Equilibrium Powerpoint Part 1
Transcript of Equilibrium Powerpoint Part 1
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Reversible Reactions & Equilibrium
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In a closed system, reactions are reversible!
The conversion of reactants to products
(forward reaction) & the conversion of products to reactants (reverse reaction) occur simultaneously.
Reversible Reactions:
A + B AB
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Equilibrium:
When RATE of forward reaction is the same as the RATE of reverse reaction EQUILIBRIUM
Only REVERSIBLE reactions can achieve equilibrium
Must be closed system & constant temp
A + B AB
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Analogy:Hockey players on a team & equilibrium:
Players on bench rotate with players on ice
Player leave bench at same rate as players come back on bench
Same players on the team (never any new or different players on the team)
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Ex.Heat + 2HI H2 + I2
Initially: only HI At Equi: decomp of HI equals rate of
synthesis of HI
Time
Con
cent
ratio
n
HI
H2 + I2
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Ex.2SO2 + O2 2SO3 ∆H = -197kJ
Initially: only SO2 & O2
At Equi: syn of SO3 equals rate of syn of SO2 & O2
Time
Con
cent
ratio
n
SO3O2
SO2
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Equilibrium Constant – Keq
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Keq: expresses the concentrations of reactants &
products at equilibrium
constant for every reversible reactions at equilibrium at a given temp & pressure
aA + bB cC + dD
Keq = [C]c[D]d
[A]a[B]b
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Given:
2NO2 N2O4
Calculate Keq, if the conc of N2O4 at equi is 0.00140 M & the conc of NO2 at equi is 0.0172M
Keq = [N2O4][NO2]2
Keq = [0.0014][0.0172]2
Keq = 4.73
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Given:
2NO2 N2O4
Calculate Keq, if the conc of N2O4 at equi is 0.00452 M & the conc of NO2 at equi is 0.031M
Keq = [N2O4][NO2]2
Keq = [0.00452][0.031]2
Keq = 4.70 Same Keq!!!
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The value of Keq tells you exactly what is happening with a reaction
If: Keq = 1
At equi, conc of products & reactants are the same
Keq > 1 At equi, greater conc of products, than reactants
Keq < 1 At equi, greater conc of reactants, than products
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Keq is the same for a given reaction at equilibrium at the same temp no matter what the initial conc were.
Keq does change with temp
Do not include pure solids and pure liquids in the Keq expression because their concentrations vary little.
Everything else must be included
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Calculating
Keq
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Consider:H2 (g) + I 2 (g) 2HI (g)
Find the Keq if the conc of H2 is 0.46, I2 is 0.39 M & HI is 3.0 M at equilibrium
Keq = [HI]2__ [H2][I2]
Keq = __[3]2__ [0.46][0.39]
Keq = 50
Conc of products greater at equi
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Consider:PCl5 (g) PCl3 (g) + Cl2 (g)
Find the Keq if the initial conc of PCl5 is 0.70M & the final conc of Cl2 is 0.15M
Keq expression for equi conc only!!!
Use an ICE table
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PCl5 PCl3 Cl2
I
C
E
0.7
0.55
0 0
+ 0.15+ 0.15- 0.15
0.15 0.15
Keq = [PCl3][Cl2]__ [PCl5]
Keq = [0.15][0.15]__ [0.55]
Keq = 0.041
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Consider:2NH3 (g) 3H2 (g) + N2 (g)
Initially a 5.0L flask contains 0.2M NH3 & 0.08M N2. After equi [NH3] is 0.156M.
NH3 H2 N2
I
C
E
0.2 0.080
0.156
- 0.044 + 0.066 + 0.022
0.066 0.102
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Keq = [H2]3[N2]__ [NH3]2
Keq = [0.066]3[0.102]__ [0.156]2
Keq = 0.0012
At equi, conc of reactants greater
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Consider:H2 (g) + I2 (g) 2HI (g)
Find the [HI] at equi, if [H2] at equi is 0.50M and the [I2] at equi is 0.50M and Keq = 50.
H2 I2 HI
I
C
E
----- ----------
0.5
----- ----- -----
0.5 ?
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Keq = [HI]2
[H2] [I2]
50 = [x]2__ [0.5][0.5]
[HI] = 3.54 M at equi
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Consider:H2 (g) + I2 (g) 2HI (g)
What are the equi conc of each substance if a flask initially contains only 0.5M of H2 and 0.5M of I2? Keq = 50.
H2 I2 HI
I
C
E
0.5 00.5
0.5 - x
- x - x + 2x
0.5 - x 2x
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Keq = [HI]2
[H2] [I2]
50 = [2x]2__ [0.5-x][0.5-x]
At Equi:
[H2] = 0.11M[I2] = 0.11M[HI] = 0.78M
7.07 = [2x]_ [0.5-x]
7.07(0.5 – x) = 2x
3.54 - 7.07x = 2x
x = 0.39
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Consider:H2 (g) + I2 (g) 2HI (g)
What are the equi conc of each substance if a 0.5L flask initially contains 2 moles H2 and 2 moles of I2?
H2 I2 HI
I
C
E
4 04
4 - x
- x - x + 2x
4 - x 2x
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Keq = [HI]2
[H2] [I2]
50 = [2x]2__ [4-x][4-x]
At Equi:
[H2] = 0.88M[I2] = 0.88M[HI] = 6.24M
7.07 = [2x]_ [4-x]
7.07(4 – x) = 2x
28. 28 - 7.07x = 2x
x = 3.12
If not same, multiply out then
use quadratic formula
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Le Chatelier’s Principle
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Le Chatelier’s Principle:
• If a closed system at equilibrium is subject to a change (stress), processes will occur to counteract the stress.
Factors that will affect Equi:
1. Concentration
2. Temperature
3. Pressure & Volume
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Given:N2O4 (g) + 59.0kJ 2NO2 (g)
1. Concentration:• Adding more reactant pushes reaction in
direction of product• Removing reactant pushes reaction in direction
of reactants• Adding more product pushes reaction in
direction of reactants• Removing product pushes reaction in direction
of products
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Given:N2O4 (g) + 59.0kJ 2NO2 (g)
NO2
More products, so shift equi to counteract, so more N2O4 produced
Time
Con
cent
ratio
n
What would happen if added more NO2?
N2O4
Add NO2
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Given:N2O4 (g) + 59.0kJ 2NO2 (g)
1. Temperature:• Increasing temp causes the equi to favor the
endothermic side• Why? Increasing temp will increase the rate of
the endo reaction so excess heat is used up.• Decreasing temperature causes the equi to
favor the exothermic side (so that heat is produced/given off)
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Given:N2O4 (g) + 59.0kJ 2NO2 (g)
NO2
Increase temp favors endo side, so equi shifts to products, producing more NO2
Time
Con
cent
ratio
n
What would happen if added heat?
N2O4
Increase Temp
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Given:N2O4 (g) + 59.0kJ 2NO2 (g)
1. Pressure & Volume:• Affects an equi with an unequal number of
moles of gaseous reactants and products
• If increase press (decrease vol), equi shifts to side with less moles
• If decrease press (increase volume), equi shifts to side with more moles
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Given:N2O4 (g) + 59.0kJ 2NO2 (g)
NO2
Increase press favors side with less moles, so equi shifts to reactants, producing more N2O4
Time
Con
cent
ratio
n
What would happen if increased the press / lowered vol?
N2O4
Increase Pressure
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• CATALYSTS:– Increase RATE– But have NO affect on equilibrium
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• Adding an inert gas to a system will not affect the equi.
• Solids do not affect an equi, because their concentration doesn’t change
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Factors Affecting Keq
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• Only TEMPERATURE will affect the value of Keq
• Why?– When you increase temp, rate increases– But endo rate will increase more than the exo
rate. – Changing the ratio of products to reactants &
changing Keq!
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2C6H6 + 15O2 12CO2 + 6H2O + heat
What would happen to Keq if we increased the temp?
Increase temp, favors endo reaction So, more reactants produced Keq = [products] / [reactants]
Reactants will ↑ & products will ↓ Keq will decrease