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Chapter 6: Chemical Equilibrium
6.1 The Equilibrium Condition6.2 The Equilibrium Constant6.3 Equilibrium Expressions Involving Pressures6.4 The Concept of Activity6.5 Heterogeneous Equilibria6.6 Applications of the Equilibrium Constant6.7 Solving Equilibrium Problems6.8 LeChatelier’s Principle6.9 OMIT: Equilibria Involving Real Gases
2 HWs due Satd at 6 PM: HW #7: Chapter 5, part 2HW #8: Chapter 6, part 1
A new HW assignment (HW#8) over 1st part of Ch. 6 is also due this Satd..
The 2nd part of Ch. 6 (HW#9) will be due 7 AM on Friday, Nov. 19. Exam on Nov. 22.
-------------------All Discussion Sessions after 12:00 on Nov. 18 have been
rescheduled to:Wed. Nov. 17th from 2:45 to 3:45
in Bagley Hall room 154(run by Emily Sprafka, a TA)
---------------------------She will also hold a Review for Exam 2on Fri. Nov. 19th from 3:30 to 5:00 PM
in BAG 131.
Molecular Picture of Establishment of Equilibrium
CO(g) + H2O(g) CO2(g) + H2(g)
INITIAL
7 CO(g) + 7 H2O(g) + 0 CO2(g) + 0 H2(g) AFTER IT STOPS CHANGING
2 CO(g) + 2 H2O(g) + 5 CO2(g) + 5 H2(g)
Chemical Equilibrium• Previously, we assumed that a chemical reaction goes to completion as written.
H2O(g) + CO (g) → H2(g) + CO2(g)
• In general this is not correct. Instead, a stable state of the system in reached, which includes both reactants and products. It is called the equilibrium state, or simply “equilibrium”.
H2O(g) + CO (g) ⇌ H2(g) + CO2(g)
How far toward completion it goes depends on the specific reaction, and on temperature.
Concentration vs. TimeCO(g) + H2O(g) → CO2(g) + H2(g)
Ratio depends on temperature
Characteristics of Chemical Equilibrium States
• Reaching equilibrium requires reactions to occur.
• Once reached, they show no macroscopic evidence of further change.
• Reached through dynamic balance of forward and reverse reaction rates.
Figure 5.21: The collision rate of gas particles defines
the maximum blue-pink reaction rate!
Z = Collision rate (of one pink with blues)= # collisions with blues per s = 4 [Nblue/V] d2 (πRT/M)1/2
= 4 [blues] d2 (πRT/M)1/2
Equilibrium arises through dynamic balance between forward and back reactions
Forward rate = k1[NO2][NO2] = k1[NO2]2
Back rate = k-1[NO3][NO]
The rate constants k1 and k-1 reflect probabilities that one collision leads to a successful reaction.
Reverse: NO3(g) + NO(g) → 2NO2(g)k -1
Kinetics of Approach to Equilibrium
[ ] [ ][ ][ ][ ][ ]
21 2 1 3
312
1 2
NO NO NO
NO NO
NO
k k
kk
−
−
=
= = K
At eqbm.: Forward rate = back rate
⇑“Equilibrium constant”
The Equilibrium Constant - DefinitionConsider the generalized chemical reaction:
a A + b B ⇌ c C + d DA, B, C and D represent chemical species and a, b, c, and d are their stoichiometric coefficients in the balanced chemical equation. At equilibrium,
The square brackets indicate the concentrations of the species in equilibrium.
K is a constant called the equilibrium constant.
K depends only on T , and not on concentrations.
[ ] [ ][ ] [ ]
c d
a b
C DK
A B= Note: The “units” for K are
concentration units raised to some power = c+d–(a+b)
Reaching Equilibrium on the Macroscopic and Molecular Level
N2O4 (g) 2 NO2 (g)Colorless Brown
K = [NO2]2 / [N2O4] Units are mol/L.
Equilibrium from Different Starting Points
CO(g) + 2 H2(g) ⇌ CH3OH(g)
[ ][ ][ ]
32
2
has same value at equil.for all 3 starting points
CH OHK
CO H=
N2(g) + 3 H2(g) 2 NH3(g)Fe
K = [NH3]2
[N2][H2]3
Key Stages in the Haber Synthesis of Ammonia
Gerhard Ertl,2007 Nobel Prize in Chemistry
Small K N2 (g) + O2 (g) 2 NO(g) K = 1 x 10 –30
Essentially only reactants at eqbm. (1015 x products)
Intermediate K 2 BrCl(g) Br2 (g) + Cl2 (g) K = 5
Comparable amounts of products and reactants at eqbm.
Large K 2 CO(g) + O2 (g) 2 CO2 (g) K = 2.2 x 1022
Essentially only products at eqbm.
Equilibrium Constants can have a wide range of values
CS2(g) + 3 O2(g) ⇌ CO2(g) + 2 SO2(g)
• The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original expression.
CO2(g) + 2 SO2(g) ⇌ CS2(g) + 3 O2(g)
• If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n.
2 CS2(g) + 6 O2(g) ⇌ 2 CO2(g) + 4 SO2(g)
[ ][ ][ ][ ]322
222
1 OCSSOCO
=K
[ ][ ][ ][ ] 1
222
322
21
SOCOOCS
KK ==
[ ] [ ][ ] [ ]
( )2 4
22 23 12 6
2 2
CO SOCS O
K K= =
Example: Calculation of the Equilibrium Constant from equilibrium amounts
At 454 K, the following reaction takes place:
3 Al2Cl6(g) ⇌ 2 Al3Cl9(g)
At this temperature, the eqbm concentration of Al2Cl6(g) is 1.00 Mand the equilibrium concentration of Al3Cl9(g) is 1.02 x 10-2 M. Compute the equilibrium constant at 454 K.
Strategy: Substitute values into K
[ ][ ]
( )( )
143
22
362
293 1004.1
00.11002.1
ClAlClAl −−
−
×=×
== MM
MK
Like Example 6.1 (P 201) - IThe following equilibrium concentrations were observed for the reaction between CO and H2 to form CH4 and H2O at 927oC:
CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g)
[CO] = 0.613 mol/L [CH4] = 0.387 mol/L[H2] = 1.839 mol/L [H2O] = 0.387 mol/L
a) Calculate the value of K at 927oC for this reaction.b) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g)c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g)Solution:a) For the first balanced reaction above:
K = =[CO] [H2]3[CH4] [H2O]
Like Example 6.1 (P 201) - IThe following equilibrium concentrations were observed for the Reaction between CO and H2 to form CH4 and H2O at 927oC.
CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g)
[CO] = 0.613 mol/L [CH4] = 0.387 mol/L[H2] = 1.839 mol/L [H2O] = 0.387 mol/L
a) Calculate the value of K at 927oC for this reaction.b) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g)c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g)Solution:a) For the first balanced reaction above:
K = = = 0.0393 L2/mol2[CO] [H2]3[CH4] [H2O] (0.387 mol/L) (0.387 mol/L)
(0.613 mol/L) (1.839 mol/L)3
Like Example 6.1 (P 201) - IIb) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g)
K = =[CO] [H2]3
[CH4][H2O]
Easier way: This K is just the reciprocal of K from part a:1Ka
K =1
c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g)
K = =[CO]1/3 [H2]
[H2O]1/3[CH4]1/3
Easier way: K = (K from part a)1/3
Like Example 6.1 (P 201) - IIb) Calculate the value of the equilibrium constant at 927oC for:
H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g)
K = = = 25.45 mol2/L2[CO] [H2]3
[CH4][H2O](0.613 mol/L) (1.839 mol/L)3
(0.387 mol/L) (0.387 mol/L)
1Ka
K = = = 25.45 mol2/L21
0.0393 L2/mol2
c) Calculate the value of the equilibrium constant at 927oC for:
1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)
K = =[CO]1/3 [H2]
[H2O]1/3[CH4]1/3 (0.387mol/L)1/3 (0.387 mol/L)1/3
(0.613 mol/L)1/3 (1.839 mol/L)
Easier way: K = (Ka)1/3 = (0.0393L2/mol2)1/3
Easier way: This K is just the reciprocal of K from part a:
Determining Equilibrium Concentrations from KExample: Methane can be made by reacting carbon disulfide with hydrogen gas. K for this reaction is 27.8 (L/mol)2 at 900°C.
CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2S (g)
At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol CS2, 1.10 mol of H2, and 0.45 mol of H2S. How much methane was formed?
Strategy:(1) Calculate the equilibrium concentrations from the moles given and the volume of the container. (2)Use the value of K to solve for the concentration of methane.(3) Calculate the number of moles of methane from M and V.
[ ]20.250 molCS 0.0532 M
4.70 L= =
[ ]21.10 molH 0.234 M
4.70 L= =
[ ]20.450 molH S 0.0957 M
4.70 L= =
[ ][ ][ ][ ]
( )2
24 24
2 2
CH H S27.8 L/mol
CS HK = =
[ ] [ ][ ][ ]
( ) [ ] [ ][ ]
4 422 2
4 2 22
CS H 0.0532 0.234CH 27.8 L/mol 0.484
H S 0.0957M M
K MM
×= = =
4-1
m o le s C H
0 .4 8 4 m o le s L 4 .7 0 L = 2 .2 7 m o le s
M V= ×
= ×
CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g)
KP: Equilibrium Constant Involving PressuresFor a reaction of the type
a A + b B ⇌ c C + d D where the species are gases,
It is sometimes convenient to write the equilibrium expression in terms of partial pressures:
P indicates the partial pressures of the species in equilibrium and KP is a constant called the equilibrium constant in terms of partial pressures. KP depends only on T , and not on pressure.
( ) ( )( ) ( )
c dC D
P a bA B
P PK
P P=
Expressing K with Pressure UnitsFor gases, PiV = niRT can be rearranged to give: Pi = RTni
Vor: =ni Pi
V RT = Molar concentration of gas i = [i]niV
For an equilibrium between gaseous compounds, there is a relationship between the equilibrium constants K and KP.
Example: 2 NO (g) + O2 (g) 2 NO2 (g)
Kp = P 2
NO2
P 2NO PO2
[NO2]2
[NO]2 [O2]K = and
Using concentrations Using pressures
How is KP related to K?Answer: Through the use of the ideal gas law. For all species i:
or ii i i i
nPV n RT P RT C RTV⎛ ⎞= = =⎜ ⎟⎝ ⎠
( ) ( )( ) ( )
( ) ( )( ) ( )
( ) ( )( ) ( )
( )( )
( )( ) ( )
c d c dC D C D
P a b a bA BA B
c d c dC D c d a b n
a b a b
A B
P P C RT C RTK
C RT C RTP P
C C RTK RT K RT
RTC C
++ − + Δ
+
× ×= =
× ×
= × = =
Where Δn = c+d-(a+b) Note that K =KP when Δn = 0 NOTE: Δn here refers ONLY to the gases in rxn.
For reaction 2NO(g) + O2(g) ⇌ 2NO2(g)
Sum of gas coefficients (+ for products, – for reactants)
Δn = 2 – 2 – 1 = – 1
[ ][ ] [ ]
22
22
NO
NO OK =
2
2
2NO
2NO O
P
PK
P P=
1( )PKK K RT
RT−= =
Unitless method for expressing K and KP
• Units for K are concentration raised to power Δn, and those for KP are pressure units raised to power Δn.
• In later chapters, we will divide each concentration in K by a reference concentration of 1 molar, and each partial pressure in KP by a reference pressure of 1 atm. This renders all concentrations and pressures unitless, at which point we will call them both activities (ai).
• Just replace all [ i ] or Pi with ai in K or KP.
Use of Activities in Equilibrium Constant Expressions
ref
Activity ( th component)
where partial pressure of the th gaseous componenta
For Pressure
nd 1 atm
s:
(exactly)
ii
ref
i
Pi aP
P iP
= =
=
=
ref
Activity ( th component)
where molarity of the th comp
F
o
or Conce
nentand
nt
1 M (exac
ration
)
:
tly
ii
ref
i
Mi aM
M iM
= =
=
=
: Reference state with unit activity
: Reference state with unit activity
CaCO3 (s) CaO (s) + CO2 (g)
KP = aCaO(s)PCO2 /aCaCO3(s) = PCO2 (The 2 solids do NOT appear in KP!)
The position of an equilibrium involving a pure solid or liquiddoes not depend upon the amounts of pure solid or liquids present. The activity of a pure solid or liquid is always equal to 1 if present, soits “concentration” does not appear in K expression.
For pure solids and liquids, activity = 1.0
1.0 1.0
Equilibrium involving pure solids or liquids:
set their activity = 1
Example: NH4NO2(s) ⇌ N2(g) + 2 H2O(g)
The equilibrium constant for this reaction would normally be expressed as: [ ][ ]
[ ]
22 2
4 2
N H O'
NH NOK =
4 2
However, a pure solid or liquid retains the same activityduring the reaction. Thus we set the activity of NH NO ( )to one.
s
[ ][ ] ( )( )2 2
222 2 N H ON H O and P PpK K= =1.0000 1.0000
Equilibrium involving pure solids or liquids:
set their activity = 1
Example: NH4NO2(s) ⇌ N2(g) + 2 H2O(g)
The equilibrium constant for this reaction would normally be expressed as:
[ ][ ][ ]
22 2
4 2
N H O'
NH NOK =
4 2
However, a pure solid or liquid retains the same activityduring the reaction. Thus we set the activity of NH NO ( )to one.
s
[ ][ ] ( )( )2 2
222 2 N H ON H O and P PpK K= =
The Reaction Quotient, Q
[ ] [ ][ ] [ ]
Consider the reaction: a A( ) B( ) C( ) D( ) The reaction quotient, Q is defined as
where the subscripts indicate momentary concentrations at some time t before (or after) equi
c d
t ta b
t t
g b g c g dm g
C DQ
A B
t
+ = +
=
librium has established.
Q has same form as K, but the concentrations are the actual concentrations at any time t rather than
the concentrations after equilibrium is reached.
Reaching Equilibrium on the Macroscopic and Molecular Level
N2O4 (g) 2 NO2 (g)Colorless Brown
Q = [NO2]2 / [N2O4] Units are mol/L.
TIME
[ ][ ]
2 4 22
2
2 4
N O (g) 2NO (g)
NOQ =
N O
⇌
Reaction Direction and the Relative Sizes of Q and K
“Excess reactantsinitially”
“Excess productsinitially”
Which Direction?
• If Q < K, the system will shift to the right by converting reactants to products.
• If Q = K, the system is at equilibrium; the concentrations will not change.
• If Q > K, the system will shift to the left by converting products back to reactants.
Example: Predict Reaction Direction using Q
For the following reaction:
CH4(g) + 2 H2S(g) ⇌ CS2(g) + 4 H2(g)
1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S and 2.00 mol H2 are mixed in a 250 mL vessel at 960oC. At this temperature, K = 0.036. In which direction will the reaction proceed in order to reach equilibrium?
Strategy:
1) Calculate the actual concentrations
2) Calculate Q.
3) Compare Q and K
Solution: Predict Reaction Direction using Q1) [CH4]0= 1.00 mol/0.250 L = 4.00 M.
[H2S]0 = 2.00 mol/0.250 L = 8.00 M,
[CS2]0 = 1.00 mol/0.250 L = 4.00 M, and
[H2]0 = 2.00 mol/0.250 L = 8.00 M.
2) Calculate the value of Q:
[ ] [ ][ ] [ ]
( )4 42 20 0
2 24 20 0
CS H 4.00 8.0064.00 0.036
4.00 (8.00)CH H SQ
×= = = >>
×
(3) Comparing Q and K: Q > K, so the reaction goes to the left. Therefore reactants concentrations increase and products concentrations decrease.
Writing the Reaction Quotient from the Balanced Equation
Problem: Write the reaction quotient for each of the following reactions:(a) The thermal decomposition of potassium chlorate:
KClO3 (s) KCl(s) + O2 (g)
(b) The combustion of butane in oxygen:C4H10 (g) + O2 (g) CO2 (g) + H2O(g)
Plan: We first balance the equations, then construct the reaction quotient
Solution:
(a) 2 KClO3 (s) ⇌ 2 KCl(s) + 3 O2 (g) Qc = = [O2]3[KCl]2[O2]3
[KClO3]2
(b) 2 C4H10 (g) + 13 O2 (g) ⇌ 8 CO2 (g) + 10 H2O(g)
Qc =[CO2]8 [H2O]10
[C4H10]2 [O2]13
Reaction Direction and the Relative Sizes of Q and K
“Excess reactantsinitially”
“Excess productsinitially”
Example: Calculating Equilibrium pressures and concentrations from K
and initial conditions.Consider the equilibrium: CO(g) + H2O(g) CO2(g) + H2(g)
0.250 mol CO and 0.250 mol H2O are placed in a 125 mLflask at 900 K. What is the composition of the equilibrium mixture if K = 1.56?
The original reactant concentrations are:
[CO]0 = [H2O]0 = 0.250 mol/ 0.125 L = 2.00 M
Q = 0.Therefore, Q < K, so reactants are consumed and products made.
CO(g) + H2O(g) CO2(g) + H2(g)
Construct the reaction table:
Conc. (M)
CO(g) H2O(g) CO2(g) H2(g)
Init. [i] 2.00 2.00 0 0
Conc. (M)
CO(g) H2O(g) CO2(g) H2(g)
Init. [i] 2.00 2.00 0 0
Change = x times coefficient= neg. for reactants
-x -x +x +x
CO(g) + H2O(g) CO2(g) + H2(g)
Conc. (M)
CO(g) H2O(g) CO2(g) H2(g)
Init. [i] 2.00 2.00 0 0
Change= x times coefficient= neg. for reactants
-x -x +x +x
Eqbm. [i]= Init.+Change
2.00 - x 2.00 - x 0+x = x 0+x = x
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Called an “I.C.E.” table.
Taking the square root of both sides:
= 1.56 1.25 2.00
xx
= ±−
Substitute [i]eqbm into the equilibrium expression:
[ ][ ][ ][ ]
( )( )( )( ) ( )
22 2
22
CO H1.56
CO H O 2.00 2.00 2.00
x x xKx x x
= = = =− − −
Since only the positive root meaningful, ignore the negative root:
1.252.00
1.11 M
xx
x
=−
=
[ ] [ ][ ] [ ]
2
2 2
Calculating equilibrium concentrations:CO H O 2.00 2.00 M - 1.11 M 0.89 M
CO H 1.11 M
x
x
= = − = =
= = =
[ ][ ][ ][ ]
2 2
2
Check results:CO H 1.11 1.11 1.6 1.56CO H O 0.89 0.89
K ×= = = ≈
×
Solving Equilibrium Problems
• Write the balanced equation for the reaction.• Write the equilibrium expression K.• List the initial concentrations, [i]initial.• Calculate Q and determine the direction of shift to
equilibrium.• Define change in [i] to reach equilibrium:
Change = x times the stoich. coefficient• Express equilibrium concentrations in terms of x. • Substitute the equilibrium concentrations into the
equilibrium expression for K.• Solve for x and calculate [i]eqbm for all i. • Check the solution by calculating K and making sure
it is identical to the original K.
Like Example 6.3 (P210) - ICalculate the equilibrium concentrations when Hydrogen Chloride gas ismade from Hydrogen gas and Chlorine gas. Initially 4.000 mol of H2, and4.000 mol of Cl2, are added to 2.000 mol of gaseous HCl in a 2.000 liter flask.
H2 (g) + Cl2 (g) 2 HCl(g)
K = 2.76 x 102 = [Cl2] = [H2] = 4.000mol/2.000L = 2.000M[HCl] = 2.000 mol/2.000L = 1.000M
Initial Concentration Change Equilibrium Conc. (mol/L) (mol/L) (mol/L)
[H2]o = 2.000M -x [H2] =
[Cl2]o = 2.000M [Cl2] =
[HCl]o = 1.000M [HCl] =
[HCl]2
[H2] [Cl2]
Like Example 6.3 (P210) - ICalculate the equilibrium concentrations when Hydrogen Chloride gas ismade from Hydrogen gas and Chlorine gas. Initially 4.000 mol of H2, and4.000 mol of Cl2, are added to 2.000 mol of gaseous HCl in a 2.000 liter flask.
H2 (g) + Cl2 (g) 2 HCl(g)
K = 2.76 x 102 = [Cl2] = [H2] = 4.000mol/2.000L = 2.000M[HCl] = 2.000 mol/2.000L = 1.000M
Initial Concentration Change Equilibrium Conc. (mol/L) (mol/L) (mol/L)
[H2]o = 2.000M -x [H2] = 2.000-x
[Cl2]o = 2.000M -x [Cl2] = 2.000-x
[HCl]o = 1.000M +2x [HCl] = 1.000 + 2x
[HCl]2
[H2] [Cl2]
Like Example 6.3 (P210) - II
K = 2.76 x 102 = =
Solve for x:
[HCl]2
[H2] [Cl2]
Like Example 6.3 (P210) - II
K = 2.76 x 102 = = =[HCl]2
[H2] [Cl2](1.000 + 2x)2
(2.000 –x)(2.000 – x)(1.000 + 2x)2
(2.000 – x)2
Take the square root of each side:
16.61 =(1.000 + 2x)(2.000 – x)
33.22 – 16.61x = 1.000 + 2x Therefore: [H2] = 0.269 M32.22 = 18.61x [Cl2] = 0.269 M
x = 1.731 [HCl] = 4.462 M
Check:
= = 276 OK![HCl]2
[H2] [Cl2](4.462)2
(0.269)(0.269)
Example: Calculate equilibrium partial pressures using exact solution of Quadratic Equation
-42 2
The reaction between nitrogen and oxygen to form nitric acidproceeds according to the reaction:
N (g) + O (g) 2NO(g) K 4.10 10 at 2000. K= ×
2 20.500 moles of N and 0.860 moles of O are put into a 2.00 L vessel initially. Calculate the partial pressure of all the species at equilibrium.
⇌
2
2
Initial concentration of N = 0.500 mol 2.00 L = 0.250 M,Initial concentration of O = 0.860 mol 2.00 L = 0.430 M,Initial concentration of NO 0Q =0; therefore the reaction proceeds to the rightConstruct
=
the reaction table:
Molarity NO(g)
Initial 0.250 0.430 02N (g) 2O (g)
2 2N (g) + O (g) 2NO(g)⇌
Molarity N2(g) O2(g) NO(g)
Initial 0.250 0.430 0
change -x -x 2x
2 2N (g) + O (g) 2NO(g)⇌
2 2N (g) + O (g) 2NO(g)
Molarity N2(g) O2(g) NO(g)
Initial 0.250 0.430 0
change -x -x 2x
At equib 0.250-x 0.430-x 2x
⇌
[ ][ ][ ]
( )( )( )
2 24
2 2
Substituting the equilibrium concentrations from the table into the equilibrium expression:
NO 24.10 10
N O 0.250 0.430x
Kx x
−= = = ×− −
2
2
This is a quadratic equation of the general form0
where the (two) roots can be obtained from the quadratic formula:
4x2
ax bx c
b b aca±
+ + =
− ± −=
2 4 5
This expression simplifies to4.00 2.78 10 4.41 10 0x x− −+ × − × =
[ ]
-3 -3
We obtain two possible solutionsx 3.35 10 and x 3.28 10Since only the positive root leads to all positive concentrations, we ignore the negative root.Calculating equilibrium concentrations:NO
= − × = ×
[ ][ ]
[ ][ ][ ]
( )
-3
2
2
22 -34 4
2 2
2 6.56 10
N 0.250 0.247
O 0.430 0.427Check:
6.56 10NO4.09 10 4.10 10
N O 0.247 0.427
x M
M x M
M x M
K − −
= = ×
= − =
= − =
×= = = × ≈ ×
×
Example: Solving equilibrium problems with simplifying assumptions
Phosgene decomposes into CO and Cl2 when heated according to the equation.
Calculate the concentration of all species at equilibrium if 5.00 moles of phosgene is placed into a 10.0 L flask
-42 2COCl (g) CO(g) + Cl (g) K = 8.3 10 at 360 C× o
[ ]25.00 molesCOCl (g) 0.500
10.0 LM= =
⇌
MolarityInit. 0.500 0 0
Change
Equil.
-42 2COCl (g) CO(g) + Cl (g) K = 8.3 10 at 360 C× o
2COCl (g) CO(g) 2Cl (g)
⇌
MolarityInit. 0.500 0 0
Change -x +x +x
Equil.
-42 2COCl (g) CO(g) + Cl (g) K = 8.3 10 at 360 C× o
2COCl (g) CO(g) 2Cl (g)
⇌
MolarityInit. 0.500 0 0
Change -x +x +x
Equil. 0.500 - x x x
-42 2COCl (g) CO(g) + Cl (g) K = 8.3 10 at 360 C× o
2COCl (g) CO(g) 2Cl (g)
⇌
[ ][ ][ ]
242
2
2
2 24 2 2
2 2
CO Cl8.3 10
COCl 0.500
For K to be so small, 0.500 and so 0.500 0.500
8.3 10 2.037 10 2.04 100.500 0.500Exact solution is 1.996 10 2.00 10Approximation gives error o
xKx
x x xx x x
xx
−
− − −
− −
= = = ×−
− >> − ≈
≈ = × = × → ×−
= × → ×f 2%
Calculating K from Concentration Data–IProblem: Hydrogen iodide decomposes at moderate temperatures by thereaction below:
When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibriummixture was found to contain 0.442 mol I2. What is the value of Kc ?Plan: First we calculate the molar concentrations, and then put them intothe equilibrium expression to find it’s value.Solution: To calculate the concentrations of HI and I2, we divide the amounts of these compounds by the volume of the vessel.
2 HI(g) H2 (g) + I2 (g)
Starting conc. of HI = = 0.800 M4.00 mol5.00 L
Equilibrium conc. of I2 = = 0.0884 M0.442 mol5.00 L
Conc. (M) 2HI(g) H2 (g) I2 (g)
Starting 0.800 0 0ChangeEquilibrium 0.0884
Calculating K from Concentration Data–IProblem: Hydrogen iodide decomposes at moderate temperatures by thereaction below:
When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibriummixture was found to contain 0.442 mol I2. What is the value of Kc ?Plan: First we calculate the molar concentrations, and then put them intothe equilibrium expression to find it’s value.Solution: To calculate the concentrations of HI and I2, we divide the amounts of these compounds by the volume of the vessel.
2 HI(g) H2 (g) + I2 (g)
Starting conc. of HI = = 0.800 M4.00 mol5.00 L
Equilibrium conc. of I2 = = 0.0884 M0.442 mol5.00 L
Conc. (M) 2HI(g) H2 (g) I2 (g)
Starting 0.800 0 0Change - 2x x xEquilibrium 0.800 - 2x x x = 0.0884
Calculating K from Concentration Data–II
[HI] =
[H2] =
Kc = = =[H2] [I2]
[HI]2
Calculating K from Concentration Data–II
[HI] = M = (0.800 - 2 x 0.0884) M = 0.623 M
[H2] = x = 0.0884 M = [I2]
Kc = = = 0.0201[H2] [I2]
[HI]2
( 0.0884)(0.0884)(0.623)2
Therefore the equilibrium constant for the decomposition of HydrogenIodide at 458°C is only 0.0201 meaning that the decomposition does notproceed very far under these temperature conditions. We were given the initial concentrations, and that of one at equilibrium, and found the others that were needed to calculate the equilibrium constant.
Using the Quadratic Formula to Solve for the UnknownGiven the Reaction between CO and H2O:Concentration (M) CO(g) + H2O(g) CO2(g) + H2(g)
Initial 2.00 1.00 0 0Change -x -x +x +xEquilibrium 2.00-x 1.00-x x x
Qc = = = = 1.56[CO2][H2][CO][H2O]
(x) (x)(2.00-x)(1.00-x)
x2
x2 - 3.00x + 2.00We rearrange the equation: 0.56x2 - 4.68x + 3.12 = 0
ax2 + bx + c = 0quadratic equation:
x = - b + b2 - 4ac2a
x = = 7.6 Mand 0.73 M
4.68 + (-4.68)2 - 4(0.56)(3.12)2(0.56)
[CO] = 1.27 M[H2O] = 0.27 M[CO2] = 0.73 M
[H2] = 0.73 M
Le Chatelier’s Principle
If a change in conditions (a ‘stress’ such as change in P, T, or concentration) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions.
Henri Le Chatelier, 1884
The Effect of a Change in Concentration–IGiven an equilibrium equation such as :
CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)
If one adds ammonia to the reaction mixture at equilibrium, it will force the reaction to go to the right producing more product. Likewise, if one takes ammonia from the equilibrium mixture, it will force the reaction back to produce more reactants by recombining H2 and HCN to givemore of the initial reactants, CH4 and NH3.
CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)
CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)
Add NH3
Forces equilibrium to produce more product.
Forces the reaction equilibrium to go back to the left and produce more of the reactants.
Remove NH3
The Effect of a Change in Concentration–II
CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)
If to this same equilibrium mixture one decides to add one of the products to the equilibrium mixture, it will force the equilibrium backtoward the reactant side and increase the concentrations of reactants.Likewise, if one takes away some of the hydrogen or hydrogen cyanidefrom the product side, it will force the equilibrium to replace it.
CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)
CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g)
Add H2Forces equilibrium to go toward the reactant direction.
Remove HCNForces equilibrium to make more produce and replace the lost HCN.
The Effect of a Change in Concentration
• If a gaseous reactant or product is added to a system at equilibrium, the system will shift in a direction to reduce the concentration of the added component.
• If a gaseous reactant or product is removed from a system at equilibrium, the system will shift in a direction to increase the concentration of the removed component.
[ ] [ ][ ][ ][ ][ ]
2 32
1 2 1 3
312
1 2
2NO (g) NO(g) + NO (g)
NO NO NO
NO NO
NO
k k
kk
−
−
=
=
⇌
Examples: The Effect of a Change in Concentration
Consider the following reaction:
2 H2S(g) + O2(g) ⇌ 2 S(s) + 2 H2O(g)
What happens to:
(a) [H2O] if O2 is added?
The reaction proceeds to the right so H2O increases.
(b) [H2S] if O2 is added?
Some H2S reacts with the added O2 to move the reaction to the right, so [H2S] decreases.
2 H2S(g) + O2(g) ⇌ 2 S(s) + 2 H2O(g)
(c) [O2] if H2S is removed?
The reaction proceeds to the left to re-form H2S, more O2 is formed as well, O2 increases.
(d) [H2S] if S(s) is added?
S is a solid, so its activity does not change. Thus, [H2S] is unchanged.
The Effect of a Change in Pressure (Volume)Pressure changes are mainly involving gases as liquids and solidsare nearly incompressible. For gases, pressure changes can occur in three ways:
Changing the concentration of a gaseous component
Adding an inert gas (one that does not take part in the reaction)
Changing the volume of the reaction vessel
When a system at equilibrium that contains a gas undergoes a changein pressure as a result of a change in volume, the equilibrium position shifts to reduce the effect of the change.
If the volume is lower (pressure is higher), the total number of gas molecules decrease.
If the volume is higher (pressure is lower), the total number of gasmolecules increases.
Examples: The Effect of a Change in Pressure
How would you change the total pressure or volume in the following reactions to increase the yield of the products:
(a) CaCO3(s) ⇌ CaO(s) + CO2 (g)
The only gas is the product CO2. To move the reaction to the right increase the volume.
(b) S(s) + 3 F2 (g) ⇌ SF6 (g)
With 3 moles of gas on the left and only one on the right, we increase the pressure to form more SF6.
(c) Cl2(g) + I2(g) ⇌ 2 ICl (g)
The number of moles of gas is the same on both sides of the equation, so a change in pressure or volume will have no effect.
Figure 6.9: Brown NO2(g) and colorless N2O4(g) at equilibrium in a syringe
Source: Ken O’Donoghue
2 NO2 (g) N2O4 (g)
Brown Colorless
The Effect of a Change in Temperature
Only temperature changes will alter the equilibrium constant, and that iswhy we always specify the temperature when giving the value of Kc.
The best way to look at temperature effects is to realize that energy is a component of the equation, the same as a reactant, or product. For example, if you have an exothermic reaction, heat (energy) is on the product side of the equation, but if it is an endothermic reaction, it will be on the reactant side of the equation.
O2 (g) + 2 H2 (g) 2 H2O(g) + Energy = Exothermic
Electrical energy + 2 H2O(g) 2 H2 (g) + O2 (g) = Endothermic
A temperature increase favors the endothermic direction and a temperature decrease favors the exothermic direction.
Shifting the N2O4(g) and 2NO2(g) equilibrium by changing the temperature
Exothermic Reactions• Release energy upon reaction. Treat energy as a
reaction product. Use Le Chatelier’s principle.
• Increasing T adds energy to system. The equilibrium shifts towards the reactants. (System absorbs energy)
• The value of K decreases in consequence.
• Note that change in concentration at constant T changes equilibrium position, but not value of K.
Endothermic Reactions
• Absorb energy upon reaction. Treat energy as a reactant. Use Le Chatelier’s principle.
• Increasing T adds energy to system. The equilibrium shifts towards the products. (System absorbs energy)
• The value of K increases in consequence.
Example: How does an increase in temperature affect the equilibrium concentration of the indicated substance and K for the following reactions:
(a) CaO(s) + H2O (l) ⇌Ca(OH)2 (aq) + energy
Increasing T adds thermal energy. This shifts the system to the left, where it absorbs energy. [Ca(OH)2] and K decrease.
(b) CaCO3 (s) + energy ⇌ CaO(s) + CO2 (g)Increasing T adds thermal energy. This shifts the system to the
right, where it absorbs energy.[CO2] and K increase.
(c) SO2 (g) + energy ⇌ S(s) + O2(g)
Increasing T adds thermal energy. This shifts the system to the right, where it absorbs energy.[SO2] will decrease and K increases.
Effect of Various Stresses on an Equilibrium SystemStress Net Direction of Reaction Effect on Value of K
ConcentrationIncrease [reactant] Toward formation of products NoneDecrease [reactant] Toward formation of reactants None
Volume (of closed container holding reaction at constant T)Increase V Toward formation of larger
amount (mol) of gas unless Δn = 0 None
Decrease V Toward formation of smalleramount (mol) of gas unless Δn = 0 None
Temperature - Adding energy in form of heat (Opposite if T decrease…):Increase T Toward formation of products if energy K increases
is a “reactant” (endothermic reaction) with T increase
Toward formation of reactants if energy K decreasesis a “product” (exothermic reaction) with T increase
CoCl42-(aq) + 6 H2O(l) ⇌ Co(H2O)6
2+(aq) + 4 Cl-(aq)blue pink
[Co(H2O)62+][Cl-]4
[CoCl42-]Q = , which initially = K
DEMO!
Ways of Expressing the Reaction Quotient, Q
Form of Chemical Equation Form of Q Value of K
Reference reaction: A B Q(ref) = K(ref) =
Reverse reaction: B A Q = = K =
Reaction as sum of two steps:
[B][A]
[B]eq
[A]eq1 [A]Q(ref) [B]
1K(ref)
(1) A C
(2) C BQoverall = Q1 x Q2 = Q(ref) Koverall = K1 x K2
= x = = K(ref)[C] [B] [B][A] [C] [A]
Coefficients multiplied by n Q = Qn(ref) K = Kn
(ref)
[A] [C][C] [B]Q1 = ; Q2 =
Reaction with pure solid or Q’ = Q(ref)[A] = [B] K’ = K(ref)[A] = [B]liquid component, such as A(s)
Predicting the Effect of a Change in Concentration on the Position of the Equilibrium
Problem: Carbon will react with water to yield carbon monoxide andand hydrogen, in a reaction called the water gas reaction that was used to convert coal into a fuel that can be used by industry.
C(s) + H2O (g) CO(g) + H2 (g)What happens to:(a) [CO] if C is added? (c) [H2O] if H2 is added?(b) [CO] if H2O is added? (d) [H2O] if CO is removed?Plan: We either write the reaction quotient to see how equilibrium will be effected, or look at the equation, and predict the change in direction of the reaction, and the effect of the material desired.Solution: (a) No change, as carbon is a solid, and not involved in the equilibrium, as long as some carbon is present to allow the reaction.(b) The reaction moves to the product side, and [CO] increases.(c) The reaction moves to the reactant side, and [H2O] increases.(d) The reaction moves to the product side, and [H2O] decreases.
