the equilibrium constant, k, and the reaction quotient, q
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The Equilibrium Constant, K, and The Reaction Quotient, Q. SCH 4U. Chemical Systems at Equilibrium. For reversible reactions in closed systems , the reaction mixture always contains both reactants and products. At equilibrium, the forward and reverse reaction rates are ____________ - PowerPoint PPT PresentationTRANSCRIPT
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The Equilibrium Constant, K,and The Reaction Quotient, QSCH 4U
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Chemical Systems at EquilibriumFor reversible reactions in closed
systems, the reaction mixture always contains both reactants and products.
At equilibrium, the forward and reverse reaction rates are ____________and the reactant and product concentrations are ____________.
EQUAL
CONSTANT
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Just like we learned yesterday that the percentage reaction is always constant (under what conditions?)...
...These two smart Norwegian dudes noticed, for a system at equilibrium,
the following relationship
exists...
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Equilibrium LawFor the general rxn:
aA + bB cC + dD
(a-d are coefficients; A-D are chemical species)
The following Equilibrium Law expression can be written:
K = [C]c[D]d
[A]a[B]b
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K is a constant called...The Equilibrium Constant
Note: This is for a system AT EQUILIBRIUM, therefore conc’s are equilibrium conc’s!
Also, species are in aqueous or gas phase...
...if liquids or solids, they do NOT appear in the expression (b/c conc. unchanging)
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No matter what reactant/product concentrations we start at, once equilibrium is established, the value of K is always the same!*
Ahem.... Equilibrium Constant...(see p.439 if you don’t believe me)
*It is still temperature-dependent though!
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Writing an Expression for KEx. 1 N2(g) + 3 H2(g) 2 NH3(g)
K = __________[NH3][N2 ] [H2 ]
2
3
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Type 1: Calculating KFor the same system, given the following
equilibrium concentrations, find K:[N2(g)] = 1.50 x 10-5 mol/L[H2(g)] = 3.45 x 10-1 mol/L[NH3(g)] = 2.00 x 10-4 mol/L
K = (2.00 x 10-4)2 . (1.5 x 10-5) (3.45 x 10-1)3
K = 6.49 x 10-2 at 500°C
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Type 2: Calculating Equil. Conc.
H2(g) + I2(g) 2 HI(g) K= 49.70 at 458°C
Given: [H2]equil = 1.07 mol/L[I2]equil = 1.07 mol/L
Calculate [HI] at equilibrium
K = [HI]2
[H2] [I2]
[HI] = √K[H2][I2] Rearranging:
= 7.54 mol/L
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The Magnitude of KK = [C]c[D]d
[A]a[B]b
reaction proceeds toward completion (much more products)
concentrations of reactants and products are approx. equal at equilibrium
very small amounts of products formed (much more reactants)
productsreactants
If K >> 1
If K ≈ 1
If K ≈ 1
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Comparing K’s
H2(g) + I2(g) 2HI(g) K = 50 (at
450°C)
CO2(g) + H2(g) CO(g) K = 1.1 (at
900°C)
How do these systems at equilibrium compare?
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Type 3: Using K and ICE Table(knowing initial conc’s and K)
H2(g) + I2(g) 2 HI(g) K= 64 (at partic. temp)
Given: [H2]initial = [I2]initial = 3.0 M
Find all equilibrium concentrations.H2(g) + I2(g) = 2 HI(g)
Initial 3.0 M 3.0 M 0ChangeEquilibrium
-x -x +2x3.0 M - x 3.0 M - x 2x
K = [HI]2 [H2] [I2]
= [2x]2
[3.0M – x][3.0M – x]= [2x]2
[3.0M – x]2
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Therefore at equilibrium:[H2] = 3.0M – 2.4 = 0.6 M[I2] = 3.0M – 2.4 = 0.6M[HI] = 2(2.4M) = 4.8 M
K = [2x]2
[3.0M – x]2
64 = [2x]2
[3.0M – x]2
8 = 2x3.0 M – x
x = 2.4
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Type 4: Using K and ICE Table(knowing initial and equil concs)
Given above concentrations, find K
H2(g) + I2(g) = 2 HI(g)
Initial 2.0 M 1.0 M 0ChangeEquilibrium 1.80 M
First need to find equilibrium concentrations!
K = [HI]2 [H2] [I2]
-x -x +2x2.0M -x 1.0M -x
Need to solve for x!
1.80M = 2x 0.9M = x
= 1.1M = 0.1M
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Now can plug in equilibrium conc’s to find value of K:
K = (1.80)2 (1.10)(0.1)K = 29
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The Reaction Quotient, Q
Q = [C]c[D]d
[A]a[B]b
K is calculated using conc’s at equilibrium
Q is calculated using conc’s that may or may not be at equilibrium
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... Tells us if system is at equilibrium or, if not, which way system must shift to reach equilibrium!
If Q = K
If Q < K
If Q > K
System at equilibrium
Have more products than system at equilibrium, must shift to left (toward reactants)
Have less products than system at equilibrium, must shift to right (toward products)
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ExampleN2(g) + 3H2(g) 2NH3(g) K= 0.064 at 450°C
Given: [N2] = 4.0 mol/L[H2] = 2.0 x 10-2 mol/L[NH3] = 2.2 x 10-4 mol/L
Is system at equilibrium?Q = [NH3]2
[N2] [H2]3
Q < K, not at equilibrium, needs to shift right, toward products
= 0.0015
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Type 5: Calculating Q FirstCalculate the equilibrium concs of
H2, I2, and HI if initial concs are 0.5M, 0.5M, and 4.0M respectively and K is 50.
Set up ICE Table...H2(g) + I2(g) = 2 HI(g)
Initial 0.5 M 0.5 M 4.0 MChangeEquilibrium
-x -x +2x
WATCH OUT!!
None of the reactant conc’s are zero – have to first det. in which direction system is shifting!
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How do we do that?→ Find Q!
Q = [HI]2= (4.0)2 = 64 [H2][I2] (0.5)(0.5)Now compare Q to K...Q > K therefore system shifts to
left (toward reactants)H2(g) + I2(g) = 2
HI(g)
Initial 0.5 M 0.5 M 4.0 MChangeEquilibrium
+ x + x - 2x0.5M + x 0.5M + x 4.0M - 2x
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How do we proceed? (Don’t know x yet) Use K to find value of x...
50 = (4.0M – 2x)2
(0.5M + x)(0.5M + x)7.07 = (4.0M – 2x) x = 0.05
(0.5M + x)Then can calculate equil. conc’s...(Do it!)
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Note: If an equil. problem cannot be solved by taking the square root of both sides of the equil. equation, the quadratic equation must be used.
x = – b + √b2 – 4ac2a
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Using the 100 Rule2CO2(g) 2CO(g) + O2(g) K = 6.40 x
10-7 Calc. equil. concentrations if 0.250
mol of CO2 is placed in a 1L closed container and heated to 2000°C
Since there are initially no products, Q=0 and reaction will proceed to right...
at 2000°C
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Use K to find x...K = [CO]2[O2] = 6.40 x 10-7
[CO2]2
= (2x)2(x) (0.25 – 2x)2
= 4x3
(0.25 – 2x)2
2 CO2(g) = 2 CO(g) + O2(g)
Initial 0.250 M 0 0Change - 2x +2x +xEquilibrium 0.250 – 2x 2x x
This is a cubic (x3) function, hard to solve directly...
... Can we simplify it?
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Yes!K is very small (10-7) in comparison
to initial conc of CO2(g)
...which means very little CO2(g) decomposes at this
temp
Therefore we can assume x to be very small (2x as well) and negligible...
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i.e. 0.250 – 2x ≈ 0.250
So our expression becomes...
4x3 ≈ 6.40 x 10-7
(0.25)2
x ≈ 2.15 x 10-3
(can check validity of assumption)
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100 Rule
If concentration to which x is added or subtracted from is 100x or more greater than the value of K, then we can ignore x
e.g. [CO2]initial = 0.250 = 3.91 x 105
K 6.40 x 10-7
See Example p. 472Which is >> 100!
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No Class TomorrowIn total 2 lessons behind the other class...It is your responsibility to make sure you are okay with ALL the homework questions that follow... you have 4 nights/ 3 days!
Come see me before school or at lunch on Monday if there are any problems.
We will be having an assessment on Monday
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Practice Problems*See (do!) sample problem on p. 472 (100
rule)p. 444 # 2, 3 p. 445 # 5, 6p. 447 # 7p. 449 # 9p.472 # 5, 6p. 476 # 8p. 480 # 9, 10p. 481 # 2, 3a, 3c, 4-8