In Class Problems and Notes
General Equilibrium
AP Chemistry
Chemical _________________ is the condition reached, in a chemical reaction, when the _______________
of reactants and products no longer change, or are ___________. This condition is attained when the rates of
the forward and reverse reactions are_____________. Only _______ and _______ are included in the
equilibrium equation, pure liquids and solids are ______________!!!
aA(g) + bB(g) → cC(g) + dD(g)
[C]c[D]d
K=
[A]a[B]b
[C]c[D]d
Kc =
[A]a[B]b
d
PCc PD
Kp = a b
PAPB
Kc is the equilibrium constant in which the ________________ of reactant and products are expressed
in terms of ____________. Kp is the equilibrium constant in which the ______________ of reactant and
products are expressed in terms of ____________ _______________ measured in atm or mm Hg.
The equilibrium constant is a number whose value reflects whether the amount of products is greater or
less than the amount of reactants at equilibrium. When K is _____, the amount of products is greater
than reactants; the reaction favors _____________. When K is ______, the amount of products is less
than reactants; the reaction favors __________________.
The equilibrium constant expression is the ____________ of product equilibrium concentration to
reactant equilibrium concentrations as determined by the _______________ chemical equation.
1. The following reaction is at equilibrium at a particular temperature
H2(g) + I2(g) <--> 2HI(g)
and the [H2]eq = 0.012 M, [I2]eq = 0.15 M and [HI]eq = 0.30 M. Write out the equilibrium expression and
calculate the magnitude of Kc for the reaction.
2. A vessel initially has a partial pressure of NO equal to 0.526 atm and a partial pressure of Br2 equal to
0.329 atm. At equilibrium the partial pressure of Br2 is 0.203 atm. Calculate Kp for the reaction
2NO(g) + Br2(g) <--> 2NOBr(g)
Relationship of Kp to Kc:
Consider the reaction
A(g) + B(g) <--> 3C(g)
The Kp expression for this would be:
Remember that concentration is ____/____ and if you rearrange the ideal gas law: PV=nRT, then P =
_____ RT = _______ RT
This reduces to the general:
Kp = ______(RT) _____
The Dn is the change in the # of moles of a gas (___________ – ___________). Don’t memorize it, it’s on
your sheet but remember to use it!
3. At 1100 K, Kp= 0.25 for the reaction: 2SO2(g) + O2 (g) <--> 2SO3
Calculate K at this temperature.
Changing reaction form
I f the reaction is changed in anyway, the value of the ____________ _____________ will change, but
this changed can be determined algebraically.
Consider the following reaction
H2(g) + I2(g) <--> 2HI(g)
[HI]2
Kc = [H ][I ] = 50.
2 2
If the reaction is written as
2HI(g) <--> H2(g) + I2(g)
[H2][I2]
1
1
Kc' =
= K = 50 = _______
2
[HI]
c
If the reaction is written as
1
1
H
(g)
+
2
2
2 I2(g) <--> HI(g)
When the coefficients in the reaction are changed, the new equilibrium constant expression becomes;
[HI]
Kc'' =
=
[H2]1/2[I2]1/2
Kc = _______
Adding equations
If two reactions are added, the resulting constant of the equation will be the products of the two equations
that are added.
Consider the following reactions:
A <--> B
C <--> D
[B]
Kc 1 =
[A]
[D]
Kc 2 =
[C]
[B] [D]
A + C <--> D + B
Kc 3 =
[A] [C]
= Kc 1 x Kc 2
4.
Equilibrium constants for the following reactions have been determined at 550 °C:
CoO(s) + H2(g) <--> Co(s) + H2O(g)
K1 = 67
CoO(s) + CO(g) <--> Co(s) + CO2(g)
K2 = 490
Calculate K (at the same temperature) for the commercially important water gas shift reaction
H2O(g) + CO(g) <--> H2(g) + CO2(g)
K3 = ?
5.
Calculate Kc for the reaction
N2(g) + O2(g) <--> 2NO(g)
if Kc for the reaction
1
1
2 N2(g) + 2 O2(g) <--> NO(g)
is 1.3 x 104.
RICE Tables (AKA Ice AKA Ice Box?):
6. 1.00 liter container holds 1.06 moles of H2 and 1.57 moles of CO at a temperature of 162 °C. At this
temperature, the following reaction occurs,
2H2(g) + CO(g) <--> CH3OH(g)
After equilibrium is established, analysis shows 0.200 moles of CH3OH in the container.
Calculate the [CO]eq, [H2]eq and Kc.
7.The following reaction,
2HI(g) <--> I2(g) + H2(g)
occurs at 298K. If 2.00 mol of HI are placed into a 1.00 liter container and permitted to react, at
equilibrium it is found
that 20.0 % of the HI has decomposed. Calculate Kc and Kp.
8. The equilibrium constant, Kc, for the reaction PCl5(g) <--> PCl3(g) + Cl2(g)
is 33.3 at 760 °C. If 0.400 mol of PCl5 are placed in a 2.00 liter container, calculate the equilibrium
concentrations of all species.
9. The equilibrium constant, Kc, for the reaction
PCl5(g) <--> PCl3(g) + Cl2(g)
is 33.3 at 760 °C. If 0.400 mol of PCl5 and 1.0 mol of Cl2 are placed in a 2.00 liter container, calculate the
equilibrium concentrations of all species.
Le Chatelier's principle.
If a chemical reaction at equilibrium is subjected to a change in conditions that displaces it from
equilibrium, then the reaction proceeds toward a new equilibrium state in the direction that offsets the
change in conditions.
Three factors which can affect a reaction at equilibrium.
•1.Concentration of a reactant or product
When a system is at equilibrium, increasing the concentration of a reactant or product will cause the
reaction to re-establish equilibrium by shifting the reaction in a direction which relieves the stress. For
example, in the simulation reaction,
A2(g) +
B2(g) <--> 2AB(g)
When the reaction is at equilibrium and the [B2] is increased the reaction proceeds from left-to-right to
relieve the stress. When B2 is added the reaction shifts in a direction to relieve the stress, by trying to
decrease the amount of B2. So the reactions proceeds from left-to-right. Similarly when the reaction is at
equilibrium, addition of AB shifts the reaction from right-to-left to relieve the stress.
•2.Reaction volume or pressure
To understand the effect of volume on a reaction, consider the following reaction.
PCl5(g) → PCl3(g) + Cl2(g)
Assume the reaction is at equilibrium in a sealed container. If the volume of the reaction container is
lowered, molecules in the container are pushed closer together and the internal pressure increases. To
relieve this stress the reaction shifts in the direction to decrease the crowding of the molecules. It
proceeds from right to left because the left side of the reaction has fewer gas molecules compared to
the right.
If the volume is increased the reaction proceeds from left-to-right. The increase in volume means there
are fewer molecules per unit volume and the reaction proceeds in a direction to increase the number of
molecules per unit volume and thus maintain constant pressure.
•3.Temperature
To understand how temperature effects a reaction at equilibrium consider the equilibrium of NO2 and
N2O4. The reaction is endothermic;
heat + N2O4(g) → 2NO2(g)
colorless
orange-brown
When the reaction is cooled the color of the sample becomes lighter, indicating more N2O4(g) was
formed. Heat is removed from the reaction system when the reaction is cooled. As the reaction is
cooled it proceeds in the direction to offset the stress, in a direction to produce heat, from right to left.
In general for endothermic reactions decreasing the temperature shifts the equilibrium towards the left
and adding heat shifts the equilibrium to the right. It is just the opposite for exothermic reactions.
10. The reaction
2H2S(g) +3O2(g) <--> 2H2O(l) + 2SO2(g)
has a ∆H = -1036 kJ. Given the reaction is at equilibrium, predict the direction the reaction will shift when
disrupted by each of the following
i) the amount of H2O is increased
ii) the temperature of the reaction is increased
iii) the volume of the container is decreased
iv) the amount of H2S is decreased
The non-equilibrium reaction quotient, Q
The reaction quotient for the chemical reaction,
aA(g) + bB(g) <--> cC(g) + dD(g)
is defined as,
Q=
[C]c[D]d
[A]a[B]b
The concentration used in the quotient, can be nonequilibrium concentrations. When equilibrium
concentrations are used Q = Kc.
The direction the reaction will proceed to establish equilibrium depends on the comparison of the
magnitudes of Q and Kc.
A guide line for predicting the direction a reaction will proceed to reach equilibrium again is as follows:
IF Q<Kc , the concentration of products will increase so that Q will increase
IF Q>Kc, the concentration of reactants will increase so that Q will decrease
IF Q = Kc, then no concentrations will change as the system is already at equilibrium
11. The reaction
1
NOBr(g) <--> NO(g) + 2 Br2(g)
has been carefully studied at 350 °C and the Kc is 0.079. Which direction will the reaction proceed to
establish equilibrium under each of the following initial conditions?
i)
[NOBr]o = 0.100 M : [NO]o = 0 : [Br2]o = 0
ii)
[NOBr]o = 0 : [NO]o = 0.100 M : [Br2]o = 0.100 M
iii)
[NOBr]o = 0.100 M : [NO]o = 0.100 M : [Br2]o = 0.100 M
iv)
[NOBr]o = 0.200 M : [NO]o = 0.0500 M : [Br2]o = 0.100 M
12.
Kc = 19.9 for the reaction:
Cl2 (g) + F2 (g) ↔ 2ClF (g)
Find the equilibrium concentrations of all species if the initial concentration of [Cl2] is 0.4M, [F2] is 0.2 M,
and [ClF] is 7.3 M.
13. Kc = 170 at 298K for the following reaction: 2NO2 (g) ↔ N2O4 (g)
a) Suppose the initial concentration of NO2 is 0.015 M and the concentration of N2O4 is 0.025 M.
Determine the equilibrium concentrations of all species.
14. Kc for the decomposition of ammonium hydrogen sulfide is 1.8 x 10-4 at 298K.
NH4HS (s) ↔ NH3(g) + H2S (g)
a) When the pure salt decomposes in a flask, what are the equilibrium concentration of NH3 and
H2S?
b) If NH4HS is placed in a flask already containing 0.020 mol/L of NH3 and then the system is
allowed to come to equilibrium, what are the equilibrium concentrations of NH3 and H2S?
Relationship of Kinetics to equilibrium
case I
energy
energy
case II
progress of rxn.
progress of rxn.
energy
case III
progress of rxn.
Consider the following above situations:
Case I: The energy of activation of the forward reaction (the amount of energy necessary to go from
reactants to the top of the hill) is much smaller than the energy of activation of the reverse reaction.
This means that for the reverse reaction to happen at the same rate as the forward reaction (the
condition necessary for equilibrium to occur, you must have a large concentration of products, and a
much smaller concentration of reactants. Since Keq is calculated as products over reactants case I will
result in a equilibrium constant larger than 1.
Case II: In this case, the activation energy of the reverse reaction is much smaller, meaning you will
need more reactants to make the reactions occur at an equal rate. Since you will have more reactants
at equilibrium, Keq will be less than one.
Case III: In this case the activation energies for the forward and the reverse reactions are about the
same, therefore you would expect approximately equal concentrations for the two, and Keq will be
close to a value of 1.
Now lets look at how equilibrium can affect the measured rate law. Consider the following mechanism:
(1) A + B <--> C equilibrium, fast
(2) C --> D
slow
(3) D --> E
fast
Since we know that rate law for a given mechanism is based on the slow step, the rate law expression
for the above reaction would be:
rate = k2[C]
But notice here that C is an intermediate and is not easily measured. To find measureable species, we
need to look at the equilibrium step:
rate1f = rate1r
kf1[A][B] = kr1[C]
[C] =
solving for C,
kf1
[A][B]
kr1
now we can sub this back in our original rate law expression, and get a rate in terms of only A and B.
rate = k
2
kf1
[A][B]
kr1
Normally all the k's are lumped into one big constant, since they are all constants themselves.
15. The formation of nitrosyl bromide, NOBr, has the following proposed mechanism:
(1)
NO + Br2  NOBr2
fast, eq.
(2)
NOBr2 + NO  2NOBr
slow
Determine the Rate law!
16. The synthesis of HBr proceeds via the following steps in the gaseous state:
(1)
(2)
(3)
Br2  2Br
Br + H2  HBr + H
H + Br2  HBr + Br
Determine the Rate law!
fast, eq.
slow
fast
17. Determine the Rate law!
(1)
(2)
(3)
A + B  C
C  D
D  E
fast, eq.
slow
fast
Notes on Solubility of Salts Ksp
KNOW YOUR SOLUBILITY RULES by ♥!!!!!!!!!!!!
Precipitation is the act of precipitating or forming an insoluble solid. It is formed by the reaction of two or
more solutions of soluble compounds.
Solubility is the measure of the of the extent to which a compound will dissolve in a given solvent. In
most cases, we are interested in the solubility of a compound in water. The following equation represents the
dissolution of an insoluble solid, which is fact is soluble to a very very small extent:
AgCl (s) ↔ Ag+(aq) + Cl- (aq)
Notice the reaction is in an equilibrium situation where only a very small amount of the silver chloride will break
up.
The equilibrium expression for a reaction is described by the general solubility equation below:
MnXm (s) ↔ nMm+ + mXnThe solubility constant, Ksp, is a measure of the extent to which a compound will dissolve in water. Very small
values of Ksp suggest the compound is insoluble. The bigger the Ksp, the more the solid dissolves. The Ksp
expression for the above reaction is:
Ksp= [nMm+]n[mXn-]m
Notice that the solids, as in all equilibrium expressions, is NOT included!
18)Calculate Ksp for the following salts using the information provided.
a)The concentration of CrO42–(aq) in a saturated solution of Ag2CrO4 is 6.50 x 10–5 M.
b)The solubility of AgBrO3 in water is 7.2 x 10–2 g/L.
19. Calculate the solubility of the following compounds in water.
(Use a table of solubility product constants in your text or some other reference book.)
a)
BaCO3
b)
Cr(OH)3
20.Calculate the solubility of;
a)BaCO3 in 0.500 M Ba(NO3)2
b)PbCl2 in 0.0250 M CaCl2
21.A 45 mL sample of 0.015 M calcium chloride is added to 55 mL of 0.010 M sodium sulfate. Is
a precipitate expected? Explain. (Your answer must include a calculation!)
Equilibrium HW
1977 D
For the system 2 SO2(g) + O2(g)  2 SO3(g) , H is negative for the production of SO3. Assume that
one has an equilibrium mixture of these substances. Predict the effect of each of the following
changes on the value of the equilibrium constant and on the number of moles of SO 3 present in
the mixture at equilibrium. Briefly account for each of your predictions. (Assume that in each case all
other factors remain constant.)
(a) Decreasing the volume of the system.
(b) Adding oxygen to the equilibrium mixture.
(c) Raising the temperature of the system.
1980 D
NH4Cl(s)  NH3(g) + HCl(g)
H = +42.1 kilocalories
Suppose the substances in the reaction above are at equilibrium at 600K in volume V and at
pressure P. State whether the partial pressure of NH3(g) will have increased, decreased, or
remained the same when equilibrium is reestablished after each of the following disturbances
of the original system. Some solid NH4Cl remains in the flask at all times. Justify each answer
with a one-or-two sentence explanation.
(a) A small quantity of NH4Cl is added.
(b) The temperature of the system is increased.
(c) The volume of the system is increased.
(d) A quantity of gaseous HCl is added.
(e) A quantity of gaseous NH3 is added.
1983 A
Sulfuryl chloride, SO2Cl2, is a highly reactive gaseous compound. When heated, it decomposes
as follows: SO2Cl2(g)  SO2(g) + Cl2(g). This decomposition is endothermic. A sample of 3.509
grams of SO2Cl2 is placed in an evacuated 1.00 litre bulb and the temperature is raised to 375K.
(a) What would be the pressure in atmospheres in the bulb if no dissociation of the SO 2Cl2(g)
occurred?
(b) When the system has come to equilibrium at 375K, the total pressure in the bulb is found to
be 1.43 atmospheres. Calculate the partial pressures of SO2, Cl2, and SO2Cl2 at equilibrium at
375K.
(c) Give the expression for the equilibrium constant (either Kp or Kc) for the decomposition of
SO2Cl2(g) at 375K. Calculate the value of the equilibrium constant you have given, and specify
its units.
(d) If the temperature were raised to 500K, what effect would this have on the equilibrium
constant? Explain
1985 A
At 25ºC the solubility product constant, Ksp, for strontium sulfate, SrSO4, is 7.610-7. The
solubility product constant for strontium fluoride, SrF2, is 7.9 10-10.
(a) What is the molar solubility of SrSO4 in pure water at 25ºC?
(b) What is the molar solubility of SrF2 in pure water at 25ºC?
(c) An aqueous solution of Sr(NO3)2 is added slowly to 1.0 litre of a well-stirred solution
containing 0.020 mole F- and 0.10 mole SO42- at 25ºC. (You may assume that the added
Sr(NO3)2 solution does not materially affect the total volume of the system.)
1.
2.
Which salt precipitates first?
What is the concentration of strontium ion, Sr2+, in the solution when the first
precipitate begins to form?
(d) As more Sr(NO3)2 is added to the mixture in (c) a second precipitate begins to form. At that
stage, what percent of the anion of the first precipitate remains in solution?
1988 A
At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown by the
following equation: SbCl5(g)  SbCl3(g) + Cl2(g)
(a) An 89.7 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0 litre
container at 182ºC.
1. What is the concentration in moles per litre of SbCl5 in the container before any
decomposition occurs?
2. What is the pressure in atmospheres of SbCl5 in the container before any
decomposition occurs?
(b) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182ºC, calculate
the value for either equilibrium constant Kp or Kc, for this decomposition reaction. Indicated
whether you are calculating Kp or Kc.
(c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an empty 2.00
litre container maintained at a temperature different from 182ºC. At this temperature, K c,
equals 0.117. How many moles of Cl2 must be added to this container to reduce the number
of moles of SbCl3 to 0.700 mole at equilibrium?
1988 D
NH4HS(s)  NH3(g) + H2S(g)
Hº = +93 kilojoules
The equilibrium above is established by placing solid NH4HS in an evacuated container at 25ºC.
At equilibrium, some solid NH4HS remains in the container. Predict and explain each of the
following.
(a) The effect on the equilibrium partial pressure of NH3 gas when additional solid NH4HS is
introduced into the container
(b) The effect on the equilibrium partial pressure of NH3 gas when additional solid H2S is
introduced into the container
(c) The effect on the mass of solid NH4HS present when the volume of the container is
decreased
(d) The effect on the mass of solid NH4HS present when the temperature is increased.
1992 A
2 NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation
above.
(a) A sample of 100. grams of solid NaHCO3 was placed in a previously evacuated rigid 5.00-liter
container and heated to 160ºC. Some of the original solid remained and the total pressure
in the container was 7.76 atmospheres when equilibrium was reached. Calculate the
number of moles of H2O(g) present at equilibrium.
(b) How many grams of the original solid remain in the container under the conditions
described in (a)?
(c) Write the equilibrium expression for the equilibrium constant, KP, and calculate its value for
the reaction under the conditions in (a).
(d) If 110. grams of solid NaHCO3 had been placed in the 5.00-liter container and heated to
160ºC, what would the total pressure have been at equilibrium? Explain.
1994 A
MgF2(s)  Mg2+(aq) + 2 F-(aq)
In a saturated solution of MgF2 at 18ºC, the concentration of Mg2+ is 1.21 10-3 molar. The
equilibrium is represented by the equation above.
(a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18ºC.
(b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at
18ºC to which 0.100 mole of solid KF has been added. The KF dissolves completely. Assume
the volume change is negligible.
(c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.00 10-3-molar
Mg(NO3)2 solution is mixed with 200.0 milliliters of a 2.00 l0-3-molar NaF solution at 18ºC.
Calculations to support your prediction must be shown.
(d) At 27ºC the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 10-3 molar. Is the
dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation
to support your conclusion.
1995 A
CO2(g) + H2(g)  H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation
above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
(a) What is the mole fraction of CO(g) in the equilibrium mixture?
(b) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium
constant for the reaction.
(c) Determine Kp in terms of Kc for this system.
(d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO (g)
is converted back to CO2(g). Calculate the value of Kc at this lower temperature.
(e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter
reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of
CO(g) at this temperature.
1998 D
C(s) + H2O(g)  CO(g) + H2(g)
Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at equilibrium.
State whether the number of moles of CO(g) in the container will increase, decrease, or remain
the same after each of the following disturbances is applied to the original mixture. For each
case, assume that all other variables remain constant except for the given disturbance. Explain
each answer with a short statement.
(a) Additional H2(g) is added to the equilibrium mixture at constant volume.
(b) The temperature of the equilibrium mixture is increased at constant volume.
(c) The volume of the container is decreased at constant temperature.
(d) The graphite pellets are pulverized.
2000 A Required
1. 2 H2S(g)  2 H2(g) + S2(g)
When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g
sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is
heated to 483 K, and 3.7210–2 mol of S2(g) is present at equilibrium.
(a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction
represented above.
(b) Calculate the equilibrium concentration, in molL-1, of the following gases in the container at
483 K.
(i) H2(g)
(ii) H2S(g)
(c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K.
(d) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K.
(e) For the reaction H2(g) +
constant, Kc.
1
2
S2(g)  H2S(g) at 483 K, calculate the value of the equilibrium