The Equilibrium Constant, K, and The Reaction Quotient, Q
29
The Equilibrium Constant, K, and The Reaction Quotient, Q SCH 4U
description
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 Presentation
Transcript of The Equilibrium Constant, K, and The Reaction Quotient, Q
The Equilibrium Constant, K,and The Reaction Quotient, QSCH 4U
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
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...
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
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)
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!
Writing an Expression for KEx. 1 N2(g) + 3 H2(g) 2 NH3(g)
K = __________[NH3][N2 ] [H2 ]
2
3
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
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
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
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?
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
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
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
Now can plug in equilibrium conc’s to find value of K:
K = (1.80)2 (1.10)(0.1)K = 29
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
... 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)
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
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!
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
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!)
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
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
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?
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...
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)
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!
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
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
