Ch. 13: Fundamental Equilibrium Concepts Dr. Namphol Sinkaset Chem 201: General Chemistry II

I. Chapter Outline Introduction Chemical Equilibria Equilibrium Constants Le Châtelier’s Principle Equilibrium Calculations

I. Introduction

II. Chemical Equilibria Equilibrium will be the focus for the next several chapters. Most reactions are reversible, meaning they can proceed in both forward and reverse directions. This means that as products build up, they will react and reform reactants. At equilibrium, the forward and backward reaction rates are equal.

II. Example Equilibrium

II. Equilibrium Concentrations Equilibrium does not mean that concentrations are all equal!! However, we can quantify concentrations at equilibrium.

III. The Reaction Quotient The reaction quotient is a number that expresses the extent of a reaction at any given point in time. For the reaction mA + nB  xC + yD: What is the value of Qc when pure reactants are mixed (at time zero)?

III. Units in Equilibrium Expressions Square brackets mean molarity concentration, e.g. [NaOH] = 1.00 M. But, in equilibrium expressions, we use activity, which is a substance’s effective concentration under specific conditions. Activity has no units, and for dilute solutions, activity is similar to molarity. Activities of pure solids and liquids = 1.

III. Sample Problem Write Qc expressions for the following equilibrium reactions. N2(g) + 3H2(g)  2NH3(g) 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

III. Value of Qc Over Time What happens to the value of Qc as a reaction proceeds and eventually reaches equilibrium?

III. The Equilibrium Constant When calculated at equilibrium, Qc becomes Kc, the equilibrium constant. This is the mathematical statement of the law of mass action: when a reaction reaches equilibrium at a given T, the reaction quotient for the reaction always has the same value.

III. Equilibrium [ ]’s vs. Kc For any reaction, the final equilibrium [ ]’s will depend on the initial [ ]’s of reactants or products. However, no matter how you set up the reaction, the value of the equilibrium constant will be the same if the temperature is the same.

III. Equilibrium [ ]’s vs. Kc

III. Physical Meaning of Kc Kc is a measure of the yield of the reaction at equilibrium, i.e. how far the reaction goes to completion. Large values of Kc mean that the equilibrium favors products. Why?* Small values of Kc mean that the equilibrium favors reactants. Why?*

III. Sample Problem Write the equilibrium constant expression for the following reaction. What does it mean if its listed Kc states “approaches infinity”? C3H8(g) + 5O2(g)  3CO2(g) 4H2O(g)

III. Using Qc With Kc The value of Qc relative to Kc tells you whether the reaction will form more products or more reactants to reach equilibrium. Qc < Kc means reaction forms products. Qc > Kc means reaction form reactants. Qc = Kc means reaction is at equilibrium.

III. Sample Problem Consider the reaction N2O4(g)  2NO2(g) with Kc = 5.85 x 10-3. If a reaction mixture contains [NO2] = 0.0255 M and [N2O4] = 0.0331 M, which way will the reaction proceed?

III. Homogeneous Equilibria As can be guessed, these systems have all reactants and products in a single solution. These can be liquid-based solutions or gas-phase solutions. In either case, we can calculate Qc or Kc as shown on the next slide.

III. Homogeneous Equilibria

III. K in Terms of Pressure Up to this point, we’ve been using concentration (activity) exclusively in the equilibrium expressions. Partial pressures are proportional to concentration via PV = nRT. Thus, for gas reactions, partial pressures can be used in place of concentrations.

III. Two Different K’s For the reaction 2SO3(g)  2SO2(g) + O2(g), we can write two equilibrium expressions.

III. Relationship Between Concentration and Pressure To be able to convert between Kc and Kp, we need a relationship between concentration and pressure.

III. Converting Between Kc and Kp

III. Converting Between Kc and Kp The Δn is the change in the number of moles of gas when going from reactants to products. When does Kp equal Kc?*

III. Sample Problem Methanol can be synthesized via the reaction CO(g) + 2H2(g)  CH3OH(g). If Kp of this reaction equals 3.8 x 10-2 at 200 °C, what’s the value of Kc?

III. Heterogeneous Equilibria If an equilibrium contains pure solids or pure liquids, they are not included in the equilibrium constant expression because their activities are equal to 1.

III. Rules for Manipulating K If the equation is reversed, the equilibrium constant is inverted.

III. Rules for Manipulating K If the equation is multiplied by a factor, the equilibrium constant is raised to the same factor.

III. Rules for Manipulating K When chemical equations are added, their equilibrium constants are multiplied together to get the overall equilibrium constant.

III. Sample Problem Predict the equilibrium constant for the first reaction given the equilibrium constants for the second and third reactions. CO2(g) + 3H2(g)  CH3OH(g) + H2O(g) K1 = ? CO(g) + H2O(g)  CO2(g) + H2(g) K2 = 1.0 x 105 CO(g) + 2H2(g)  CH3OH(g) K3 = 1.4 x 107

IV. Predicting Qualitative Changes to Equilibrium If a system is at equilibrium, what happens when it is disturbed? Le Châtelier’s Principle allow us to make qualitative predictions about changes in chemical equilibria: When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance.

IV. Three Ways to Disturb an Equilibrium We look at three ways that disturb a system at equilibrium: Adding/removing a reactant or product. Changing the volume/pressure in gaseous reactions. Changing the temperature.

IV. Adding Reactant/Product If we add or remove a reactant or product, we’re changing the [ ]’s. The equilibrium will shift in a direction that will partially consume a reactant or product that is added, or partially replace a reactant or product that has been removed.

IV. Adding Product

IV. Adding Reactant/Product

IV. Changing Volume/Pressure Via PV = nRT, pressure and volume are inversely related. For example, decreasing volume of a container increases the pressure of all gases in the container. Reducing container volume of a gaseous reaction mixture shifts the equilibrium in the direction that will, if possible, decrease the # of moles of gas.

IV. Changing Pressure

IV. Changing Temperature To determine the effect of changing the temperature, we need to know the heat of reaction, ΔHrxn. Once we know if a reaction is exo or endo, we can write “heat” as a product or reactant. We then apply Le Châtelier’s Principle in the same manner as when we considered the add product/reactant case.

IV. Changing T Changes K! In previous disturbances, K didn’t change. What did? However, changes in T change K which leads to shifts in the equilibrium.

IV. Visual Example

IV. Sample Problem For the reaction N2(g) + 3H2(g)  2NH3(g), ΔHrxn = -46.19 kJ/mole. Which way will the equilibrium shift when each of the following occurs? NH3 is removed via reaction w/ HCl. The reaction vessel is opened. The reaction vessel is cooled by 25 °C. Some Ar gas is added to the reaction vessel. A catalyst is added.

V. Equilibrium Problems 2 main categories of equilibrium problems: Finding Kc or Kp from known equilibrium concentrations or partial pressures Finding one or more equilibrium [ ]’s or partial pressures w/ a known Kc or Kp

V. Finding Kc or Kp To find equilibrium constants, we must be given information as to what [ ]’s to use in the equilibrium expression. Often, we will need to set up a table to organize the information.

V. Sample Problem An equilibrium was established for the reaction CO(g) + H2O(g)  CO2(g) + H2(g). At equilibrium, the following [ ]’s were found: [CO] = 0.180 M, [H2O] = 0.0411 M, [CO2] = 0.150 M, [H2] = 0.200 M. What’s the value of Kc?

V. Sample Problem In the reaction 2CO(g) + O2(g)  2CO2(g), it was found the concentration of O2 dropped by 0.030 mole/L. When the reaction reached equilibrium, how had the [CO] and [CO2] changed?

V. Sample Problem A student placed 0.200 mole PCl3(g) and 0.100 mole Cl2(g) into a 1.00 L container at 250 °C. After the reaction PCl3(g) + Cl2(g)  PCl5(g) came to equilibrium, 0.120 mole PCl3 was found. What’s the value of Kc for this reaction?

V. Key Points Only use equilibrium [ ]’s in the Kc/Kp expression. Initial [ ]’s should be in molarity if using Kc. Changes in [ ]’s always occur in agreement with stoichiometry in the balanced equation. When entering the “Change” row, make sure all reactants change in one direction and all products change in the opposite direction.

V. Finding Equil. [ ]’s This second type of problem is more challenging than the first. Key is interpreting information given and organizing it into a concentration table!

V. Sample Problem At 25 °C, Kc = 4.10 for the reaction CH3COOH(l) + CH3CH2OH(l)  CH3C(O)OCH2CH3(l) + H2O(l). At equilibrium, it was found that [CH3COOH] = 0.210 M, [H2O] = 0.00850 M, [CH3C(O)OCH2CH3] = 0.910 M. What’s the equilibrium [ ] of CH3CH2OH?

V. Sample Problem The reaction CO(g) + H2O(g)  CO2(g) + H2(g) has a Kc of 4.06 at 500 °C. If 0.100 mole CO and 0.100 mole H2O are placed in a 1.00 L reaction vessel at this temperature, what are the equilibrium concentrations of reactants and products?

V. Sample Problem A reaction mixture at 25 °C initially contains PI2 = 0.100 atm, PCl2 = 0.100 atm, and PICl = 0.100 atm. If the reaction I2(g) + Cl2(g)  2ICl(g) has a Kp of 81.9 at 25 °C, find the equilibrium partial pressures of each species.

V. Sample Problem At a certain temperature, Kc = 4.50 for the reaction N2O4(g)  2NO2(g). If 0.300 mole N2O4 is placed in a 2.00 L flask at this temperature, what will be the equilibrium [ ]’s of both gases?

V. Simplifications When Kc or Kp is very small or very big, we can treat x as being insignificant. As such, we discard x in any addition/subtraction operation. Why?* We must verify our assumption by comparing the value we obtain for x to the number it was discarded from. Assumption is valid if value of x is less than 5% of the number it would have been subtracted from.

V. Sample Problem In air at 25 °C and 1 atm, [N2] = 0.033 M and [O2] = 0.00810 M. The reaction between molecular nitrogen and molecular oxygen to form nitrogen monoxide has Kc = 4.8 x 10-31 at 25 °C. What is the natural [NO] in the atmosphere?

V. Sample Problem What are the concentrations of all dissolved species at equilibrium of a 0.15 M HClO solution given that its Kc in water is 2.9 × 10-8?