Chapter Thirteen: CHEMICAL EQUILIBRIUM p578
Chapter Thirteen:
CHEMICALEQUILIBRIUM
p578
Contents p578
13-1 The Equilibrium Condition p579
The violet solution in the center is at 25 ℃ and contains
significant quantities of both pink Co(H2O)62+ and blue
CoCl42-. When the solution is cooled, it turns pink because
the equilibrium is shifted to the left. Heating the solution
favors the blue CoCl42-.
The effect of temperature on the endothermic,
aqueous equilibrium:
Co(H2O)62+ + 4Cl- CoCl4
2- + 6H2O
The concept of chemical equilibrium is analogous to theflow of cars across a bridge connecting two island cities.The traffic flow is the same in both directions. So, theresult is no net change in the car population on the bridge
Figure 13.1
A molecular representation of the reaction 2NO2(g) → N2O4(g) over
time in a closed vessel. Note that the number of NO2 and N2O4 in the
container become constant (c and d) after sufficient time has passed.
p580
Chemical Equilibrium
The state where the concentrations of all reactants
and products remain constant with time.
On the molecular level, there is frantic activity.
Equilibrium is not static, but is a highly dynamic
situation.
p580
Equilibrium is: Macroscopically static.
Microscopically dynamic.
The changes in concentrations with time p580
Figure 13.3
(a) H2O and CO are mixed in equal numbers and begin to
react to form CO2 and H2. After time has passed,
equilibrium is reached (c) and the numbers of reactant
and product molecules then remain constant over time (d).
p581
The system has reached equilibrium.
p581The Changes with Time in the Rates ofForward and Reverse Reactions
The Characteristic of Chemical Equilibrium p582
Changes in Concentration
N2(g) + 3H2(g) 2NH3(g)
13-2 The Equilibrium Constant p582
Law of mass action
The square brackets indicate the concentrations of the chemical
species at equilibrium and K is a constant called the equilibrium
constant.
j A + k B lC + mD
)(6)(4)(7)(4 2223 gOHgNOgOgNH
P583Ex 13.1 Writing Equilibrium Expressions
Write the equilibrium expression for the following reaction:
solution
Applying the law of mass action gives
72
43
62
42
][O][NHO][H][NO
KCoefficient of O2
Coefficient of H2O
Ex 13.2 Calculating the Values of K
)(H3)(N)(2NH 223 ggg
P583The following equilibrium concentrations were observedfor the Haber process at 127℃:
[NH3] = 3.1 × 10 -2 mole/L; [N2] = 8.5 × 10 -1 mole/L
[H2] = 3.1 × 10 -3 mole/L
a. Calculate the value of K at 127℃ for this reaction.
b. Calculate the value of the equilibrium constant at
127℃ for the reaction.
c. Calculate the value of the equilibrium constant at
127℃ for the reaction given by the equation.
)(NH)(H23
)(N21
322 ggg
solutiona. The balanced equation for the Haber process is
N2(g) + 3H2(g) 2NH3(g)
b.
c.
p584
We can summarize theseconclusions about the equilibriumexpression as follows:
p584
Ex 13.3 Equilibrium Positions P585
The following results were collected for twoexperiments involving the reaction at 600℃ betweengaseous sulfur dioxide and oxygen to form gaseoussulfur trioxide:Show that the equilibrium constant is the samein both cases.
For Experiment 1,
For Experiment 2,
p586
13-3 Equilibrium Expressions InvolvingPressures
p586
For the ammonia synthesis, the equilibrium expressioncan be written in terms of concentrations, that is,
or in terms o the equilibrium partial pressures ofthe gases, that is,
3
2
))((22
3
HN
NHp PP
pK
)(2)()(2 2 gNOClgClgNO
P587Ex 13.4 Calculating Values of KP
The reaction for the formation of nitrosyl chloride
was studied at 25℃. The pressures at equilibrium
were found to be PNOCl = 1.2 atm; PNO = 5.0 × 10 -2
atm; PCl2= 3.0 × 10 -1 atm. Calculate the value
of KP for this reaction at 25℃.
Solution
3122
2
Cl2
NO
2NOCl 109.1
)100.3()100.5()2.1(
)()(22
PPP
k p
The relationship between K and Kp
p587
When the powers in the numerator is different fromthat in the denominator, and K does not equal Kp.
p587
p588
j A + k B lC + mD
For the general reaction
The relationship between K and Kp isn
p RTKK )(
P588Ex 13.5 Calculating K from KPUsing the value of KP obtained in SampleExercise 13.4, calculate the value of K at 25 ℃for the reaction
)(2)()(2 2 gNOClgClgNO Solution
Δn = 2 - (2 + 1)
Sum of product
Sum of reactant
13-4 Heterogeneous Equilibria
2
12'
C][CO C
K )(CO)CaO()(CaCO 23 gss
][CaCO][CaO][CO
3
2' K
p589
]CO[ 21
'2 KC
KC
P590Ex 13.6 Equilibrium Expressions forHeterogeneous Equilibria
Write the expressions for K and KP for thefollowing process:
a. Solid phosphorus pentachloridedecomposes to liquid phosphorustrichloride and chlorine gas.
b. Deep blue solid copper (II) sulfatepentahydrate is heated to drive off watervapor to from white solid copper (II) sulfate.
p590
13-5 Applications of the Equilibrium Constantp591
p592
The extent of a Reactionp592
The inherent tendency for a reaction to occur is indicated by
the magnitude of the equilibrium constant. A value of K
much than 1 means that equilibrium the reaction system will
consist of mostly products-the equilibrium lies to right.
Another way of saying this is that reactions with very large
equilibrium to completion. On the other hand, a very small
value of K means that the system at equilibrium will consist
of mostly reactants- the equilibrium position is far to the left.
The given reaction does not occur to any significant extent.
Reaction Quotientp593
Q is equal to K. The system is at equilibrium; no shiftwill occur.Q is greater than K. In this case, the ratio of initialconcentrations of products to initial concentrations ofreactants is too large. To reach equilibrium, a netchange of products to reactants must occur. The systemshifts to the left, consuming products and formingreactants, until equilibrium is achieved.Q is less than K. In this case, the ratio of initialconcentrations of products to initial concentrations ofreactants of reactants is too small. The system mustshift to the right, consuming reactants and formingproducts, to attain equilibrium.
P593Ex 13.7 Using the Reaction Quotient
For the synthesis of ammonia at 500℃, theequilibrium constant is 6.0 × 10 -2. Predictthe direction in which the system will shift toreach equilibrium in each of the followingcases:
a. [NH3]o = 1.0 × 10 -5 M;[N2]o = 1.0 × 10 -5 M;
[H2]o = 2.0 × 10 -3 M
b. [NH3]o = 2.00 × 10 -4 M;[N2]o = 1.5 × 10 -5 M;
[H2]o = 3.54 × 10 -1 M
c. [NH3]o = 1.0 × 10 -4 M;[N2]o = 5.0 M;
[H2]o = 1.0 × 10 -2 M
Solution:p594
(a)
Since Q > K the system will shift to left.
(b)
(c)
Since Q < K, so the system will shift to right.
In this case Q = K, so the system is at equilibrium.No shift will occur.
Ex 13.8 Calculating equilibrium Pressures Ip594
P595Ex 13.9 Calculating Equilibrium Pressures II
At a certain temperature a 1.00-L flask initially
contained 0.298 mol PCl3(g) and 8.70 × 10 -3 mol
PCl5(g). After the system had reached equilibrium, 2.00
× 10 -3 mol Cl2(g) was found in the flask. Gaseous PCl5
decomposes according to the reaction
)()()( 235 gClgPClgPCl
Calculate the equilibrium concentrations of all species and
the value of K.
p595
Solution
The equilibrium expression for this reaction is
][PCl]][PCl[Cl
3
32K
p596
)(H)(CO)O(H)CO( 222 gggg
P596Ex 13.10 Calculating EquilibriumConcentrations I
Carbon monoxide reacts with stream to produce
carbon. Dioxide and hydrogen. At 700K the
equilibrium concentrations of all species if 1.000
mol of each component is mixed in a 1000-L flask.Solution
The balanced equation for the reaction is
5.0O][CO][H
]][H[CO
2
22 Kand
p597
p598
Check:
Assume that the reaction for the formation of
gaseous hydrogen and fluorine has an equilibrium
constant of 1.15 × 102 at a certain temperature. In a
particular experiment, 3.000 mol of each component
was added to a 1.500-L flask. Calculate the
equilibrium concentrations of all species.
P598
Ex 13.11 Calculating EquilibriumConcentrations II
Solution: p598
p599
13-6 Solving Equilibrium Problems p600
P601Ex 13.12 Calculating Equilibrium Pressures
Assume that gaseous hydrogen iodide is synthesized
from hydrogen gas and iodine vapor at a temperature
where the equilibrium constant is 1.00 × 102. Suppose
HI at 5.000 × 10 -1 atm, H2 at 1.000 × 10 -2 atm, and I2 at
5.000 × 10 -3 atm are mixed in a 5.000-L flask.
Calculate the equilibrium pressures of all species.
p602Solution:
Le Châtelier’s Principle
If a change is imposed on a system at equilibrium,
the position of the equilibrium will shift in a
direction that tends to reduce that change.
Equilibrium Decomposition of N2O4
Treating Systems That Have Small EquilibriumConstants
)(Cl)(NO2 2 gg )(NOCl2 g
p603
and
The initial concentrations are
0][Cl0[NO]0.50L2.0
mol1.0NOCl][ 0200 M
522
2
106.1[NOCl]
][Cl[NO] K
P604
p604
Table 13.2 The percent by mass of NH3at
equilibrium in a mixture of N2, H2, NH3 as a
function of temperature and total pressure
13-7 LeChâtelier’s Principle
Effects of Changes on the System1. Concentration: The system will shift away from
the added component.
2. Temperature: K will change depending upon thetemperature (treat the energy change as areactant).
3. Pressure:(a) Addition of inert gas does not affect the
equilibrium position.(b) Decreasing the volume shifts the equilibrium
toward the side with fewer moles.
p605
LeChâtelier’s Principle
The Effect of a Change in Concentration p605
As expected, Q is less than K because the concentration ofN2 was increased. The system will shift to come to theequilibrium position. Rather than do the calculations, wesimply summarizes the results.
p605
Figure 13.8
(a) The initial equilibrium mixture of N2, H2, and NH3. (b)The new equilibrium position for the system containingmore N2 (due to addition of N2), less H2, and more than in(a).
P606Ex 13.13 Using Le Chatelier’s Principle IArsenic can be extracted from its ores by first reacting the
ore with oxygen (called roasting) to form solid As4O6,
which is then reduced using carbon: )(6)()(6)( 464 gCOgAssCsOAs
Predict the direction of shift of the equilibrium position
in response to each of the following changes in
conditions. a. Addition of carbon monoxide
b. Addition or removal of carbon ortetraarsenic hexoxide (As4O6)
c. Removal of gaseous arsenic (As4)
Solution:p607
The effect of a Change in Pressurep607
The volume of the container is changed p607
The volume of a gas is directly proportional tothe number of moles of gas present.
(a) (b) (c)
p608
)(4)(6)( 324 lPClgClsP
)()()( 523 gPClgClgPCl
P608Ex 13.14 Using Le Chatelier’s Principle II
Predict the shift in equilibrium position that will occur for
each of the following processes when the volume is reduced.
(a) The preparation of liquid phosphorus trichloride by the
reaction.
(b) The preparation of gaseous phosphorus pentachloride
according to the equation.
(c) The reaction of phosphorus trichloride with ammonia:
)(3)()()(3)( 3233 gHClgNHPgNHgPCl
Solution: p609
The Effect of a Change in Temperature p609
Shifting the N2O4(g) → 2NO2(g) equilibrium by changingthe temperature. (a) At 100 ℃ the flask is definitely red-dishbrown due to a large amount of NO2 present. (b) at 0℃ theequilibrium is shifted toward colorless N2O4(g).
p610
For an endothermic reaction:
As T Cause the equilibrium to shift to theright and the value of K to increase.
p610
)(2)()( 22 gNOgOgN
P610Ex 13.16 Using Le Chatelier’s Principle II
For each of the following reactions, predict how the value
of K changes as the temperature is increased.
a. ΔHo = 181 KJ
)(2)()(2 322 gSOgOgSO b. ΔHo = - 198 kJ
Solution: