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Section 13.4: Equilibrium Calculations

Overview of Equilibrium Calculations

  • The focus is on specific calculations surrounding equilibrium constants and reactions.

  • Importance of recognizing different types of equilibrium questions and utilizing appropriate methods for calculations.

Calculating Equilibrium Constants

  • Equilibrium constants ($K$) are derived from the concentrations or pressures of reactants and products at equilibrium.

  • K Expression: The general form of the equilibrium constant expression is:
    K=rac[products][reactants]K = rac{[products]}{[reactants]}

  • Direct Calculation: If equilibrium constants are given, plug into the $K$ expression directly.

  • Use of ICE Tables: If less information is available, an ICE (Initial, Change, Equilibrium) table is necessary to calculate equilibrium concentrations.

ICE Tables

  • Definition: An ICE table represents changes in concentrations as a reaction approaches equilibrium.

  • Components:

    • Initial concentrations of reactants and products.

    • Changes in concentrations based on stoichiometry.

    • Equilibrium concentrations as the sum of initial concentrations and changes.

Example of ICE Table Usage

  • Reaction example: Ammonia decomposition.

    • Initial concentrations may be provided. For example, if only ammonia is present, then:

    • Initial

      • [NH₃] = $x$

      • [N₂] = 0

      • [H₂] = 0

    • Stoichiometry details lead to:

    • Change:

      • Consume NH₃: -2x

      • Produce N₂: +x

      • Produce H₂: +3x

    • Equilibrium concentrations are calculated as:

    • [NH₃] = $x - 2x = -x$

    • [N₂] = $0 + x = x$

    • [H₂] = $0 + 3x = 3x$

Example Calculation

  • Iodine (I₂) and Iodide (I⁻) Reaction:

  • Given:

    • Initial [I₂] = [I⁻] = 1imes10−31 imes 10^{-3} M

    • Equilibrium [I₂] = 6.61imes10−46.61 imes 10^{-4} M

  • Construct the ICE Table:

    • Change for each species:

    • For I₂: 1imes10−3−x1 imes 10^{-3} - x

    • For I⁻: 1imes10−3−x1 imes 10^{-3} - x

    • For I₃: 0+x0 + x

  • Based on equilibrium concentration:

    • 6.61imes10−4=1imes10−3−x6.61 imes 10^{-4} = 1 imes 10^{-3} - x

    • Thus, solve for x=3.39imes10−4x = 3.39 imes 10^{-4} M.

  • Final Equilibrium Concentrations:

    • [I₂] = 6.61imes10−46.61 imes 10^{-4} M

    • [I⁻] = 6.61imes10−46.61 imes 10^{-4} M

    • [I₃] = 3.39imes10−43.39 imes 10^{-4} M

  • Calculate Equilibrium Constant $K_c$: Theory:

    • Kc=rac[I3][I2][I−]K_c = rac{[I₃]}{[I₂][I⁻]}

Incomplete Concentration Information

  • Example Scenario: Shifting focus from concentration calculations to solving for missing data with given $K_c$.

    • Reaction of Nitrogen and Oxygen yielding NO (Nitrogen Oxide).

    • Given:

    • Kc=4.1imes10−4K_c = 4.1 imes 10^{-4}

    • Equilibrium [N₂] = 0.036 M

    • Equilibrium [O₂] = 0.0089 M

  • Goal: Calculate equilibrium [NO].

    • Write the K expression for this reaction:

    • Kc=rac[NO]2[N2][O2]K_c = rac{[NO]^2}{[N₂][O₂]}

    • Rearrage to solve for [NO]:

    • [NO]2=Kcimes[N2][O2][NO]^2 = K_c imes [N₂][O₂]

    • Plugging values yields:

      • [NO]=3.6imes10−4[NO] = 3.6 imes 10^{-4} M

Calculation Challenges and Approximations

  • Most real-world problems involve complete initial concentrations and a known equilibrium constant, often asking for equilibrium concentrations.

  • Four-step Problem-Solving Method:

    1. Identify direction of reaction (left to right or vice versa).

    2. Develop the ICE table.

    3. Calculate concentration changes to find equilibrium concentrations.

    4. Confirm results are consistent with the $K_c$ obtained.

Example of Step-By-Step Calculation: Phosphorus Pentachloride Decomposition

  • Given:

    • Kc=0.0211K_c = 0.0211 for the reaction:

    • PCl5<br>ightleftharpoonsPCl3+Cl2PCl₅ <br>ightleftharpoons PCl₃ + Cl₂

    • Initial [PCl₅] = 1 M, and no other species present initially.

  • Develop ICE Table:

    • [PCl₅] = 1, [PCl₃] = 0, [Cl₂] = 0

    • Change based on stoichiometry:

    • [PCl₅]: -x,

    • [PCl₃]: +x,

    • [Cl₂]: +x

  • Substitute into Kc expression:

    • Kc=rac[PCl3][Cl2][PCl5]K_c = rac{[PCl₃][Cl₂]}{[PCl₅]}

    • Kc=racx21−x=0.0211K_c = rac{x^2}{1-x} = 0.0211

  • Rearranging yields a quadratic; solve to find x.

  • Maintain the sign conditions for concentrations; select the physically realistic root.

  • Substitute x back to find equilibrium concentrations:

    • [PCl₅] = 0.879 M, [PCl₃] = 0.135 M, [Cl₂] = 0.135 M

Special Approximations for Simplified Calculations

  • When the change in concentration (x) is much less than the initial concentration, simplifications can be applied, especially when $K$ is small (e.g., K<1K < 1).

  • Valid approximation leads to simplified algebraic relationships, reducing the quadratic problem to linear growth.

  • Example of Using Approximations: Equilibrium concentrations of HCN reaction:

    • Starting concentrations of HCN = 0.15 M, Kc = 10−1010^{-10} indicates minimal reaction shift towards products.

  • Set up the calculation for equilibrium concentrations, validating assumptions that x is negligible compared to the initial HCN concentration.

  • Apply derived relationships efficiently to realize true concentrations of products.

Conclusion

  • Comprehensive understanding of multiple approaches and methodologies for equilibrium calculations enhances the capacity to tackle varying problem types effectively.

  • Recognition of when to apply shortcuts or approximations ensures a more streamlined problem-solving process in real-world scenarios.