Page 1
AP* Chemistry CHEMICAL EQUILIBRIA: GENERAL CONCEPTS
*AP is a registered trademark of the College Board, which was not involved in the production of this product.© 2013 by René McCormick. All rights
reserved.
THE NATURE OF THE EQUILIBRIUM STATE: Equilibrium is the state where the rate of the forward
reaction is equal to the rate of the reverse reaction. At these conditions, concentrations of all reactants and
products remain constant with time once equilibrium has been established at constant temperature.
In stoichiometry, we assumed chemical systems “went to completion”. A chemist would be quick to point out
that the vast majority of chemical reactions do not go to completion; they just have very large K values! Never,
ever forget that equilibrium is TEMPERATURE DEPENDENT.
Reactions are reversible. This is indicated by double arrows.
Dynamic equilibrium-- indicates that the reaction is proceeding in the forward and in the reverse
direction simultaneously and once equilibrium is established, the rate of each direction is equal. This
also keeps the concentration of reactants and products constant (which is not to be confused with
“equal”).
The nature and properties of the equilibrium state are the same, no matter what the direction of
approach.
Examples: Look at the following plot of the reaction between steam and carbon monoxide in a closed
vessel at a high temperature where the reaction takes place rapidly.
THE EQUILIBRIUM POSITION
Whether the reaction lies far to the right (favors products) or
to the left (favors reactants) depends on three main factors:
Initial concentrations (more collisions--faster
reaction)
Relative energies of reactants and products (nature
goes to minimum energy)
Degree of organization of reactants and products
(nature goes to maximum disorder)
The significance of K:
K > 1 means that the reaction favors the products at equilibrium
K < 1 means that the reaction favors the reactants at equilibrium
THE EQUILIBRIUM EXPRESSION: A general description of the equilibrium condition proposed by
Gudberg and Waage in 1864 is known as the Law of Mass Action. Equilibrium is temperature dependent,
however, it does not change with concentration or pressure.
equilibrium constant expression--for the general reaction
aA + bB cC + dD
Equilibrium constant: K = [C]c[D]
d * Note* K, Kc, Keq may all be used here!
[A]a[B]
b
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Chemical Equilibria: General Concepts
2
The product concentrations appear in the numerator and the reactant concentrations in the denominator.
Each concentration is raised to the power of its stoichiometric coefficient in the balanced equation.
- [ ] indicates concentration in Molarity (mol/L)
- Kc--is for concentration (aqueous)
- Kp--is for partial pressure (gases)
- “K” values are often written without units
USING EQUILIBRIUM CONSTANT EXPRESSIONS
Pure solids--do not appear in expression—you’ll see this in Ksp problems soon!
Pure liquids--do not appear in expression—H2O(l) is pure, so leave it out of the calculation
Water--as a pure liquid or reactant, does not appear in the expression. (55.5 M will not change
significantly)
o Weak acid and weak base equations are heterogeneous [multi-states of matter; pure liquid and
aqueous components] equilibria.
o Solubility of salts also fits into this category. The initial solid component has a constant
concentration and is therefore left out of the equilibrium expression.
Exercise 1 Writing Equilibrium Expressions
Write the equilibrium expression for the following reaction:
4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O(g)
K = [NO2]4[H2O]
6
[NH3]4[O2]
7
Exercise 2 Equilibrium Expressions for Heterogeneous Equilibria
Write the expressions for K and Kp for the following processes:
a. The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas.
b. Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II) sulfate.
A: K = [Cl2]
Kp = PCl2
B: K = [H2O]5
Kp = PH2O5
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Chemical Equilibria: General Concepts
3
CHANGING STOICHIOMETRIC COEFFICIENTS
when the stoichiometric coefficients of a balanced equation are multiplied by some factor, the K is
raised to the power of the multiplication factor (Kn). Thus, 2x is K squared; 3x is K cubed; etc.
REVERSING EQUATIONS
take the reciprocal of K ( 1/K)
ADDING EQUATIONS
multiply respective Ks (K1 × K2 × K3 …)
Exercise 3 Calculating the Values of K
The following equilibrium concentrations were observed for the Haber process at 127°C:
[NH3] = 3.1 × 10 2 mol/L
[N2] = 8.5 × 10 1 mol/L
[H 2] = 3.1 × 10 3 mol/L
a. Calculate the value of K at 127°C for this reaction.
b. Calculate the value of the equilibrium constant at 127°C for the reaction:
2 NH3(g) N2(g) + 3 H2(g)
c. Calculate the value of the equilibrium constant at 127°C for the reaction given by the equation:
1
2 N2(g) + 3
2 H2(g) NH3(g)
A: K = 3.8 × 104
B: K’ = 2.6 × 10-5
C: K” = 1.9 × 10
2
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Chemical Equilibria: General Concepts
4
Exercise 4 Equilibrium Positions
The following results were collected for two experiments involving the reaction at 600°C between gaseous sulfur dioxide and oxygen
to form gaseous sulfur trioxide:
Show that the equilibrium constant is the same in both cases.
K1 = 4.36
K2 = 4.32
Kc & Kp--NOT INTERCHANGEABLE! Kp = Kc(RT)n
where n is the change in the number of moles of gas going from reactants to products:
n = total moles gas produced total moles gas reacting
o R = universal gas law constant 0.0821 L atm/ mol K
o T = temperature in Kelvin
Kc = Kp if the number of moles of gaseous product = number of moles of gaseous
reactant since (RT)n = (RT)
0 = 1
Kp = Kc(RT)n is often referred to as the “politically correct” (pc) equation to help you remember the
order of the Ks in the equation!
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Chemical Equilibria: General Concepts
5
Exercise 5 Calculating Values of Kp
The reaction for the formation of nitrosyl chloride
2 NO(g) + Cl2(g) 2 NOCl(g)
was studied at 25°C. 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°C.
= 1.9 × 103
Exercise 6 Calculating K from Kp
Using the value of Kp obtained in Sample Exercise 13.4, calculate the value of K at 25° C for the reaction:
2 NO(g) + Cl2(g) 2 NOCl(g)
= 4.6 × 104
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Chemical Equilibria: General Concepts
6
MAGNITUDE OF K--what does it mean anyway? When greater than one, formation of products is
favored. When less than one, formation of reactants is favored.
Can you...
1. …write an equilibrium constant expression?
2. …tell how K is changed if the stoichiometric coefficients are changed on an equation?
3. … tell how to find K for a summary equation?
4. …tell how K depends on the way equilibrium concentrations are expressed and how to
convert K in terms of Kc vs. Kp?
5. …explain what K is telling you about a reaction?
THE REACTION QUOTIENT
For use when the system is NOT at equilibrium.
For the general reaction
aA + bB cC + dD
Reaction quotient = Qc = [C]c[D]
d
[A]a[B]
b
Qc has the appearance of K but the concentrations are not necessarily at equilibrium.
1. If Q < K, the system is not at equilibrium: Reactants products to make Q = K at equil.
2. If Q = K, the system is at equilibrium.
3. If Q > K, the system is not at equilibrium: Reactants products to make Q = K at equil.
Quite useful for predicting what will happen under special conditions.
Exercise 7 Using the Reaction Quotient
For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 × 10-2. Predict the direction in which the system
will shift to reach equilibrium in each of the following cases:
a. [NH3]0 = 1.0 × 10-3 M; [N2]0 = 1.0 × 10-5 M; [H2]0 = 2.0 × 10-3 M
b. [NH3]0 = 2.00 × 10-4 M; [N2]0 = 1.50 × 10-5 M; [H2]0 = 3.54 × 10-1 M
c. [NH3]0 = 1.0 × 10-4 M; [N2]0 = 5.0 M; [H2]0 = 1.0 × 10-2 M
A: shift left
B: no shift
C: shift right
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Chemical Equilibria: General Concepts
7
SOME CALCULATIONS WITH THE EQUILIBRIUM CONSTANT
General steps for solving equilibrium problems.
1. Set up “RICE” TABLE—write a balanced reaction, place initial concentrations into the
table, determine the change in initial concentrations in terms of x, calculate the
equilibrium concentration expressions (0.25 M – x).
R = write a balanced reaction for the predominant reacting species
I = fill in the initial concentrations
C = determine the change that is taking in place in terms of x
E = express the equilibrium concentrations in terms of x
2. Set up the equilibrium expression and set it equal to its value, if given.
3. Celebrate if you are given equilibrium concentrations! Just skip down to the “E” line and
fill them in. You may be asked to work backwards to determine the “change” in
equilibrium.
4. If you are given a K value, then use it to solve for x and use x to calculate the equilibrium
concentrations.
Hints:
Look for very small K values (where K < 105), "x" may be negligible.
If "x" is large enough to impact the equilibrium values, then you must subtract it
from the initial concentration. Your math will be simplified if the problem is a
perfect square. If not, you must use the quadratic formula. You are allowed to use
your calculator with a solver or program for solving the quadratic.
If none of the initial concentrations are zero, then Q must be calculated first to
determine the direction of the shift before following the above general steps.
Exercise 8 Calculating Equilibrium Pressures I
Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions.
In the gas phase it decomposes to gaseous nitrogen dioxide:
N2O4(g) 2 NO2(g)
Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature
where Kp = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of
NO2(g).
= 0.600 atm
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Chemical Equilibria: General Concepts
8
Exercise 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
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all species and the value of K.
[Cl2] = 2.00 × 10-3
M
[PCl3] = 0.300 M
[PCl5] = 6.70 × 10-3
M
K = 8.96 × 10-2
Exercise 10 Calculating Equilibrium Concentrations I
Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10.
Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000-L flask.
[CO] = [H2O] = 0.613 M
[CO2] = [H2] = 1.387 M
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Chemical Equilibria: General Concepts
9
Exercise 11 Calculating Equilibrium Concentrations II
Assume that the reaction for the formation of gaseous hydrogen fluoride from 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.
[H2] = [F2] = 0.472 M
[HF] = 5.056 M
Exercise 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.
PHI = 4.29 × 10-1
atm
PH2 = 4.55 × 10-2
atm
PI2 = 4.05 × 10-2
atm
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Chemical Equilibria: General Concepts
10
EXTERNAL FACTORS AFFECTING EQUILIBRIA
Le Chatelier’s Principle: If a stress is applied to a system at equilibrium, the position of the
equilibrium will shift in the direction which reduces the stress.
Shifts occur to reestablish equilibrium positions. Think about K! Generally, products
reactants
Adding or removing a reagent causes the equilibrium to shift to reestablish K.
As long as you milk a “mamma” cow, she keeps making more milk!
Increasing pressure causes the equilibrium to shift to the side containing the fewest number of
moles of gas. The converse is also true.
Adding a catalyst to a reaction has causes no shift and has NO EFFECT on K. It just causes
equilibrium to be established faster!
Changing the temperature is a lot like adding or removing a “reactant or product”. Well, as
long as you think of heat energy as a “reactant” or “product”.
Think of it this way:
ENDOTHERMIC: A + B C + D + H value (heat is added into the system heat is a reactant)
Rewrite: A + B + heat C + D
EXOTHERMIC: A + B C + D H value (heat is lost from the system heat is a product)
Rewrite: A + B C + D + heat
Exercise 13 Using Le Chatelier’s Principle I
Arsenic 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:
As4O6(s) + 6 C(s) As4(g) + 6 CO(g)
Predict the direction of the 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 or tetraarsenic hexoxide (As4O6)
c. Removal of gaseous arsenic (As4)
A: shift left
B: no shift
C: shift right
Page 11
Chemical Equilibria: General Concepts
11
Exercise 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:
P4(s) + 6 Cl2(g) 4 PCl3(l)
b. The preparation of gaseous phosphorus pentachloride according to the equation:
PCl3(g) + Cl2(g) PCl5(g)
c. The reaction of phosphorus trichloride with ammonia:
PCl3(g) + 3 NH3(g) P(NH2)3(g) + 3 HCl(g)
A: shift right
B: shift right
C: no shift
Exercise 15 Using Le Chatelier’s Principle III
For each of the following reactions, predict how the value of K changes as the temperature is increased.
a. N2(g) + O2(g) 2 NO(g) H° = 181 kJ
b. 2 SO2(g) + O2(g) 2 SO3(g) H° = 198 kJ
A: increases
B: decreases
Page 12
AP* Chemistry
Solubility Equilibrium
*AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.© 2008 by René
McCormick. All rights reserved
SOLUBILITY EQUILIBRIA (The Solubility-Product Constant, Ksp)
We’ve got good news and we’ve got bad news… Which do you want first?
The good news: Solubility equilibrium is really simple.
The bad news: You know all those solubility rules that state a substance is insoluble? They are actually a little
bit soluble after all. Only the future attorneys among you read the fine print. Soluble is often defined as
“greater than 3 grams dissolving in 100 mL of water”. So, there is a lot of wiggle room for solubility up to 3
grams! This type of equilibria deals with that wiggle room. Apologies for “lying” to you.
If you can actually see that a salt is insoluble, usually evidenced by solid “stuff” sitting on the bottom of the
container, then the solution is actually saturated. Saturated solutions of salts present due to a chemical reaction
taking place present yet another type of chemical equilibria.
Slightly soluble salts establish a dynamic equilibrium with the hydrated
cations and anions in solution. Examine the formation of AgCl (s) as a
solution of AgNO3(aq) is squirted into a solution of NaCl(aq).
When the solid is first added to water, no ions are initially present.
As dissolution proceeds, the concentration of ions increases until
equilibrium is established. This occurs when the solution is
saturated.
The equilibrium constant, the Ksp, is no more than the product of the
ions in solution. (Remember, solids do not appear in equilibrium
expressions.)
For a saturated solution of AgCl, the equation would be:
AgCl(s) Ag+(aq) + Cl (aq)
The solubility product expression for the AgCl(s) precipitate would be:
Ksp = [Ag+][Cl ]
The AgCl(s) does not appear in the equilibrium expression since solids are left out. Why?
Because, the concentration of the solid remains relatively constant.
A table of Ksp values follows on the next page.
Page 13
Solubility Equilibria 2
Write the Ksp expression for each of the following reactions and find its value in the table above.
CaF2(s) Ca+2
+ 2 F Ksp =
Ag2CrO4(s) 2 Ag+ + CrO4
2 Ksp =
CaC2O4(s) Ca+2
+ C2O42
Ksp =
DETERMINING Ksp FROM EXPERIMENTAL MEASUREMENTS
In practice, Ksp values are determined by careful laboratory measurements using various spectroscopic
methods.
Remember STOICHIOMETRY? Surely, you’ve made peace with the concept by now…
Page 14
Solubility Equilibria 3
Example: Lead(II) chloride dissolves to a slight extent in water according to the equation below.
PbCl2(s) Pb+2
+ 2Cl
Calculate the Ksp if the lead ion concentration has been found to be 1.62 × 102M.
If lead’s concentration is x, then chloride’s concentration is 2x. So. . . .
Ksp = (1.62 × 102)(3.24 × 10
2)
2 = 1.70 × 10
5
Exercise 1 Calculating Ksp from Solubility I
Copper(I) bromide has a measured solubility of 2.0 × 104 mol/L at 25°C. Calculate its
Ksp value.
Ksp = 4.0 × 108
Exercise 2 Calculating Ksp from Solubility II
Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of
1.0 × 1015
mol/L at 25°C.
Ksp = 1.1 × 1073
Page 15
Solubility Equilibria 4
ESTIMATING SALT SOLUBILITY FROM Ksp
Relative solubilities can be deduced by comparing values of Ksp BUT, BE CAREFUL!
These comparisons can only be made for salts having the same ION:ION ratio.
Please don’t forget solubility changes with temperature! Some substances become less soluble in cold water
while other increase in solubility! Aragonite is an example.
Example: The Ksp for CaCO3 is 3.8 × 109 @ 25 C. Calculate the solubility of calcium carbonate in pure
water in (a) moles per liter & (b) grams per liter:
Exercise 3 Calculating Solubility from Ksp
The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4 × 107 at 25°C. Calculate its solubility at 25°C.
= 3.3 × 103 mol/L
Exercise 4 Solubility and Common Ions
Calculate the solubility of solid CaF2 (Ksp = 4.0 × 1011
) in a 0.025 M NaF solution.
= 6.4 × 10-8
mol/L
Page 16
Solubility Equilibria 5
Ksp AND THE REACTION QUOTIENT, Q
With some knowledge of the reaction quotient, we can decide
whether a precipitate (ppt) will form AND
what concentrations of ions are required to begin the precipitation of an insoluble salt.
1. Q < Ksp, the system is not at equil. (unsaturated)
2. Q = Ksp, the system is at equil. (saturated)
3. Q > Ksp, the system is not at equil. (supersaturated)
Precipitates form when the solution is supersaturated!!!
Precipitation of insoluble salts
Metal-bearing ores often contain the metal in the form of an insoluble salt, and, to complicate
matters, the ores often contain several such metal salts.
Dissolve the metal salts to obtain the metal ion, concentrate in some manner, and ppt. selectively
only one type of metal ion as an insoluble salt.
Exercise 5 Determining Precipitation Conditions
A solution is prepared by adding 750.0 mL of 4.00 × 103 M Ce(NO3)3 to 300.0 mL of 2.00 × 10
2 M KIO3.
Will Ce(IO3)3 (Ksp = 1.9 × 1010
) precipitate from this solution?
yes
Page 17
Solubility Equilibria 6
Exercise 6 Precipitation
A solution is prepared by mixing 150.0 mL of 1.00 × 10-2
M Mg(NO3)2 and
250.0 mL of 1.00 × 101 M NaF. Calculate the concentrations of Mg
2+ and F
at equilibrium with solid MgF2 (Ksp = 6.4 × 109).
[Mg2+
] = 2.1 × 10-6
M
[F-] = 5.50 × 10
-2 M
SOLUBILITY AND THE COMMON ION EFFECT
Experiments show that the solubility of any salt is always less in the
presence of a “common ion”.
Why? LeChâtelier’s Principle, that’s why! Be reasonable and use
approximations when you can.
The pH can also affect solubility. Evaluate the equation to see which
species reacts with the addition of acid or base.
Would magnesium hydroxide be more soluble in an acid or a base? Why?
Mg(OH)2(s) Mg2+
(aq) + 2 OH (aq) (milk of magnesia)
Page 18
Solubility Equilibria 7
SOLUBILITY, ION SEPARATIONS, AND QUALITATIVE ANALYSIS
This section will introduce you to the basic chemistry of various ions.
It also illustrates how principles of chemical equilibria can be applied in the laboratory.
Objective: Separate the following metal ions from an aqueous sample containing ions of silver, lead(II),
cadmium(II) and nickel(II).
From your knowledge of solubility rules, you know that chlorides of lead and silver will form
precipitates while those of nickel and cadmium will not. Adding dilute HCl to sample will ppt. the lead
and silver ions while the nickel and cadmium will stay in solution.
Separate the lead and silver precipitates from the solution by filtration. Heating the solution causes
some of the lead chloride to dissolve. Filtering the HOT sample will separate the lead (in the filtrate)
from the silver (solid remaining in funnel with filter paper).
Separating cadmium and nickel ions require precipitation with sulfur. Use the Ksp values to determine
which ion will precipitate first as an aqueous solution of sulfide ion is added to the portion of the sample
that still contains these ions. Which precipitates first?
Page 19
Solubility Equilibria 8
Exercise 7 Selective Precipitation
A solution contains 1.0 × 104
M Cu+ and 2.0 × 10
3 M Pb
2+. If a source of I
- is added gradually to this
solution, will PbI2 (Ksp = 1.4 × 108) or CuI (Ksp = 5.3 × 10
12) precipitate first?
Specify the concentration of I necessary to begin precipitation of each salt.
CuI will precipitate first
Concentration in excess of 5.3 × 10-8
M required
SOLUBILITY AND COMPLEX IONS
The formation of complex ions can often dissolve otherwise
insoluble salts.
Often as the complex ion forms, the solubility equilibrium
shifts to the right (away from the solid) and causes the
insoluble salt to become more soluble.
Example: If sufficient aqueous ammonia is added to silver chloride, the
latter can be dissolved as [Ag(NH3)2]+
forms, which is soluble.
That is a significant improvement with regard to the solubility of AgCl(s).
The equilibrium constant for dissolving silver chloride in ammonia is not large, but, if the concentration of
ammonia is sufficiently high, the complex ion and chloride ion must also be high, and silver chloride will
dissolve. Of course, this process is quite smelly!
Page 20
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ConcentrationsatEquilibriumEquilibrium calculations – Rice Tables
1. Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo
missions. In the gas phase it decomposes to gaseous nitrogen dioxide:
N2O4 (g) ⇄ 2 NO2 (g)
Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a
temperature where Kp = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the
equilibrium pressure of NO2(g).
2. 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
PCl5 (g) ⇄ PCl3(g) + Cl2 (g)
Calculate the equilibrium concentrations of all species and the value of K.
3. Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium
constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is
mixed in a 1.000-L flask.
4. Assume that the reaction for the formation of gaseous hydrogen fluoride from 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.
5. 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.
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