Equilibrium Calculations 1

Lecture 15.1: Equilibrium Calculations

Introduction to Equilibrium Calculations

• The lecture consists of two segments focused on equilibrium calculations, emphasizing the importance of practice through example problems.

• Students are encouraged to pause and attempt problems independently before reviewing solutions to enhance learning.

Writing Equilibrium Expressions

• Key principles include placing products in the numerator and reactants in the denominator; solids and liquids are excluded from expressions.

• Coefficients from the balanced equation become exponents in the equilibrium expression.

Example Problem: Molarity Calculation

• A typical exam question involves calculating molarity from given moles and volume, using textMolarity = fractextmolestextliters .

• An ICE (Initial, Change, Equilibrium) table is necessary for determining equilibrium amounts based on stoichiometry.

Solving for Equilibrium Concentrations

• Given an initial amount of HI at equilibrium, students calculate molarities by dividing moles by total volume.

• The relationship between changes in concentration (x values) is established through stoichiometric coefficients leading to final concentrations.

Finalizing Equilibrium Constant Calculation

• After determining molarities for all components, these values are substituted into the equilibrium constant equation to find K_c .

• Another example illustrates how to approach a problem where a product amount is provided instead of reactant amounts.

Decomposition Reaction Analysis

• In this scenario, students learn how to adjust initial amounts based on decomposition data provided.

• The process involves subtracting decomposed amounts from initial concentrations while applying stoichiometric relationships for remaining substances.

Conclusion: Plugging Values into Equilibrium Constant Equation

• With all values calculated, students plug them into the K_c equation for final results, reinforcing understanding through practical application.

Understanding Equilibrium Constants and Reactions

Calculating Kc for Reactions

• The final value for the equilibrium constant (Kc) is discussed, emphasizing the importance of understanding different wording in problems related to chemical reactions.

• A reaction involving carbon monoxide and hydrogen gas producing methanol is analyzed, with a given Kc value at 500 K in a 10-liter vessel.

• Molarity calculations are performed by dividing mole amounts by the volume of the vessel; this step is crucial for determining if the reaction is at equilibrium.

• The calculated values indicate that the system is not at equilibrium since Q (reaction quotient) exceeds Kc, suggesting a shift towards reactants.

Analyzing Changes in Concentration

• Another question presents equilibrium amounts in molarity, simplifying calculations through an ICE table (Initial, Change, Equilibrium).

• Initial concentrations of 0.1000 M are established for both reactants; changes are represented as -x and +2x based on stoichiometry.

• The equation becomes complex due to squaring terms; however, recognizing patterns can simplify solving for x.

Simplifying Complex Equations

• Instead of multiplying out all terms directly, taking square roots simplifies calculations significantly.

• By applying square roots to both sides of the equation, it reduces complexity and makes solving easier.

Finalizing Molarity Calculations

• After simplification and multiplication adjustments lead to finding x = 0.0789; this value helps determine equilibrium concentrations effectively.

• Using x allows recalculating concentrations: subtracting from initial values gives new molarities at equilibrium.

Exploring Partial Pressures in Gaseous Reactions

• A new scenario introduces an initial pressure of 3.92 atm with no initial pressure for one product; questions arise about partial pressures at equilibrium.

• The equilibrium constant expression (Kp), which relates partial pressures instead of concentrations, is set up based on stoichiometric coefficients from the balanced equation.

Solving Quadratic Equations

• Substituting expressions into Kp leads to a quadratic formula situation requiring careful mathematical manipulation to solve for unknown pressures accurately.

Quadratic Formula and Equilibrium Calculations

Understanding the Quadratic Formula in Chemical Equilibrium

• The speaker introduces the concept of using the quadratic formula for calculations related to chemical equilibrium, noting that while approximations will be discussed later, a full quadratic approach is necessary for current problems.

• A specific example is presented where a portion of an equation must be squared. The setup involves variables such as x and constants like 1.27, indicating a need to manipulate these values mathematically.

• After squaring and rearranging terms, the resulting equation simplifies to 5.08x^2 - 20.8x + 19.5153 = 0 , which is ready for solving via a calculator or solver.

• Solving this quadratic yields x = 1.43 text atm , representing the pressure of N_2O_4 at equilibrium; further calculations reveal the pressure of NO_2 .

• The total pressure in the system is calculated by adding both pressures together: 1.43 text atm + 1.06 text atm = 2.49 text atm .

Transitioning to Kc Calculations

• A new problem involving equilibrium constant ( K_c ) calculations in a one-liter flask is introduced, simplifying mole amounts into molarities.

• The reaction setup includes initial concentrations with no starting amounts for products, leading to changes represented by variable x .

• At equilibrium, expressions are formed based on initial concentrations and changes: K_c = (x)(x)/(0.1000 - x) = 0.900.

• Rearranging leads to another quadratic equation: x^2 + 0.0900x - 0.0009 = 0. This can also be solved using a calculator.

• The solution gives x = 0.06 text M , representing concentrations of both products ( PCl_3 and Cl_2), while calculating remaining concentration for reactant ( PCl_5).