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CHEM 163 Chapter 17. Spring 2009 Instructor: Alissa Agnello [email protected]. What affects reaction rates?. Chemical Equilibrium. Many reactions can go forward AND backwards. If the opposite reaction can occur,. then this reaction is a reversible reaction. What is equilibrium?. - PowerPoint PPT Presentation
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What affects reaction rates?
Chemical EquilibriumMany reactions can go forward AND backwards
3
(g) 2NO(g)O(g)N 22
If the opposite reaction can occur,
(g)O (g) N(g) 2NO 22
then this reaction is a reversible reaction
What is equilibrium?
• …in terms of reaction rates?
• …in terms of reactant and product concentrations?
• Has the forward reaction stopped?• Has the reverse reaction stopped?
Chemical Equilibrium
• Reactions continue at equal (but opposite) rates• No further changes in concentrations of reactants
or products occurs5
Chemical Equilibrium
If fwd and rev reactions are both elementary steps, how would we write their rate laws?
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(g) 2NO(g)ON 242
ratefwd = kfwd [N2O4]eq raterev = krev [NO2]2eq
kfwd [N2O4]eq krev [NO2]2eq=
42
22
rev
fwd
ON
NO
k
kK
Equilibrium Constants (K)
7
Krev
fwd
k
k
eq
eq
][reactants
[products]
Equilibrium Constants: Small K
Small K value:– Greater concentration of reactants or products?– Reaction favors reactants
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Keq
eq
][reactants
[products]
Equilibrium Constants: Large K
Large K value: – Greater concentration of reactants or products?– Reaction favors products
9
Keq
eq
][reactants
[products]
Equilibrium ConstantsWill each of the following favor reactants or products?• Reaction with Kc = 2.9 x 10-12
• Reaction with Kc = 0.001 x 105
10
Reaction Quotient (Q)• K derived from rates• Q derived from concentrations
At a given temperature, a system will always return to the same [product] : [reactant] ratio
Q
What if K = Q?
Q reactants
products
Reaction Quotients
12
D C B A dcba reactants products
coefficients
Q reactants
products ba
dc
BA
DC
(g)O(g)2SO 22 (g)2SO3
2-minute practice
Write a reaction quotient expression for the following:
Compare your answer with your neighbors!
(g)gg HI2)(I)(H 22
Q for Overall reaction
Many reactions take place in multiple steps.
• Add together the steps to get overall reaction• Multiply K for each step to get overall K • Multiply Q for each step to get overall Q
Other situations…In a reversible reaction:
revfwd Q
1Q
When the coefficients are multiplied by a common factor (n):
nQQ
(g)gg HI2)(I)(H 22
22
2
IH
HIQ fwd
2
22
HI
IHQ rev
(g)gg HI)(I)(H 221
221
2
121
22 IH
HI
21
22
2
IH
HI
fwdQ
If solids or liquids are present…
• heterogeneous equilibrium• Q and K only related to concentration that
change (gases)
(g)(s)(g)(s) H OFe OH Fe 2432
Q 42
42
OH
H
Rearrange the ideal gas law, so that concentration is isolated on one side.
What is the relationship between pressure and concentration?
2-minute exercise
Qp: using partial pressures(g)gg HI2)(I)(H 22
22
2
c IH
HIQ
22 IH
2HIQPP
Pp
concentration
pressure
2HI 2
2HI
V
n
2
2HI
RT
P
2H V
n2H
RT
P2H
2I V
n2I
RT
P2I
Shortcut for relating Qp to Qc
What is the ∆n for the reaction?
(g)gg HI2)(I)(H 22 nreactants = 2
nproducts = 2∆n = 0 Qp = Qc
If ∆n ≠ 0 Qp = Qc (RT)∆n
Kp = Kc
Kp = Kc (RT)∆n
Is a reaction at equilibrium?Compare Q and K!
• Q < K:– Too much reactant – Equilibrium “shifts” towards products
• Q > K: – Too much product– Equilibrium “shifts” towards reactants
• Q = K: – At equilibrium
Equilibrium Calculations
• If equilibrium concentrations and/or equilibrium constant is known:– Write Q expression, plug in concentrations, solve– At equilibrium Q = K
• Solving for K when concentrations not given: – Write Q expression– Set up table including:• Initial concentrations (or pressures)• Change• Equilibrium concentrations (or pressures)
Mix together graphite and carbon dioxide (P = 0.458 atm) to create carbon monoxide.
Pressure (atm) CO2 (g) + C (graphite) ↔ 2CO (g)
CO2 (g) + C (graphite) ↔ 2CO (g)
Once equilibrium is reached, the pressure in the vessel (from CO2 and CO) is 0.757 atm.
Initial
Change
Equilibrium
0.458 0
- x + 2x
0.458 - x + 2x
0.458 – x + 2x = 0.757Ptotal =
Solving for “x”To solve for x, you may need to use the quadratic formula.
Set up your equation: a*x2 + b*x + c = 0 a
acbbx
2
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You will end up with two values. Which value is right?Remember: • K can’t be negative• We can’t have a higher concentration of reactant
than our initial concentration.
Steps for Solving1. Write balanced equation2. Set up table 3. Solve for x by…
1. Setting up Q expression 2. Setting total pressures equal to final pressure
4. Solve for equilibrium concentrations or pressures(using x)
5. To check: plug in solved concentration or pressures into Q expression and compare to known K value
3-minute PracticeConsider the reaction:
0.45 mole H2S is placed in a 3.0 L container. Make a table for this situation.
Kc = 9.30 x 10-8
at 700 °C H2S (g) ↔ H2 (g) + S2 (g)2 2
Calculate the equilibrium concentration of H2 (g) at 700 °C
Le Châtelier’s Principle• At equilibrium, concentrations of substances
do not change. • If a stress is put on the reaction at equilibrium,
the equilibrium will shift to relieve the stress.
What changes count as stress?• Concentration – Adding or removing reactant or product
• Volume (Pressure)• Temperature
26
Effect of Concentration Changes2NO2 (g) NO (g) + O2 (g)
27
What is the effect on the concentration of each substance?
• Add NO2?
• Add NO?
• Add O2?
• Remove NO2?
• Remove NO?
• Remove O2?
2
Calculations: adding/removing substances
• Make a table!
Pressure (atm) CO2 (g) + C (graphite) ↔ 2CO (g)
Original Equil.
Disturbance
New Initial
0.159 0.598
+ 0.1
0.698
Change
New Equil.
0.159
+ x - 2x
0.159 + x 0.698 - 2x
Predict direction of shift
Effect of Volume Changes• Each mole of gas exerts a certain pressure• Decrease the volume…– Increase the pressure…– Shifts to side of reaction with FEWER moles of gas
29
2NO2 (g) 2NO (g) + O2 (g)
• What happens if we increase the volume?• Shifts to side with MORE moles
• What happens if we increase the pressure?• Shifts to side with FEWER moles
Effect of Volume Changes
30
With volume changes and concentration changes, Equilibrium shifts to relieve new stress…in order to return to equilibrium
Effect of Temperature Changes• If T is increased, equilibrium shifts to remove the heat• If T is decreased, equilibrium shifts to create heat• Kc changes!– Unlike for concentration and volume changes
31
A (g) B (g) + C (g) Endothermic:
Heat +
Increase the temperature?• Shifts to the products (to use up heat)
• Kc increases
Effect of T on K
van’t Hoff Equation:
• Exothermic (∆H° < 0): increasing T decreases Kc
• Endothermic (∆H° > 0): increasing T increases Kc
121
2 11 ln
TTR
H
K
K orxn R = 8.314 J/mol∙K
3-minute Practice
Consider the exothermic reaction between nitrogen gas and hydrogen gas, creating ammonia gas (NH3).
Write a balanced equation for this reaction.
Which direction will the equilibrium shift if: • T is increased?• Ammonia is removed?• Volume of the container is decreased?• A catalyst is used?• Hydrogen is added?
33
Chapter 17 Homework
Due: Tuesday, 4/14
#18, 23, 29, 34, 45, 55, 63, 76, 77, 91, 114