This is going to be a bit of a review SL, to make sure you guys know the concepts covered on the exam.
EQUILIBRIUM
At equilibrium, the forward rate = the reverse rate
At equilibrium, the relative proportions of reactant & product concentrations remain constant.
SECTION 1—Differential Rate Equations
1. Write the rate expressions for the following reactions in terms of the disappearance of the reactants and
the appearance of the products.
(a) 4NH3(g) + 5O2(g)
(b) CH4(g) + 2O2(g)
4NO(g) + 6H2O(g)
CO2(g) + 2H2O(g)
2. It was found in an investigation of the reaction, CH3CHO (g)
CH4 (g) + CO (g), that the
–2
–1
–2
–1
concentration of CH3CHO changed from 2.55 10 mole litre to 2.37 10 mole litre in 6.0
minutes. What is the average rate of reaction with respect to CH3CHO over that period?
3. Consider the following reaction
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
If at a given moment C3H8 is reacting at a rate of 0.400mol L-1s-1 , what are
the rates of formation of CO2 and H2O? What is the rate at which O2 is reacting?
SECTION 2—Differential Rate Laws
4. The following data were obtained at 650oK for the reaction,
2NO(g) + O2(g)  2NO2(g)
(a) Deduce the rate law for the reaction.
(b)
Calculate the rate constant.
(c)
Calculate the rate of disappearance of NO (g) at the instant when [NO] = 0.030M and [O2] = 0.020M.
1.8 x 10-3
–
2–
2–
(5) The reaction, 2I + S2O8
 2SO + I2, was studied by measuring the time,
2–
t, for a small increase in the I2 concentration to occur as a function of the initial concentrations of S2O8 and
–
I
Determine the rate law.
SECTION 3—Integrated Rate Law & Arrenhius
(6) The reaction SO2Cl2(g) +  SO2(g) + Cl2(g) is first order with k1 = 2.2 x 10-5 s-1 at 320oC. What percent of
SO2Cl2(g) is decomposed on heating at 320oC for 90 minutes?
Answer: 11.3%
–6 –1
7 At 500°C, the first order conversion of cyclopropane to propylene has a rate constant of 1.16 10 s . If
–2
the initial concentration of cyclopropane is 1.00 10 M at 500 °C, what will its concentration be 24.0
hours after the reaction begins?
8) The gas-phase decomposition of NO2 at 383 C gives NO and O2 according to the equation & following
data. Use the data below to deduce the rate law for the reaction.
2NO2  2NO + O2
9) Consider the data below. One plot is for a catalyzed reaction and the other plot is not a non-catalyzed
reaction. How much did the catalyst reduce the activation energy
SECTION 4—Equilibrium & Le Chatelier’s Principle
10) Predict the effect of decreasing the temperature on the position of the following equilibria.
(a) H2(g) + Cl2(g) ↔ 2HCl(g) + 49.7 kJ
(b) 2NH3(g) ↔ N2(g) + 3H2(g)
∆ H = 37.2 kJ
(c) CO(g) + H2O(g) ↔ CO2(g) + H2(g)
∆ H = -27.6 kJ
(a) Shifts right because an increase in temperature favors the endothermic reaction.
(b) Shifts left to counteract the decrease in temperature.
(c) Shifts left because an increase in temperature favors the endothermic reaction.
11) 0.60 mol of Br2 and 0.60 mol of Cl2 are placed in a 1.00 L flask and allowed to reach equilibrium. (There is no
BrCl at first.) After reaching equilibrium, the flask is found to contain 0.28 mol of BrCl. What is the value of K for
this reaction? 0.37
12) When 1.00 mol of A and 0.800 mol of B are placed in a 2.00 L container and allowed to come to equilibrium,
the resulting mixture is found to be 0.20M in D. What is the value of K at equilibrium?
K = 20
13) For the following reaction, write how the each of the changes will affect the indicated quantity, assuming a
container of fixed size. Write “increase”, “decrease”, or “no change”. (Or use an “up” arrow to indicate
“increase”, and a “down” arrow to indicate “decrease”.) (For a chemical added, write how it would respond
AFTER the addition.)
H2(g) + Br2(g) ⇋ 2 HBr(g)
Change
Some H2 Added
Some HBr Added
Some H2 Removed
Some HBr Removed
Temp is increased
Temp is deccreased
Pressure is increased
and the container
volume decreased
[H2]
ΔHo = -103.7kJ
[Br2]
[HBr]
K Value
14) Given the Keq for reaction (I) below is 4.25 x 1010, what is the Keq for reaction (II)
(I)
2 I2(g) + 2 Cl2(g)
(II)
½ I2(g) + ½ Cl2(g)
4 ICl(g) K = 4.25 x 1010
ICl(g)? K’ = ________
If the coefficients in a balanced equation are multiplied by a factor, n, the
equilibrium expression is raised to the nth power.
K’ = (K)n
Reaction I was multiplied by 0.25, so K’ = (4.25 x 1010)0.25 = 4.54 x 102
15) Calculate the value of Kc for the reaction: 2 NO(g) + Br2(g) ⇋ 2 NOBr (g) using the following information.
(I)
(II)
2 NO(g) ⇋ N2(g) + O2(g)
N2(g) + Br2(g) + O2(g) ⇋ 2 NOBr(g)
Kc1 = 1 x 1030
Kc2 = 2 x 10-27
If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of
the equilibrium constants of the equations added.
K' = K1 x K2 .
The two equations can be added to yield the desired equation. The value of Kc for the reaction will be
the product of the other two.
K'c = Kc1 x Kc2 = (1 x 1030)(1 x 10-27) = 2 x 103
16) Calculate the value of Kc for the reaction:
2 N2O(g) + 3 O2(g)
2 N2O4(g), using
the following information.
Equilibrium Constant
Equation
2 N2 (g) + O2(g) ⇋ 2 N2O(g)
N2O4(g) ⇋ 2 NO2(g)
½ N2(g) + O2(g) ⇋ NO2(g)
Kc = 1.2 x 10-35
Kc = 4.6 x 10-3
Kc = 4.1 x 10-9
These three equations can be combined to get the desired reaction.
Write the first equation backwards. The K for this reaction will be the recipricol of the forward reaction
2 N2O(g) ⇋ 2 N2(g) + O2(g)
Kc = 1/(1.2 x 10-35) = 8.3 x 1034
Write the second equation backwards and multiply the coefficients by 2. The K for this reaction will be the
recipricol of the forward reaction squared.
4 NO2(g) ⇋ 2 N2O4(g)
Kc = 1/(4.6 x 10-3)2 = 4.7 x 104
Use the third equation in the forward direction but multiplied by 4. The K for this reaction will be the K of the
given reaction raised to the fourth power.
2 N2(g) + 4 O2(g) ⇋ 4 NO2(g)
Kc = (4.1 x 10-9)4 = 2.8 x 10-34
Check to see that the three equations yield the desired equation when added together. The equilibrium
constant for the desired equation will be the product of the constants for the three equations combined.
Kc = (8.3 x 1034)(4.7 x 104)(2.8 x 10-34) = 1.1 x 106
SECTION 5—Thermodynamics
We haven’t covered much thermo stuff, so this is mainly just conceptual
understanding….
(17) Word Bank:
Work
Surroundings
Thermodynamic state
Heat
System
Temperature
________ is a form of energy that flows between two samples of matter due to their differences in
temperature.
__________ is a measure of the intensity of heat, ( i.e.,the hotness or coldness of an object.
The _________ refers to the substances of interest in a process, i.e., it is the part of the universe that
is under investigation.
The ________ refer to everything in the environment of the system of interest.
The ________ of a system refers to a set of conditions that completely specifies all of the
thermodynamic properties of the system.
_______ is the application of a force through a distance. For physical or chemical changes that occur
at constant pressure, the work done on the system is –PΔV.
(a) Heat is a form of energy that flows between two samples of matter due to their differences in
temperature.
(b) Temperatureisameasureoftheintensityofheat,i.e.,thehotnessorcoldnessofanobject.
(c) The system refers to the substances of interest in a process, i.e., it is the part of the universe that
is under investigation.
(d) Thesurroundingsrefertoeverythingintheenvironmentofthesystemofinterest.
(e) The thermodynamic state of a system refers to a set of conditions that completely specifies all of
the thermodynamic properties of the system.
(f) Work is the application of a force through a distance. For physical or chemical changes that occur
at constant pressure, the work done on the system is –PΔV.
18)
(a) Give an example of heat being given off by the system.
(b) Give an example of work being done on the system.
19) Use the First Law of Thermodynamics to describe what occurs when an incandescent light is turned on.
According to the First Law of Thermodynamics, the total amount of energy in the universe is constant. When an
incandescent light is turned on, electrical energy is converted mainly into light and heat energy. A small fraction
of the energy is converted to chemical energy, which is why the filament eventually burns out.
20) Distinguish between endothermic and exothermic processes. If we know that a reaction is endothermic in
one direction, what can be said about the reaction in the reverse direction?
An endothermic process absorbs heat energy from its surroundings; an exothermic process releases heat energy
to its surroundings. If a reaction is endothermic in one direction, it is exothermic in the opposite direction. For
example, the melting of 1 mole of ice water is an endothermic process requiring 6.02 kJ of heat:
The reverse process, the freezing of 1 mole of liquid water, releasing 6.02 kJ of heat, is an exothermic process:
21) Which of the following are examples of state functions? (a) your bank balance; (b) the mass of a candy bar;
(c) your weight; (d) the heat lost by perspiration during a climb up a mountain along a fixed path.
A state function is a variable that defines the state of a system; it is a function that is independent of the
pathway by which a process occurs. Therefore, the change in a state function depends only on the initial and the
final value, not on how that change occurred.
(a) Your bank balance is a state function, because it depends only on the difference between your deposits
and withdrawals.
(b) Themassofacandybarisastatefunction,sinceitisaconstantwhereveryouare.
(c) However, your weight is not a state function. Weight depends on the gravitational attraction of your body
to the center of the earth, which changes depending on where you are on the earth.
(d) The heat lost by perspiration during a climb up a mountain along a fixed path is not a state function,
because it depends on the person - her/his size, build, metabolism and degree of fitness.
22) For each of the following reactions…
(a) does the enthalpy increase or decrease
(b) is Hreactants > Hproducts or is Hreactants < Hproducts
(c) is ΔH positive or negative?
(i)
(ii)
Al2O3  2Al + 3/2O (endothermic)
Sn + Cl2  SnCl2 (exothermic)
(i)
Since the reaction is endothermic, (a) enthalpy increases, (b) Hproduct > Hreactant and (c) ΔH is
positive.
Since the reaction is exothermic, (a) enthalpy decreases, (b) Hproduct < Hreactant and (c) ΔH is
negative.
(ii)