ICE table, the ideal gas law, and the equilibrium partial pressure to find the equilibrium constants.

Q3a. Consider the heterogeneous reaction at 1000 K: CO2(g) + C(s) ↔ 2 CO(g). A vessel initially contains 0.464 atm of CO2(g) and excess solid C. At equilibrium, the total vapor pressure in the vessel is 0.746 atm. Determine the values of Kc and Kp for the reaction at this temperature.

Background

Topic: Chemical Equilibrium, Heterogeneous Equilibria, Equilibrium Constants

This question tests your ability to analyze a heterogeneous equilibrium (solid and gases), use partial pressures, and calculate Kc and Kp for the system.

Key Terms and Formulas

• Heterogeneous equilibrium: solids are not included in K expressions

• Kp expression: $K_p = \frac{(P_{CO})^2}{P_{CO_2}}$

• Kc expression: $K_c = \frac{[CO]^2}{[CO_2]}$

• Relationship: $K_p = K_c (RT)^{\Delta n}$

• R = 0.0821 L·atm/(mol·K)

Step-by-Step Guidance

1. Set up an ICE table using the initial and equilibrium partial pressures, noting that the total pressure at equilibrium is given.

2. Express the changes in partial pressures for CO2 and CO based on the stoichiometry of the reaction.

3. Use the total equilibrium pressure to solve for the equilibrium partial pressures of CO2 and CO.

4. Write the Kp expression and substitute the equilibrium partial pressures.

5. Set up the formula to convert Kp to Kc using the relationship above.

Try solving on your own before revealing the answer!

Final Answer:

Kp = 0.398 Kc = 0.0065

We used the total pressure to find the equilibrium partial pressures and then calculated the equilibrium constants.

Q3b. If the initial mass of C(s) is 100 mg and the vessel is 2.00 L, find the mass of C that remains at equilibrium.

Background

Topic: Chemical Equilibrium, Limiting Reactant, Stoichiometry

This question tests your ability to relate the change in moles of solid reactant to the equilibrium shift and calculate the remaining mass of solid carbon.

Key Terms and Formulas

• Stoichiometry of the reaction: 1 mol C consumed per 2 mol CO formed

• Molar mass of C = 12.01 g/mol

• Change in moles of C = related to change in moles of CO2 and CO

Step-by-Step Guidance

1. Calculate the initial moles of C using its mass and molar mass.

2. Determine the change in moles of C based on the equilibrium shift (from the previous part).

3. Subtract the moles of C consumed from the initial moles to find the remaining moles.

4. Convert the remaining moles of C back to mass (grams or mg).

Try solving on your own before revealing the answer!

Final Answer:

Mass of C remaining = 88 mg

We calculated the moles of C consumed using stoichiometry and subtracted from the initial mass.

Q4. Initially, 0.500 mol·L-1 NOCl(g) is placed in a vessel. For the reaction, Kc = 1.6 × 10-5 at 35°C. Find the equilibrium concentrations and partial pressures of all species. 2 NOCl(g) ↔ 2 NO(g) + Cl2(g)

Background

Topic: Chemical Equilibrium, Small Equilibrium Constant

This question tests your ability to set up and solve an equilibrium problem with a small Kc, which often allows for simplifying assumptions.

Key Terms and Formulas

• Kc expression: $K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2}$

• ICE table for concentrations

• Partial pressure: $P = nRT/V$ or $P = CRT$

• R = 0.0821 L·atm/(mol·K)

Step-by-Step Guidance

1. Set up an ICE table for the reaction, using the initial concentration of NOCl and assuming no products initially.

2. Let x be the change in concentration of NOCl that reacts; express equilibrium concentrations in terms of x.

3. Write the Kc expression in terms of x and set it equal to the given value.

4. Because Kc is very small, consider if the change x is negligible compared to the initial concentration (the small-x approximation).

5. Set up the expressions for equilibrium partial pressures using $P = CRT$ for each species.

Try solving on your own before revealing the answer!

Final Answer:

[NOCl] = 0.500 mol/L [NO] = 0.0057 mol/L [Cl2] = 0.0029 mol/L

Partial pressures can be found by multiplying each concentration by RT.

Q5. Initially, 0.490 atm SO2(g) and 0.245 atm O2(g) are placed in a vessel. For the reaction, Kp = 8.24 × 104 at 700 K. Find the equilibrium concentrations and partial pressures of all species. 2 SO2(g) + O2(g) ↔ 2 SO3(g)

Background

Topic: Chemical Equilibrium, Large Equilibrium Constant

This question tests your ability to solve for equilibrium partial pressures and concentrations when Kp is very large, indicating the reaction proceeds nearly to completion.

Key Terms and Formulas

• Kp expression: $K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2 P_{O_2}}$

• ICE table for partial pressures

• Relationship between partial pressure and concentration: $C = \frac{P}{RT}$

• R = 0.0821 L·atm/(mol·K)

Step-by-Step Guidance

1. Set up an ICE table for the reaction, using the initial partial pressures of SO2 and O2.

2. Let x be the change in pressure as the reaction proceeds toward equilibrium; express equilibrium pressures in terms of x.

3. Write the Kp expression in terms of x and set it equal to the given value.

4. Because Kp is very large, consider if the reaction goes nearly to completion (i.e., limiting reactant approach).

5. Set up the expressions for equilibrium concentrations using $C = \frac{P}{RT}$ for each species.

Try solving on your own before revealing the answer!

Final Answer:

PSO2 = 0.000 atm PO2 = 0.000 atm PSO3 = 0.735 atm

Almost all reactants are converted to product due to the large Kp value.