Upload
others
View
3
Download
0
Embed Size (px)
Citation preview
Equilibrium Constant (Keq)
Keq:# expressing ratio of [products] & [reactants] at equilibrium
constant for every reversible rxn at equilibrium
(@ constant T & P)
aA + bB cC + dD Keq = [C]c[D]d
[A]a[B]b
** pure liquids & solids are not included in the rate
expression because their effective concentrations
do not change
Magnitude of Keq
Keq = 1
Keq > 1
Keq < 1
means products = reactants
Forward rxn is favoured, products > reactants
reverse rxn is favoured, products < reactants
Keq = [Products]
[Reactants]
Given: 2NO2 N2O4 + E
To calculate Keq, the concentrations must be given at
equilibrium!
What is the Keq if the [N2O4] is 0.00140 M & [NO2] is 0.0172M
at equilibrium?
Keq = [N2O4]
[NO2]2
= [0.0014]
[0.0172]2
Keq = 4.73
Given:
2NO2 N2O4 + E
Calculate Keq, if the [N2O4 ] is 0.00455 M & [NO2] is
0.031M at equilibrium.
Keq = [N2O4]
[NO2]2
= [0.00455]
[0.031]2 Keq = 4.73
Same Keq!!!
Important!
Keq is the same for a given reaction that is:
at equilibrium
at the same temperature
no matter what the initial concentrations were.
Keq changes with temperature!
H2 (g) + I2 (g) ⇄ 2HI (g)
Find the Keq if the [H2] is 0.46M, [I2] is 0.39M & [HI] is 3.0M
at equilibrium .
Keq = [HI]2__
[H2][I2]
= __[3.0]2__
[0.46][0.39] Keq = 50
Ex.1
PCl5 (g) PCl3 (g) + Cl2 (g)
Find the Keq, if the initial [PCl5]is 0.70M & the equilibrium [Cl2] is
0.15M.
Keq expression for equilibrium concentrations only!!!
Use an ICE table
Ex.2
[PCl5] [PCl3] [Cl2]
IInitial
CChange
EEquilibrium
0.70
0.55
0 0
+ 0.15+ 0.15- 0.15
0.15 0.15
Keq = [PCl3][Cl2]__
[PCl5]
= [0.15][0.15]__
[0.55]= 0.041
Ex.2PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)
Based on
ratio from
balanced
rxn!
P318 #1 to #7
NH3 (g) H2 (g) + N2 (g)
Initially a 5.0L flask contains 0.2M NH3 & 0.08M N2.
At equilibrium [NH3] is 0.156M. Find Keq
NH3 H2 N2
I
C
E
0.2 0.080
0.156
- 0.044 + 0.066 + 0.022MOLE
RATIO
0.066 0.102
Ex.3 2 3 1
Keq = [H2]3[N2]__
[NH3]2
= [0.066]3[0.102]__
[0.156]2
Keq = 0.0012
H2 (g) + I2 (g) 2HI (g)
Find the [HI] at equilibrium, if [H2] is 0.50M &
[I2] is 0.50M at equilibrium (Keq = 50).
Keq = [HI]2
[H2] [I2]
50 = [x]2__
[0.50][0.50]
[HI] = 3.5 M at equi
Ex.4
H2 (g) + I2 (g) 2HI (g)
What are the equilibrium concentrations of each
substance if a flask initially contains only 0.5M H2 & 0.5M I2?
Keq = 50.
H2 I2 HI
I
C
E
0.5 00.5
0.5 - x
- x - x + 2x
0.5 - x 2x
Ex.5
Keq = [HI]2
[H2] [I2]
50 = [2x]2__
[0.5-x][0.5-x]
At Equilibrium:
[H2] = 2x = 0.5 - 0.39 = 0.11M
[I2] = 0.5 – x = 0.5 - 0.39 = 0.11M
[HI] = 2x = 2 x 0.39 = 0.78M50 = [2x] 2_
[0.5-x] 2
7.07(0.5 – x) = 2x
3.54 - 7.07x = 2x
x = 0.39
7.07 = [2x]_
[0.5-x]
√ / /
H2 (g) + I2 (g) 2HI (g)
What are the equilibrium concentrations of
each substance if a 0.5L flask initially contains
2 moles H2 and 2 moles of I2? Keq = 50
H2 I2 HI
I
C
E
4 04
4 - x
- x - x + 2x
4 - x 2x
Ex.6
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
At 1100K the Keq = 25.
H2(g) + I2(g) 2HI(g)
2 moles of H2 and 3 moles of I2 are placed in a 1 L
container. Find the concentration of each substance at
equilibrium?
Ex.7
H2 I2(g) HI
I
C
E
2 M 3 M 0 M
2x M
+2x-x
2-x M 3-x M
-x
H2(g) + I2(g) 2HI(g)
Continuing the Problem
]][[
][
22
2
IH
HIKeq So..
)3)(2(
)2(25
2
xx
x
2)2()3)(2(25 xxx 22 4)56(25 xxx
22 425125150 xxx
015012521 2 xx Carried over to next slide
Using the Quadratic formula
015012521 2 xx
a
acbbx
2
42 Quadratic formula, where:
a=21, b=-125, c=150
)21(2
)150)(21(4)125()125( 2 x
29.4x 67.1xTwo solutions to formula
But only one of the solutions will give a real answer!
Choosing the Best Answers
29.4x 67.1xTwo solutions to formula
Try finding the correct concentrations using the solutions
xH E 2][ 2
xI E 3][ 2
xHI E 2][
29.229.42][ 2 H 33.067.12][ 2 H
33.167.13][ 2 I
34.3)67.1(2][ HI
At equilibrium:
[H2] = 0.33 mol/L,
[I2] = 1.33 mol/L and
[HI] = 3.34 mol/L
Concentration
cannot be
negative!
Solve using 4.29 Solve using 1.67
Simplifying I.C.E. tables(the 5% rule)
When doing an ICE table you may have to subtract a very small
value from a relatively large value…
for example 2.0 mol/L – 1.0x10-4 mol/L.
In this case, don’t bother doing the subtraction, since by the time
you change it to show significant digits, the result will be the
same:
2.0 mol/L – 0.0001 mol/L = 1.9999 mol/L ≈ 2.0 mol/L.
If the number you subtract is less than 5% of the original number,
you can usually skip the subtraction.