equilibrium student 2014 2
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TRANSCRIPT
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Chemical Equilibrium
When a system is at equilibrium, the forward and reverse reaction are proceeding at the same rate
The concentrations of all species remains constant over time, but both the forward and reverse reaction never cease
That means this is a dynamic equilibrium, not like a beach ball on a seal’s nose.
Equilibrium is signified with double arrows or the equal sign
Symbol found in “Arrows” section of Insert equations in Microsoft Office.
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⇌Equilibrium is signified with double arrows as above Symbol found in “Arrows” section of Insert “Equations” in Microsoft Office.If you cannot find that symbol in your software an acceptable alternative is:
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Dynamic Equilibrium
N2O4 2 NO2
Initially forward reaction rapid
As some reacts [N2O4] so rate forward
Initially Reverse reaction slow No products
As NO2 forms Reverse rate
molecules collide more often as [NO2]
Eventually rateforward =ratereverse this is equilibrium
From this point on concentrations will not change
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Reversibility N2O4 2 NO2
For given overall system compositionAlways reach same equilibrium concentrationsWhether equilibrium is approached from forward or reverse direction
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Mass Action Expression
Relates [reactants] and [products]
Forward reaction A → B Rate = kf[A]
Reverse reaction B → A Rate = kr[B]
Equilibrium A B: kf[A] = kr[B]
Rearranging [A] = kf = constant
[B] kr
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Mass Action ExpressionUses stoichiometric coefficients as exponent for
each reactant
For reaction: aA + bB cC + dD
Reaction quotientNumerical value of mass action expression
Equals “Q ” at any time, and
Equals “K ” only when reaction is known to be at equilibrium
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Calculation Results
Q = __[HI]2_ = same for all = K if all
[H2][I2] experiments at equilibrium
Expt [H2] [I2] [HI] Kc
I .0222 .0222 .156 49.4
II .0350 .0450 .280 49.8
III .0150 .0135 .100 49.4
IV .0442 .0442 .311 49.5
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H2(g) + I2(g) 2 HI(g)
Equilibrium established
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Equilibrium Law
At 440oC equilibrium law:
Kc = _[HI]2_ = 49.5
[H2][I2]
Equilibrium constant = Kc at given T
Use Kc as normally working with molarity
At equilibrium Q = Kc
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Predicting Equilibrium Law
aA + bB cC + dD
Q = [C]c[D]d
[A]a[B]b
Exponents are stoichiometric coefficients of balanced equation.
Law is:
Kc = [C]c[D]d
[A]a[B]b
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Predicting Equilibrium Law
Where only concentrations that satisfy this equation are equilibrium concentrations
Numerator
Multiply [products] raised to their stoichiometric coefficients
Denominator
Multiply [reactants] raised to their stoichiometric coefficients
Kc = [products]f
[reactants]d
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Check
3H2(g) + N2(g) 2NH3(g)
Kc = 4.26 × 108 at 25 °C
What is equilibrium law?
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Check
3H2(g) + N2(g) 2NH3(g)
Kc = 4.26 × 108 at 25 °C
What is equilibrium law?
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Example
In the reaction:
N2O4 2 NO2
Q = [NO2]2
[N2O4]
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CheckWhich of the following is the correct mass action expression for the reaction:
Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?
15
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Manipulating Equations
Reverse direction, new Kc is 1/Kc
A + B C + D Kc = [C][D]
[A][B]
C + D A + B K’c = [A][B} = _1_
[C][D] Kc
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Reverse Equilibrium
3 H2(g) + N2(g) 2 NH3(g)
Kc = _[NH3]2_ = 4.26 x 108
[H2]3[N2]
2 NH3(g) 3 H2(g) + N2(g)
Kc = [H2]3[N2] = ___1____ = 2.35 x 10-9
[NH3]2 4.26 x 108
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Kc Related to Equations II
With aA + bB cC + dD
Kc = [C]c[D]d
[A]a[B]b
If multiply equation, raise Kc to that power
2aA + 2bB 2cC + 2dD
Knew = [C]2c[D]2d
[A]2a[B]2b
Here Knew = Kc2
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Kc Related to Equations III
Add equations, multiply Kc values
aA bB K1 = [B]b
[A]a
bB cC K2 = [C]c
[B]b
Add equations: aA cC
Multiply K: Kc = [B]b x [C]c = K1 x K2
[A]a [B]b
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Example
3 H2(g) + N2(g) 2 NH3(g)
Kc = _[NH3]2_ = 4.26 x 108
[H2]3[N2]
Multiply by 3
9 H2(g) + 3 N2(g) 6 NH3(g)
Kc = _[NH3]6_ = Kc3 7.73 x 1025
[H2]9[N2]3
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Adding Equations
A + B C + D Kc1 = [C][D]
[A][B]
C + E F + G Kc2 = [F][G]
[C][E]
A + B+ E D + F + G
Kcc = [C][D] x [F][G] = [D][F][G]
[A][B] [C][E] [A][B][E]
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Adding Equations
For related reactions in series, the overall Kc is obtained:-
Kcc = Kc1 x Kc2
A. 2CO + O2 2CO2 KcA = 3.3 x 1091
B. 2H2 + O2 2H2O KcB = 9.1 x 1080
Find Kc for H2O + CO CO2 + H2
CO + ½O2 CO2
H2O H2 + ½O2 Add:
H2O + CO CO2 + H2
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Adding Equations II
Divide A by 2 CO + ½O2 CO2 Take sq. root of K cA
Kc1 = [CO2] = (3.3 x 1091)½ = 5.7 x 104
[CO][O2]½
Divide B by 2 and reverse:H2O H2 + ½O2
Kc2 = [H2][O2]½ = (9.1 x 1080)-½ 3.3 x 10-41 [H2O]
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Adding Equations III
Adding equations gives the overall equation and multiplying K values gives the overall Kc:-Kc = [CO2] x [H2][O2]½ = 1.9 x 105
[CO][O2]½ [H2O]
Kc = (5.7 x 1045) x (3.3 x 10-41) = 1.9 x 105
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Practice 1
For: N2(g) + 3 H2(g) 2 NH3(g)
Kc = 500. at a particular temperature.
What would be Kc for following?
2 NH3(g) N2(g) + 3 H2(g)
½ N2(g) + 3/2 H2(g) NH3(g)
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Practice 2
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Equilibrium Constant, Kc
Constant value equal to ratio of product concentrations to reactant concentrations raised to their respective exponents
Kc = [products]f
[reactants]d
Changes with temperature (van’t Hoff Equation)
Depends on solution concentrations
Assumes reactants and products are in solution
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Equilibrium Constant, Kp
Based on reactions in which substances are gaseous
Assumes gas quantities are expressed in atmospheres in mass action expression
Use partial pressures for each gas in place of concentrations
Ex. N2 (g) + 3 H2 (g) 2 NH3 (g)
Kp = _P2NH3
PN2.P3
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How are Kp and Kc Related?Start with ideal gas law
PV = nRT
Rearranging gives
Substituting P/RT for molar concentration into Kc results in pressure-based formula
∆n = moles of gas in product – moles of gas in reactant
Kp = Kc when Δn = 0
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Check
Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25ºC, what would be the Kp?
A. 0.99
B. 2.0
C. 24.
D. 2400
E. None of these
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Solution
Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25ºC, what would be the Kp?
n = nproducts – nreactants = 4 – 3 = 1
Kp = Kc(RT)Δn = 0.99 x (0.08206 x 298)1 = 24.2
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Check 2
Consider the reaction: 2NO2(g) N2O4(g)
If Kp = 0.480 for the reaction at 25°C, what is value of Kc at same temperature?
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Solution 2
Consider the reaction: 2NO2(g) N2O4(g)
If Kp = 0.480 for the reaction at 25°C, what is value of Kc at same temperature?
n = nproducts – nreactants = 1 – 2 = –1
Kp = Kc(RT)Δng
Kc = _Kp__ = _____0.480___ = 19.6
(RT)Δn (0.08206 x 298)-1
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Practice 3
Calculate the value of Kp or Kc for each of the following at 27 °C2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025
N2(g) + 2 O2(g) 2 NO2(g) Kp = 3 x 10−17
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Heterogeneous Equilibrium
Homogeneous reaction/equilibrium
All reactants and products in same phase
Can mix freely
Heterogeneous reaction/equilibrium
Reactants and products in different phases
Can’t mix freely
Solutions are expressed in M
Gases are expressed in M
Governed by Kc
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Heterogeneous Equilibrium II
Pure solids and pure liquids changing concentration would be changing density, cannot do that.Therefore solids and liquids are not included in the equilibrium constant expression
For aA(s) + bB(aq) cC(l) + dD(aq)
the equilibrium constant expression is
Kc = [C]c[B]b = [D]d
[A]a[D]d [B]b
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Heterogeneous Equilibrium III
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
Kc = [Na2CO3(s)] [H2Og][CO2(g)] = [H2Og][CO2(g)]
[NaHCO3(s)]2
1 mol NaHCO3 , V = 38.9 mL
M = 1 mol/.0389 L = 25.7 M
2 mol NaHCO3 , V = 77.8 mL
M = 2 mol/.0778 L = 25.7 M
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Check
Write equilibrium laws for the following:
Ag+(aq) + Cl–
(aq) AgCl(s)
Kc = ____1___
[Ag+][Cl-]H3PO4(aq) + H2O(ℓ) H3O+
(aq) + H2PO4(aq)
Kc = [H3O+][H2PO4-]
[H3PO4]
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CheckGiven the reaction:
3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)
What is the mass action expression?
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Large Kc
Kc >> 1
Rich in product
Goes toward complete
2SO2(g) + O2(g) 2SO3(g)
Kc = 7.0 1025 at 25 ° C
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Small Kc
Kc << 1
Rich in reactant
Few products
H2(g) + Br2(g) 2HBr(g)
Kc = 1.4 10–21 at 25 °C
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K near 1
K 1
Products and reactants
concentratrions close
2NO2(g) N2O4(g)
Kp = 0.480 at 25°C
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Check
Consider the reaction of 2NO2(g) N2O4(g)
If Kp = 0.480 at 25 °C, does the reaction favor product or reactant?
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Equilibrium and Changes
Equilibrium positions Combination of concentrations that allow Q = K
Infinite number of possible equilibrium positions
Le Châtelier’s principleSystem at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress
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Reaction Direction
Can us value of Q to predict reaction direction1. Qc > Kc reaction goes to left; reverse
2. Qc < Kc reaction goes to right; forward
3. Qc = Kc reaction at equilibrium.4. If only reactants Q = 0; forward5. If only reactants Q = ∞; reverse
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Relationship Between Q and K
Q = K reaction at equilibrium
Q < K reactants go to products
Too many reactants
Must convert some reactant to product to move reaction toward equilibrium: shift to right
Q > K products go to reactantsToo many products
Must convert some product to reactant to move reaction toward equilibrium: shift to left
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Le Châtelier’s Principle
1. Concentration
2. Pressure and volume
3. Temperature
4. Catalysts
5. Adding inert gas to system at constant volume
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1. Effect of Change in ConcentrationCu(H2O)4
2+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O
blue yellow
Equilibrium mixture is blue-green
Add excess Cl– (conc HCl)Equilibrium shifts to products
Makes more yellow CuCl42–
Solution becomes green
4)()(
242
42)(
24
]Cl][O)Cu(H[
]OH][CuCl[
aqaq
aqcK
4)()(
242
)(24
42 ]Cl][O)Cu(H[
]CuCl[
]OH[ aqaq
aqcc
KK
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1. Effect of Change in ConcentrationCu(H2O)4
2+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O
blue yellow
Add Ag+ Removes Cl–: Ag+(aq) + Cl–(aq) AgCl(s)
Equilibrium shifts to reactants
Makes more blue Cu(H2O)42+
Solution becomes bluer
Add H2O?
4)()(
242
)(24
]Cl][O)Cu(H[
]CuCl[
aqaq
aqcK
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Addition of Product
Add NO2(g) to N2O4(g) 2NO2(g)
System is disturbed so Q now > K; system responds by converting more NO2(g) into N2O4(g) until again Q = K but with larger values for all concentrations.
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Check
2SO2(g) + O2(g) → 2SO3(g)
Kc = 2.4 x 10-3 at 700 oC
Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture ?
A. Towards the products
B. Towards the reactants
C. No change will occur
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Concentration Effect Summary
When changing concentrations of reactants or products
Equilibrium shifts to remove reactants or products that have been added
Equilibrium shifts to replace reactants or products that have been removed
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Effect of Change in P & V
Gaseous system at constant T and n
3H2 (g) + N2(g) 2NH3 (g) Kp = __P2NH3__
PN2.P3H2
If Vexpect PTo reduce P, look at each side of reaction
Which has less moles of gas
Reactants = 3 + 1 = 4 moles gas
Products = 2 moles gas
Reaction favors products (shifts to right)
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Effect of Change in P & V
If Vexpect PTo reduce P, look at each side of reaction
Which has less moles of gas
Reactants = 3 + 1 = 4 moles gas
Products = 2 moles gas
Reaction favors products (shifts to right)
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Effect of Change in P & V II
2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g) Kp = PH2O.PSO2
If you V of reaction,
Reactants: no moles gas = all solids
Products: 2 moles gas
V, causes P
Reaction shifts to left (reactants), as this has fewer moles of gas
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Effect of Change in Temperature
Cu(H2O)42+
(aq) + 4Cl–(aq) + Heat CuCl4
2–(aq) + 4H2O
blue yellow
Reaction endothermic
Adding heat shifts equilibrium toward products
Cooling shifts equilibrium toward reactants56
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Temperature Change
H2O(s) H2O(ℓ) H° =+6 kJ (at 0 °C)
Energy + H2O(s) H2O(ℓ)
Energy is reactant
Add heat, shift reaction right
3H2(g) + N2(g) 2NH3(g) Hf°= –47.19 kJ
3 H2(g) + N2(g) 2 NH3(g) + energy
Energy is product
Add heat, shift reaction left
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Effect of T Change
T shifts reaction in direction that produces endothermic (heat absorbing) change
T shifts reaction in direction that produces exothermic (heat releasing) change
Changes in T change value of mass action expression at equilibrium, so K changed
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Effect of T Change II
K depends on T
T of exothermic reaction makes K smallerMore heat (product) forces equilibrium to reactants
T of endothermic reaction makes K largerMore heat (reactant) forces equilibrium to products
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Catalysts
Catalysts lower Ea
Change in Ea has
equal effect on kf & kr
Result is no effect
on K
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Addition of Inert Gas
Will not affect concentrations or partial pressures of any components
Will not affect reaction
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Learning Check:Consider:
H3PO4(aq) + 3OH–(aq) 3H2O(ℓ) + PO43–(aq)
What will happen if PO43– is removed?
Q is proportional to [PO43– ]
[PO43– ], Q
Q < K equilibrium shifts to right
]POH[]OH[
]PO[Q
433
34
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63
Learning Check:The reaction
H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq)
is exothermic.
What will happen if system is cooled?
Since reaction is exothermic, heat is product Heat is directly proportional to Q T, Q Q < K equilibrium shifts to right
heat
]POH[]OH[
]PO[Q
433
34
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Check
The equilibrium between aqueous cobalt ion and the chlorine ion is shown:[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O(ℓ)
pink blue
It is noted that heating a pink sample causes it to turn violet. The reaction is:
A. endothermic B. exothermic
C. cannot tell from the given information
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Calculating Kc from [Equilibrium]
Most direct way.
Measure all reactant & product concentrations.
Kc always be the same as long as T constant
65
Tro: Chemistry: A Molecular Approach, 2/e
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Equilibrium Calculations
For solution reactions, must use KC
For gaseous reactions, use either KP or KC
Calculate K from equilibrium [X] or PX
Find one or more equilibrium [X] or PX
from KP or KC
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Kc From [X}s at Equilibrium
Use mass action expression to relate [X] values
Can use:Equilibrium [reactant]s & [product]s
From initial and one final [product] find remaining {X}s then find Kc
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Kc from Equilibrium [X]
1 N2O4(g) 2NO2(g)
If you place 0.0350 mol N2O4 in 1 L flask at equilibrium, what is KC?
[N2O4]eq = 0.0292 M
[NO2]eq = 0.0116 M
Kc = _[NO2}2 = (0.0292)2 = 4.61 x 10-3
[N2O4] (0.0292)
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Practice 4
For the reaction: 2A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature?
A. 14
B. 0.15
C. 1.5
D. 6.75
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From [X]i and [Y]f
Set up ICE table
Changes in same ratio as coefficients
[X]eq = [X]initial - [X]change
1. Assign x as coefficient, + for materials on side it is going to or – for opposite
2. Solve for x, if 2nd order, take square root, use quadratic equation or simplify if K very large or very small.
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Generic
Use concentration tables set up as follows:
A + B 2C
[A] [B] [C]
I P Q R
C -x -x +2x
E P – x Q – x R + 2x
Kc = _ (R + 2x)2__
(P - x)(Q - x)
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Example
2SO2(g) + O2(g) 2SO3(g)
1.000 mol SO2 and 1.000 mol O2 are placed in a 1.000 L flask at 1000 K. At equilibrium 0.925 mol SO3 has formed. Calculate KC for this reaction.
[SO2]i = [O2]i = 1.00 mol = 1.00 M
1.00 L
[SO3]eq = 0.925 mol = 0.925 M
1.00 L
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ICE Table
2SO2(g) + O2(g) 2SO3(g)
I 1.00 1.00 0.000
C -0.925 -0.462 +0.925
E 0.075 0.538 0.925
Kc = __[SO3]2__ = _(0.925)2__ 2.8 x 102
[SO2]2[O2] (.075)2(.538)
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Practice 5
CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 5.67
[CH4] = 0.400 M; [H2] = 0.800M; [CO] =0.300M
What is [H2O] at equilibrium?
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Practice
H2(g) + I2(g) 2HI(g) at 425 °C
KC = 55.64
If one mole each of H2 and I2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?
Kc = __[HI]2_= 55.64
[H2][I2]
[H2]i = [I2]i = 1.00 mol = 2.00 M
0.500 L
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Solution
H2(g) + I2(g) 2HI(g)
I 2.00 2.00 0.00
C -x -x +2x
E 2.00 – x2.00 – x 2x
55.64 = ____(2x)2______ = ____(2x)2___
(2.00 – x)(2.00 – x) (2.00 – x)2
Take square root of both sides
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Solution II
55.64 = 7.459 = __(2x)__
(2.00 – x)
Rearrange: (7.459)(2.00 – x) = 2x
14.918 – 7.459x = 2x
14.918 = 9.459x
x = 14.918 = 1.58
9.459
[H2]eq = [I2]eq = 2.00 – 1.58 = 0.42 M
[HI]eq = 2x = 2(1.58) = 3.16 M
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Example 2
H2(g) + I2(g) 2HI(g) at 425 °C
KC = 55.64
If one mole each of H2, I2 and HI are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?
Now have product as well as reactants initially
Kc = __[HI]2_= 55.64
[H2][I2]
H2]i = [I2]i = [HI] 1.00 mol = 2.00 M
0.500 L
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Solution
H2(g) + I2(g) 2HI(g)
I 2.00 2.00 2.00
C -x -x +2x
E 2.00 – x2.00 – x 2.00 + 2x
55.64 = __(2.00 + 2x)2__ = (2.00 + 2x)2
(2.00 – x)(2.00 – x) (2.00 – x)2
Take square root of both sides
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Solution
55.64 = 7.459 = (2.00 + 2x)_
(2.00 – x)
Rearrange: (7.459)(2.00 – x) = 2.00 + 2x
14.918 – 7.459x = 2.00 + 2x
12.918 = 9.459x
x = 12.918 = 1.37
9.459
[H2]eq = [I2]eq = 2.00 – 1.37 = 0.63 M
[HI]eq = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M
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Practice 6
N2(g) + O2(g) → 2NO(g)
Kc = 0.0123 at 3900 oC
If 0.25 moles of N2 and O2 are placed in a 250 mL container, what are the equilibrium concentrations of all species ?
A. 0.0526 M, 0.947 M, 0.105 M
B. 0.947 M, 0.947 M, 0.105 M
C. 0.947 M 0.105 M, 0.0526 M
D. 0.105 M, 0.105 M, 0.947 M
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Practice
CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) +H2O(l)
acetic acid ethanol ethyl acetate KC = 0.11
An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?
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Solution
Kc = __[CH3CO2C2H5]__
[CH3CO2H][C2H5OH]
CH3CO2H C2H5OH CH3CO2C2H5
I 0.810 0.810 0.000
C -x -x +x
E .810 – x .810 – x x
0.11 = ________x________
(.810 – x)(.810 – x)
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Solution II
Rearranging:
0.11 x(0.6561 – 1.62x + x2) = x
Expand then write as quadratic equation
ax2 + bx + c = 0
0.07217 – 0.1782x + 0.11x2 - x = 0
0.11x2 – 1.1782x + 0.07217 = 0
x = -b (b2 – 4ac)
2a
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85
Solution III
This gives two roots: x = 10.6 and x = 0.064Only x = 0.064 is possible
x = 10.6 is >> 0.810 initial concentrations 0.810 – 10.6 = negative concentration,
which is impossible
)11.0(2)07217.0)(11.0(4)1782.1()1782.1( 2
x
22.0164.11782.1
22.0)032.0()388.1(1782.1
x
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Solution IV
[CH3CO2C2H5]eq = x = 0.064 M
[CH3CO2H] = [C2H5OH]
= 0.810 – x = 0.810 – 0.064 = 0.746 M
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Simplifications
If get trinomial use successive approximations
If K is very small x is also small can drop x in change part only
Valid if [X]I = 400x > K
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More Problems
1.Finding Kc from change in reactant concentrations.
PCl 3(g) + Cl2(g) PCl5(g)
0.200 mol PCl3 and 0.100 mol Cl2 put in 1.00 L flask, at equilibrium found 0.120 mol PCl3. Means change in [PCl3] is 0.200 - 0.120 = 0.080 mol, change must be same for other components. As volume is 1.0 L, molarity has same value.
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Example
Using Kc to calculate equilibrium concentrations:A flask contains a mixture of Br2 and Cl2 ,both with an initial concentration of 0.0250 M. Find the equilibrium concentration of all molecules from the following equilibrium information.
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Example Continued
[BrCl] [Br2] [Cl2]
I 0 .0250 .0250
C +2x -x -x
E 2x .0250 – x .0250 - x Kc = [Br2][Cl2] = (.0250 - x)2 = .145 [BrCl]2 (2x)2
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Continued
Kc = [Br2][Cl2] = (.0250 - x)2 = .145
[BrCl]2 (2x)2
Take square root:(.0250 - x) = .381
2x
.0250 - x = .762 or .0250 = 1.762 x
x = .0250/1.762 = .0142 for Br2 & Cl2
BrCl = 2x = .0284 M
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Practice
Find equilibrium concentrations with very small Kc.
N2O + NO2 3NO Kc = 1.4 x 10-10 Start with .200 mol N2O & .400 mol NO2 in
4.00 L
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Solution
[N2O] = .200 mol/4.00 L = 0.0500 M
[N2O] = .400 mol/4.00 L = 0.100 M
[N2O] [NO2] [NO]I .0500 .100 0C -x -x +3xE .0500 – x .100 – x 3x
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Problem 3 (II)
Kc = [NO]3 = (3x)3 =1.4 x10-10
[N2O][NO2] (.0500 - x)(.100 - x) Simplify reactant concentrations as x very small:
27x3 = 1.4 x 10-10
(.0500)(.100)x3 = 2.6 x 10-14 or x = 3.0 x 10 -5
[NO] = 3x = 9.0 x 10-5