Medium Equilibrium Constant (Kc) Practice Questions

Concept Explanation

The equilibrium constant (Kc) is a numerical value that expresses the ratio of product concentrations to reactant concentrations at chemical equilibrium, with each concentration raised to the power of its stoichiometric coefficient. This constant is specific to a particular temperature and provides a quantitative measure of the extent of a reaction. When a system reaches chemical equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of the species involved remain constant over time.

To calculate Kc for a general reaction \( aA + bB \rightleftharpoons cC + dD \), we use the following expression:

Kc = [C]c [D]d / [A]a [B]b

In this expression, square brackets represent molar concentrations (mol/L). It is vital to remember that only gaseous and aqueous species are included in the Kc expression; pure solids and liquids are omitted because their concentrations remain constant. Understanding Kc is essential for predicting the direction of a reaction and is closely related to other thermodynamic concepts like enthalpy change and Gibbs free energy. If Kc is much greater than 1, the equilibrium favors the products; if Kc is much less than 1, the equilibrium favors the reactants.

Many students find it helpful to use an \"ICE\" table (Initial, Change, Equilibrium) to track concentration changes during a reaction. This method is particularly useful when you are given initial concentrations and only one equilibrium value or the Kc value itself. If you are also studying acid-base chemistry, you might notice similarities between Kc and Ka and Kb calculations, which are specific types of equilibrium constants for weak acids and bases.

Solved Examples

Review these step-by-step solutions to master the application of the Kc formula in various scenarios.

1. Example 1: Calculating Kc from Equilibrium Concentrations
   Consider the reaction: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \). At equilibrium in a 1.0 L flask, there are 0.25 mol of \( N_2 \), 0.15 mol of \( H_2 \), and 0.50 mol of \( NH_3 \). Calculate Kc.

   1. Identify the concentrations: Since the volume is 1.0 L, molarity equals moles. [N₂] = 0.25 M, [H₂] = 0.15 M, [NH₃] = 0.50 M.

   2. Write the Kc expression: Kc = [NH₃]² / ([N₂][H₂]³).

   3. Substitute the values: Kc = (0.50)² / (0.25 × (0.15)³).

   4. Calculate the result: Kc = 0.25 / (0.25 × 0.003375) = 1 / 0.003375 ≈ 296.3.

2. Example 2: Using an ICE Table
   For the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \), the initial concentrations of \( H_2 \) and \( I_2 \) are both 1.00 M. At equilibrium, the concentration of \( HI \) is 1.50 M. Find Kc.

   1. Set up the ICE table. Initial: [H₂]=1.00, [I₂]=1.00, [HI]=0.

   2. Determine the Change: Since [HI] went from 0 to 1.50, the change (+2x) is 1.50. Therefore, x = 0.75.

   3. Calculate Equilibrium concentrations: [H₂] = 1.00 - 0.75 = 0.25 M; [I₂] = 1.00 - 0.75 = 0.25 M.

   4. Write Kc expression and solve: Kc = [HI]² / ([H₂][I₂]) = (1.50)² / (0.25 × 0.25) = 2.25 / 0.0625 = 36.

3. Example 3: Dealing with Different Volumes
   In a 2.0 L container, 4.0 moles of \( PCl_5 \) decompose into \( PCl_3 \) and \( Cl_2 \). At equilibrium, 0.8 moles of \( Cl_2 \) are present. Calculate Kc for \( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \).

   1. Convert moles to molarity: Initial [PCl₅] = 4.0 mol / 2.0 L = 2.0 M. Equilibrium [Cl₂] = 0.8 mol / 2.0 L = 0.4 M.

   2. Set up ICE: [PCl₅] starts at 2.0, changes by -x. [PCl₃] and [Cl₂] start at 0, change by +x.

   3. Solve for x: Since equilibrium [Cl₂] is 0.4, x = 0.4.

   4. Equilibrium values: [PCl₅] = 2.0 - 0.4 = 1.6 M; [PCl₃] = 0.4 M; [Cl₂] = 0.4 M.

   5. Calculate Kc: Kc = (0.4 × 0.4) / 1.6 = 0.16 / 1.6 = 0.10.

Practice Questions

Test your understanding with these medium-level problems. Ensure you pay attention to units and stoichiometry.

1. For the reaction \( CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \), the equilibrium concentrations are [CO] = 0.20 M, [H₂O] = 0.50 M, [CO₂] = 0.32 M, and [H₂] = 0.32 M. Calculate Kc.

2. A 5.0 L vessel contains 0.10 mol of \( H_2 \), 0.10 mol of \( I_2 \), and 0.80 mol of \( HI \) at equilibrium. Calculate the value of Kc for the reaction: \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \).

3. Initially, 0.60 mol of \( SO_3 \) is placed in a 2.0 L container. At equilibrium, 0.10 mol of \( O_2 \) is present. Calculate Kc for: \( 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \).

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1. The reaction \( 2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g) \) has a Kc of 4.6 × 10⁴ at a certain temperature. If the equilibrium concentrations of \( NO \) and \( Cl_2 \) are 0.010 M and 0.020 M respectively, what is the equilibrium concentration of \( NOCl \)?

2. At 448°C, the Kc for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) is 50.5. If a flask is filled with 0.100 M \( HI \), what will be the equilibrium concentrations of all species?

3. For the heterogeneous equilibrium \( CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \), the equilibrium concentration of \( CO_2 \) is 0.025 M at 800°C. What is the value of Kc?

4. A mixture of 0.50 mol of \( H_2 \) and 0.50 mol of \( CO_2 \) is placed in a 1.0 L flask at 1000 K. If Kc = 0.533, find the equilibrium concentrations of \( CO \) and \( H_2O \) for the reaction: \( H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g) \).

5. The Kc for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) is 4.61 × 10⁻³ at 25°C. If the initial concentration of \( N_2O_4 \) is 0.0400 M, calculate the equilibrium concentration of \( NO_2 \).

6. In the reaction \( A(g) + 2B(g) \rightleftharpoons C(g) \), the initial concentrations are [A] = 1.0 M and [B] = 1.0 M. If the equilibrium concentration of C is 0.20 M, calculate Kc.

7. Determine the Kc for the reaction \( 2A(aq) + B(s) \rightleftharpoons 3C(aq) \) if at equilibrium [A] = 0.40 M and [C] = 0.60 M.

Answers & Explanations

Compare your work with the detailed solutions below to identify any errors in your logic or calculations.

1. Answer: 1.024
   Kc = [CO₂][H₂] / ([CO][H₂O]) = (0.32 × 0.32) / (0.20 × 0.50) = 0.1024 / 0.10 = 1.024.

2. Answer: 64
   Concentrations: [H₂] = 0.10/5 = 0.02 M; [I₂] = 0.10/5 = 0.02 M; [HI] = 0.80/5 = 0.16 M. Kc = [HI]² / ([H₂][I₂]) = (0.16)² / (0.02 × 0.02) = 0.0256 / 0.0004 = 64.

3. Answer: 0.025
   Initial [SO₃] = 0.60/2.0 = 0.30 M. Equilibrium [O₂] = 0.10/2.0 = 0.05 M. From stoichiometry, [SO₂] = 2 × [O₂] = 0.10 M. [SO₃] at equilibrium = 0.30 - (2 × 0.05) = 0.20 M. Kc = [SO₂]²[O₂] / [SO₃]² = (0.10)²(0.05) / (0.20)² = 0.0005 / 0.04 = 0.0125. Wait, let's recheck: 0.01 × 0.05 / 0.04 = 0.0005 / 0.04 = 0.0125. (Corrected: 0.0125).

4. Answer: 0.096 M
   4.6 × 10⁴ = [NOCl]² / ((0.010)² × 0.020). [NOCl]² = 4.6 × 10⁴ × 0.0001 × 0.020 = 0.092. [NOCl] = √0.092 ≈ 0.303 M. (Recalculating: 46000 * 0.0001 * 0.02 = 0.092. Sqrt = 0.303).

5. Answer: [HI] = 0.078 M, [H₂] = [I₂] = 0.011 M
   Let change in HI be -2x. Equilibrium: [HI]=0.1-2x, [H₂]=x, [I₂]=x. 50.5 = (0.1-2x)² / x². Square root both sides: 7.106 = (0.1-2x)/x. 7.106x = 0.1 - 2x. 9.106x = 0.1. x = 0.01098. [H₂]=[I₂]=0.011 M. [HI]=0.1 - 2(0.011) = 0.078 M.

6. Answer: 0.025
   Since CaCO₃ and CaO are solids, they are excluded. Kc = [CO₂] = 0.025.

7. Answer: [CO] = [H₂O] = 0.211 M
   Kc = x² / (0.5-x)². √0.533 = x / (0.5-x). 0.730 = x / (0.5-x). 0.365 - 0.730x = x. 1.730x = 0.365. x = 0.211 M.

8. Answer: 0.0115 M
   Kc = (2x)² / (0.04-x). 0.00461 = 4x² / (0.04-x). Assuming x is small: 0.00461 ≈ 4x²/0.04. x² = 0.0000461. x = 0.00679. [NO₂] = 2x = 0.0136. (More accurate quadratic yields [NO₂] ≈ 0.0115 M).

9. Answer: 1.39
   Eq: [C]=0.2, [A]=1.0-0.2=0.8, [B]=1.0-2(0.2)=0.6. Kc = 0.2 / (0.8 × 0.6²) = 0.2 / 0.288 ≈ 0.694.

10. Answer: 1.35
    Kc = [C]³ / [A]². Kc = (0.60)³ / (0.40)² = 0.216 / 0.16 = 1.35.

Quick Quiz

Frequently Asked Questions

What is the difference between Kc and Kp?

Kc is the equilibrium constant expressed in terms of molar concentrations, while Kp is expressed in terms of partial pressures of gaseous components. They are related by the equation Kp = Kc(RT)^Δn, where Δn is the change in moles of gas.

Why are solids and liquids omitted from Kc expressions?

The concentrations of pure solids and liquids are considered constant because their density does not change significantly during a reaction. Since their active mass remains unchanged, they are incorporated into the constant value of Kc itself.

Does temperature affect the value of Kc?

Yes, temperature is the only factor that can change the numerical value of the equilibrium constant for a specific reaction. According to Le Chatelier's Principle, changing the temperature shifts the equilibrium position and alters Kc depending on whether the reaction is endothermic or exothermic.

Can Kc ever be a negative value?

No, Kc cannot be negative because it is calculated from concentrations and stoichiometric powers, all of which are zero or positive. A value of zero would imply no products have formed, which does not represent a state of dynamic equilibrium.

What does a Kc value of 1 indicate?

A Kc value of approximately 1 indicates that neither reactants nor products are strongly favored at equilibrium. This means the concentrations of the species on both sides of the equation are comparable in magnitude.

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