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![Page 1: 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.](https://reader036.fdocuments.in/reader036/viewer/2022081508/56649d2b5503460f94a00468/html5/thumbnails/1.jpg)
2-Day AP/Pre-AP Science ConferenceCorpus 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
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Important Web SiteParticulate Level Simulations (Supported by the
National Science Foundation)http://genchem1.chem.okstate.edu/CCLI/Startup.htmlLoginFree access to simulations and guided inquiry
activities.
Gas LawsShifting Reactions (Equilibrium)Chemical Kinetics
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Experimental BasisCharacteristics of a hypothetical chemical reaction?
Investigate the reaction: R + GB
Investigate the reaction: B + RG
Reversible reaction
Irreversible reaction
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Experimental BasisHow can concentration changes affect a
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) PredictExperiment #6: Remove product RG (to 0.5 M)) Predict
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Experimental Basis
How can concentration changes affect a chemical reaction?
Summary of Data
Exp. Stress
Change in [reactants]
Change in [products] Reaction Shifts
#2 Add R Conc decrease
Conc. Increases
Left to right
#3
#4….
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Experimental Basis
How can concentration changes affect a 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.
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Experimental BasisWhat is the relationship between reactant and product
concentrations at the end of a chemical reaction?
Exp. Initial Conditions Final Conditions
R BG RG B R BG RG B
#1 2.0 2.0 0 0
#2 1.0 1.0 0 0
#3 1.5 1.5 0 0
#4 0 0 1.0 1.0
#6 1.0 0.6 1.2 0.8
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Experimental BasisWhat is the relationship between reactant and product
concentrations at the end of a chemical reaction?
Exp. Initial Conditions Final Conditions
R BG RG B R BG RG B
#1 2.0 2.0 0 0 .64 .64 1.36 1.36
#2 1.0 1.0 0 0
#3 1.5 1.5 0 0
#4 0 0 1.0 1.0
#6 1.0 0.6 1.2 0.8
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Experimental BasisWhat is the relationship between reactant and product
concentrations at the end of a chemical reaction?
Exp. Initial Conditions Final Conditions
R BG RG B R BG RG B
#1 2.0 2.0 0 0 .64 .64 1.36 1.36
#2 1.0 1.0 0 0 .32 .32 0.68 0.68
#3 1.5 1.5 0 0
#4 0 0 1.0 1.0
#6 1.0 0.6 1.2 0.8
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Experimental BasisWhat is the relationship between reactant and product
concentrations at the end of a chemical reaction?
Exp. Initial Conditions Final Conditions
R BG RG B R BG RG B
#1 2.0 2.0 0 0 .64 .64 1.36 1.36
#2 1.0 1.0 0 0 .32 .32 0.68 0.68
#3 1.5 1.5 0 0 .48 .48 1.02 1.02
#4 0 0 1.0 1.0
#6 1.0 0.6 1.2 0.8
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Experimental BasisWhat is the relationship between reactant and product
concentrations at the end of a chemical reaction?
Exp. Initial Conditions Final Conditions
R BG RG B R BG RG B
#1 2.0 2.0 0 0 .64 .64 1.36 1.36
#2 1.0 1.0 0 0 .32 .32 0.68 0.68
#3 1.5 1.5 0 0 .48 .48 1.02 1.02
#4 0 0 1.0 1.0 .32 .32 0.68 0.68
#6 1.0 0.6 1.2 0.8
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Experimental BasisWhat is the relationship between reactant and product
concentrations at the end of a chemical reaction?
Exp. Initial Conditions Final Conditions
R BG RG B R BG RG B
#1 2.0 2.0 0 0 .64 .64 1.36 1.36
#2 1.0 1.0 0 0 .32 .32 0.68 0.68
#3 1.5 1.5 0 0 .48 .48 1.02 1.02
#4 0 0 1.0 1.0 .32 .32 0.68 0.68
#6 1.0 0.6 1.2 0.8 .79 .39 1.41 1.01
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Experimental Basis
What is the relationship between reactant and product concentrations at the end of a 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.
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Experimental Basis
Additional Investigations:Try some different reactions: R + G RG
BG B + GRB R + B
Change in temperature;Change the volume of the container;K, T and ∆HK and Q
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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.
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Introduction to Equilibrium
aA(g) + bB(g) cC(g) + dD(g)
Kp = PcC · Pd
D/PaA · Pb
B
KC = [C]c · [D]d/ [A]a · [B]b
Kp = KC·(RT)∆n
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/La) What is the mol fraction of CO(g) in the equilibrium mixture?b) Using the equilibrium concentrations given above, calculate the value of
KC, the equilibrium constant for the reaction.
c) Determine KP in terms KC 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 CO2(g). Calculate the value of KC at this lower temperature.
e) 0.50 mol of H2 is mixed with 0.50 mol of CO2 in a 3.0 liter container at 2000 K. Calculate the CO(g) at equilibrium.
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/La) 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
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
b) Using the equilibrium concentrations given above, calculate the value of KC, the equilibrium constant for the reaction.
KC = [H2O] · [CO]/ [H2] · [CO2]= [0.55] · [0.55]/ [0.20] · [0.30]
= 5.0
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
c) Determine KP in terms KC of for the system.
Kp = KC·(RT)∆n
∆n = mol product gases - mol of reactant gases∆n = 1 + 1 - (1 + 1) = 0
Kp = KC
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [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 CO2(g). Calculate the value of KC at this lower temperature.
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
initial 0.20 0.30 0.55 0.55change - 0.17 equilibrium
0.30 · 0.55 M = 0.17 M0.17 M CO reacts
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
initial 0.20 0.30 0.55 0.55change + 0.17 + 0.17 - 0.17 - 0.17 equilibrium 0.37 0.47 0.38 0.38
KC = [H2O] · [CO]/ [H2] · [CO2]= [0.38] · [0.38]/ [0.37] · [0.47]
= 0.83
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2000K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
e) 0.50 mol of H2 is mixed with 0.50 mol of CO2 in a 3.0 liter container at 2000 K. Calculate the CO(g) at equilibrium.
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
e) 0.50 mol of H2 is mixed with 0.50 mol of CO2 in a 3.0 liter container at 2000 K. Calculate the CO(g) at equilibrium.
[H2] = [CO2] = 0.50 mol/3.0 L = 0.17 M
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
initial 0.17 0.17 0 0changeequilibrium
Q = [H2O] · [CO]/ [H2] · [CO2]= [0] · [0]/ [0.17] · [0.17]
= 0
So Q < Kc (5.0)So the reaction proceeds from left to right to establish equilibrium
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1995 H2(g) + CO2(g) H2O(g) + CO(g)
initial 0.17 0.17 0 0change -x -x +x +xequilibrium 0.17 - x 0.17 - x 0 - x 0 - x
Kc = [H2O] · [CO]/ [H2] · [CO2]5.0 = [x] · [x]/ [0.17 - x] · [0.17 - x]
5.0 = x2/(0.17 - x)2
2.236 = x/(0.17 - x)x = 0.12 M = [CO]
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1998 C(s) + H2O(g) CO(g) + H2(g) Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(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.
• Additional H2(g) is added to the equilibrium mixture at constant volume.
• The temperature of the equilibrium mixture is increased at constant volume.
• The volume of the container is decreased at constant temperature.
(d) The graphite pellets are pulverized.
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1998 C(s) + H2O(g) CO(g) + H2(g) Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(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.
• Additional H2(g) is added to the equilibrium mixture at constant volume.
• The temperature of the equilibrium mixture is increased at constant volume.
• The volume of the container is decreased at constant temperature.(d) The graphite pellets are pulverized.
Kp = PCO · PH2/PH2O
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1998 C(s) + H2O(g) CO(g) + H2(g) Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(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.
• Additional H2(g) is added to the equilibrium mixture at constant volume.
Kp = PCO · PH2/PH2O
Adding H2(g) causes Q to be greater than Kp.
The reaction must shift in the direction to reduce Q, right to left. The moles of CO(g) will decrease.
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1998 C(s) + H2O(g) CO(g) + H2(g) Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(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.
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1998 C(s) + H2O(g) CO(g) + H2(g) Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(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.
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1998 C(s) + H2O(g) CO(g) + H2(g) Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(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.
Kp = PCO · PH2/PH2O
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.