2-Day AP/Pre-AP Science Conference Corpus Christi Omni-Marina Hotel January 20 & 21, 2006 Introduction to Equilibrium Concepts John I. Gelder Former Chief Reader AP Chemistry Department of Chemistry Oklahoma State University

Important Web Site Particulate Level Simulations (Supported by the National Science Foundation) html Login Free access to simulations and guided inquiry activities. Gas Laws Shifting Reactions (Equilibrium) Chemical Kinetics

Experimental Basis Characteristics of a hypothetical chemical reaction?chemical reaction Investigate the reaction: R + GB  Investigate the reaction: B + RG  Reversible reaction Irreversible reaction

Experimental Basis How can concentration changes affect a chemical reaction?chemical reaction Experiment #1: Experiment #2: Add reactant R (to 2.0 M) Experiment #3: Add reactant BG (to 2.0 M) Experiment #4: Remove reactants R and BG (to 1.0 and 0.8 M) Experiment #5: Add product RG (to 1.5 M) Predict Experiment #6: Remove product RG (to 0.5 M)) Predict

Experimental Basis How can concentration changes affect a chemical reaction?chemical reaction Summary of Data Exp.Stress Change in [reactants] Change in [products] Reaction Shifts #2Add RConc decrease Conc. Increases Left to right #3 #4 ….

Experimental Basis How can concentration changes affect a chemical reaction?chemical reaction Review the summary of your experimental observations and write a statement(s) that generalizes how stressing a reaction by adding or removing a reactant of product shifts the chemical reaction.

Experimental Basis What is the relationship between reactant and product concentrations at the end of a chemical reaction?chemical reaction Exp.Initial ConditionsFinal Conditions RBGRGBRBGRGB # # # # #

Experimental Basis What is the relationship between reactant and product concentrations at the end of a chemical reaction?chemical reaction Exp.Initial ConditionsFinal Conditions RBGRGBRBGRGB # # # # #

Experimental Basis What is the relationship between reactant and product concentrations at the end of a chemical reaction?chemical reaction Exp.Initial ConditionsFinal Conditions RBGRGBRBGRGB # # # # #

Experimental Basis What is the relationship between reactant and product concentrations at the end of a chemical reaction?chemical reaction Exp.Initial ConditionsFinal Conditions RBGRGBRBGRGB # # # # #

Experimental Basis What is the relationship between reactant and product concentrations at the end of a chemical reaction?chemical reaction Exp.Initial ConditionsFinal Conditions RBGRGBRBGRGB # # # # #

Experimental Basis What is the relationship between reactant and product concentrations at the end of a chemical reaction?chemical reaction Exp.Initial ConditionsFinal Conditions RBGRGBRBGRGB # # # # #

Experimental Basis What is the relationship between reactant and product concentrations at the end of a chemical reaction?chemical reaction Challenge students to arrive at an algebraic equation that relates the concentrations of the reactants and products in the reaction. Hints…. Use the results to predict the final concentrations for a new initial set of conditions.

Experimental Basis Additional Investigations: Try some different reactions: R + G  RG BG  B + G RB  R + B Change in temperature; Change the volume of the container; K, T and ∆H K and Q

Introduction to Equilibrium K, the equilibrium constant, measures the extent of a chemical reaction. Large K (greater than 1) reaction will proceed towards products when the initial conditions have all reactants and products at 1 M or 1 atm. Small K (less than 1) reaction will proceed towards reactants when the initial conditions have all reactants and products at 1 M or 1 atm.

Introduction to Equilibrium aA(g) + bB(g) cC(g) + dD(g) K p = P c C · P d D /P a A · P b B K C = [C] c · [D] d / [A] a · [B] b K p = K C ·(RT) ∆n

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) When H 2 (g) is mixed with CO 2 (g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H 2 ] = 0.20 mol/L [CO 2 ] = 0.30 mol/L [H 2 O] = [CO] = 0.55 mol/L a)What is the mol fraction of CO(g) in the equilibrium mixture? b)Using the equilibrium concentrations given above, calculate the value of K C, the equilibrium constant for the reaction. c)Determine K P in terms K C of for the system. d)When the system is cooled from 2000 K to a lower temperature, 30.0% of the CO(g) is converted back to CO 2 (g). Calculate the value of K C at this lower temperature. e)0.50 mol of H 2 is mixed with 0.50 mol of CO 2 in a 3.0 liter container at 2000 K. Calculate the CO(g) at equilibrium.

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) When H 2 (g) is mixed with CO 2 (g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H 2 ] = 0.20 mol/L [CO 2 ] = 0.30 mol/L [H 2 O] = [CO] = 0.55 mol/L a)What is the mol fraction of CO(g) in the equilibrium mixture? Mol fraction CO = mol CO/total mol = 0.55/1.60 = 0.34

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) When H 2 (g) is mixed with CO 2 (g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H 2 ] = 0.20 mol/L [CO 2 ] = 0.30 mol/L [H 2 O] = [CO] = 0.55 mol/L b)Using the equilibrium concentrations given above, calculate the value of K C, the equilibrium constant for the reaction. K C = [H 2 O] · [CO]/ [H 2 ] · [CO 2 ] = [0.55] · [0.55]/ [0.20] · [0.30] = 5.0

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) When H 2 (g) is mixed with CO 2 (g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H 2 ] = 0.20 mol/L [CO 2 ] = 0.30 mol/L [H 2 O] = [CO] = 0.55 mol/L c)Determine K P in terms K C of for the system. K p = K C ·(RT) ∆n ∆n = mol product gases - mol of reactant gases ∆n = (1 + 1) = 0 K p = K C

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) When H 2 (g) is mixed with CO 2 (g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H 2 ] = 0.20 mol/L [CO 2 ] = 0.30 mol/L [H 2 O] = [CO] = 0.55 mol/L d)When the system is cooled from 2000 K to a lower temperature, 30.0% of the CO(g) is converted back to CO 2 (g). Calculate the value of K C at this lower temperature.

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) initial change equilibrium 0.30 · 0.55 M = 0.17 M 0.17 M CO reacts

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) initial change equilibrium K C = [H 2 O] · [CO]/ [H 2 ] · [CO 2 ] = [0.38] · [0.38]/ [0.37] · [0.47] = 0.83

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) When H 2 (g) is mixed with CO 2 (g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H 2 ] = 0.20 mol/L [CO 2 ] = 0.30 mol/L [H 2 O] = [CO] = 0.55 mol/L e)0.50 mol of H 2 is mixed with 0.50 mol of CO 2 in a 3.0 liter container at 2000 K. Calculate the CO(g) at equilibrium.

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) e)0.50 mol of H 2 is mixed with 0.50 mol of CO 2 in a 3.0 liter container at 2000 K. Calculate the CO(g) at equilibrium. [H 2 ] = [CO 2 ] = 0.50 mol/3.0 L = 0.17 M

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) initial change equilibrium Q = [H 2 O] · [CO]/ [H 2 ] · [CO 2 ] = [0] · [0]/ [0.17] · [0.17] = 0 So Q < K c (5.0) So the reaction proceeds from left to right to establish equilibrium

1995 H 2 (g) + CO 2 (g) H 2 O(g) + CO(g) initial change -x -x +x +x equilibrium x x 0 - x 0 - x K c = [H 2 O] · [CO]/ [H 2 ] · [CO 2 ] 5.0 = [x] · [x]/ [ x] · [ x] 5.0 = x 2 /( x) = x/( x) x = 0.12 M = [CO]

1998 C(s) + H 2 O(g) CO(g) + H 2 (g)  H º = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H 2 O(g), CO(g), and H 2 (g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (a)Additional H 2 (g) is added to the equilibrium mixture at constant volume. (b)The temperature of the equilibrium mixture is increased at constant volume. (c)The volume of the container is decreased at constant temperature. (d)The graphite pellets are pulverized.

1998 C(s) + H 2 O(g) CO(g) + H 2 (g)  H º = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H 2 O(g), CO(g), and H 2 (g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (a)Additional H 2 (g) is added to the equilibrium mixture at constant volume. (b)The temperature of the equilibrium mixture is increased at constant volume. (c)The volume of the container is decreased at constant temperature. (d)The graphite pellets are pulverized. K p = P CO · P H 2 /P H 2 O

1998 C(s) + H 2 O(g) CO(g) + H 2 (g)  H º = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H 2 O(g), CO(g), and H 2 (g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (a)Additional H 2 (g) is added to the equilibrium mixture at constant volume. K p = P CO · P H 2 /P H 2 O Adding H 2 (g) causes Q to be greater than K p. The reaction must shift in the direction to reduce Q, right to left. The moles of CO(g) will decrease.

1998 C(s) + H 2 O(g) CO(g) + H 2 (g)  H º = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H 2 O(g), CO(g), and H 2 (g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (b)The temperature of the equilibrium mixture is increased at constant volume. The reaction is endothermic. Increasing the temperature corresponds to adding heat to the reactants side. The reaction will shift from left to right to relieve the stress. The moles of CO(g) will increase.

1998 C(s) + H 2 O(g) CO(g) + H 2 (g)  H º = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H 2 O(g), CO(g), and H 2 (g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (c )The volume of the container is decreased at constant temperature. Decreasing the volume of the container, increases the pressure inside, to relieve the pressure the reaction will proceed in the direction to decrease the pressure (in the direcction of the fewest moles of gas) the reaction will proceed from right to left. The moles of CO(g) will decrease.

1998 C(s) + H 2 O(g) CO(g) + H 2 (g)  H º = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H 2 O(g), CO(g), and H 2 (g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (d)The graphite pellets are pulverized. K p = P CO · P H 2 /P H 2 O Since C(s) does not appear in the equilibrium expression pulverizing the graphite has no effect of the moles of CO(g). So the moles of CO(g) remain the same.