AP Chemistry Chapter 14 Study Guide- Kinetics

1. STUDY GUIDE— AP Chemistry
CHAPTER 14- Chemical Kinetics
14.1 Factors that Affect Reaction Rates
•Chemical kinetics is the study of how fast chemical reactions occur.
•There are several important factors which affect rates of reactions:
•physical state of the reactants,
•concentration of the reactants,
•temperature of the reaction, and
•presence or absence of a catalyst.
•The goal is to understand chemical reactions at the molecular level.
14.2 Reaction Rates
•The speed of a reaction is defined as the change that occurs per unit time.
•It is often determined by measuring the change in concentration of a reactant or product with time.
•The speed of the chemical reaction is its reaction rate.
•For a reaction A  B: change in the concentration of B
Average rate with respect to B =
change in time
•Here the change in the concentration of B is defined as:
Δ(concentration of B) = (concentration of B at final time) − (concentration of B at initial time)
•Illustrate this with an example:
•Suppose A reacts to form B. Let us begin with 1.00 M A.
•At t = 0 (time zero) there is 1.00 M A and no B present.
•At t = 20 sec, there is 0.54 M A and 0.46 M B.
•At t = 40 sec, there is 0.30 M A and 0.70 M B.
•We can use this information to find the average rate with respect to B:
" (Conc B) (Conc of B at t = 20 s ) ! (Conc of B at t = 0 s)
Avg Rate = =
"t 20 s ! 0 min
0.46M ! 0.00 M M
Avg Rate = = 0.023
20 s ! 0 s s
•For the reaction A  B there are two ways of measuring rate:
•the rate of appearance of product B (i.e., change in moles of B per unit time) and
•the rate of disappearance of reactant A (i.e., the change in moles of A per unit time).
" ![ A]
Average Rate = Square brackets mean “the molar concentration of”.
!t [A] = the concentration of A in moles/L
•Note the negative sign! This reminds us that rate is being expressed in terms of the disappearance of a
reactant.
•A plot of number of moles versus time shows that as the reactants (A) disappear, the products (B) appear.
Change of Rate with Time
•In most chemical reactions we will determine the reaction rate by monitoring a change in concentration (of a reactant or
product).
•The most useful unit to use for rate is molarity.
•Since volume is constant, molarity and moles are directly proportional.
•Consider the following reaction:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
•We can calculate the average rate in terms of the disappearance of C4H9Cl.
•The units for average rate are mol/(L•s) or M/s.
•The average rate decreases with time.

2. •We can plot [C4H9Cl] versus time.
•The rate at any instant in time is called the instantaneous rate.
•It is the slope of the straight line tangent to the curve at that instant.
•Instantaneous rate is different from average rate.
•It is the rate at that particular instant in time.
•We will call the "instantaneous rate" the rate, unless otherwise indicated.
Reaction Rates and Stoichiometry
•For the reaction: C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
•The rate of appearance of C4H9OH must equal the rate of disappearance of C4H9Cl.
![ 4H9Cl] ![ 4H9OH ]
C C
Rate = " =
!t !t
•What if the stoichiometric relationships are not one-to-one?
• For the reaction:
2HI(g)  H2(g) + I2(g)
• The rate may be expressed as:
1 ![HI] ![H2] ![I2]
Rate = " = =
2 !t !t !t
•We can generalize this equation a bit.
•For the reaction:
aA + bB  cC + dD
•The rate may be expressed as:
1 ![A] 1 ![B ] 1 ![ ] 1 ![D ]
C
Rate = " =" = =
a !t b !t c !t d !t
14.3 Concentration and Rate
•In general, rates:
•increase when reactant concentration is increased.
•decrease as the concentration of reactants is reduced.
•We often examine the effect of concentration on reaction rate by measuring the way in which reaction rate at the
beginning of a reaction depends on starting conditions.
•Consider the reaction:
NH4+(aq) + NO2– (aq)  N2(g) + 2H2O(l)
•We measure initial reaction rates.
•The initial rate is the instantaneous rate at time t = 0.
•We find this at various initial concentrations of each reactant.
•As [NH4+] doubles with [NO2–] constant the rate doubles.
•We conclude that the rate is proportional to [NH4+].
•As [NO2–] doubles with [NH4+] constant the rate doubles.
•We conclude that the rate is proportional to [NO2–].
•The overall concentration dependence of reaction rate is given in a rate law or rate expression.
•For our example, the rate law is:
Rate = k[NH4+][ NO2–]
•The proportionality constant k is called the rate constant.
•Once we have determined the rate law and the rate constant, we can use them to calculate initial reaction
rates under any set of initial concentrations.
Exponents in the Rate Law
•For a general reaction with rate law:
Rate = k[reactant 1]m[reactant 2]n

3. •The exponents m and n are called reaction orders.
•The overall reaction order is the sum of the reaction orders.
•The overall order of reaction is m + n + …
•For the reaction: NH4+(aq) + NO2– (aq)  N2(g) + 2H2O(l)
•The reaction is said to be first order in [NH4+], first order in [NO2–], and second order overall.
•Note that reaction orders must be determined experimentally.
•They do not necessarily correspond to the stoichiometric coefficients in the balanced chemical equation!
•We commonly encounter reaction orders of 0, 1 or 2.
•Even fractional or negative values are possible.
Units of Rate Constants
•Units of the rate constant depend on the overall reaction order.
•For example, for a reaction that is second order overall:
•Units of rate are:
2
Units of rate = (Units of rate constant )(Units of concentration )
•Thus the units of the rate constant are:
(Units of rate) M s
Units of rate constant = 2
= = M !1 s !1
(Units of concentration) M2
Using Initial Rates to Determine Rate Laws
•To determine the rate law, we observe the effect of changing initial concentrations.
•If a reaction is zero order in a reactant, changing the initial concentration of that reactant will have no
effect on rate (as long as some reactant is present).
•If a reaction is first order, doubling the concentration will cause the rate to double.
•If a reaction is second order, doubling the concentration will result in a 22 increase in rate.
•Similarly, tripling the concentration results in a 32 increase in rate.
•A reaction is nth order if doubling the concentration causes a 2n increase in rate.
•Note that the rate, not the rate constant, depends on concentration.
•The rate constant IS affected by temperature and by the presence of a catalyst.
14.4 The Change of Concentration with Time
•Goal: Convert the rate law into a convenient equation that gives concentration as a function of time.
First-Order Reactions
•For a first-order reaction, the rate doubles as the concentration of a reactant doubles.
•Therefore we can write the differential rate law: •Integrating:
![A]
Rate = " = k [A]
!t
•We get the integrated rate law:
•Rearranging: •An alternate form:
•A plot of ln[A]t versus t is a straight line with slope −k and intercept ln[A]0.
•Note that in this equation we use the natural logarithm, ln (log to the base e).
Second-Order Reactions
•A second-order reaction is one whose rate depends on the reactant concentration to the second power or on the
concentration of two reactants, each raised to the first power.
•For a second-order reaction with just one reactant we write the differential rate law:

4. •Integrating, •We get the integrated form of the rate law:
•A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0.
•For a second-order reaction, a plot of ln[A]t vs. t is not linear.
•Note that a second-order process can have a rate constant expression of the form:
Rate = k[A][B]
•That is, the reaction is second order overall, but has first-order dependence on A and B.
Half-life
•Half-life, t½ , is the time required for the concentration of a reactant to decrease to half its original value.
•That is, half life, t½, is the time taken for [A]0 to reach ½ [A]0.
•Mathematically, the half life of a first-order reaction is:
ln
[A]t = !kt
[A]0
So, for t = t1/2 and [A]t = ½ [A]0
1 [ A] 0
ln 1 = ! kt 1
ln 2 = ! kt 1 2 2
[ A] 0 2
ln 1 2 0.693
" t1 2 = ! =
k k
•Note that the half-life of a first-order reaction is independent of the initial concentration of the reactant.
•We can show that the half-life of a second order reaction is:
1
t1 =
2 k[ A] 0
•Note that the half-life of a second-order reaction is dependent on the initial concentration of reactant.
14.5 Temperature and Rate
•Most reactions speed up as temperature increases. The higher the temperature, the faster the reaction.
•As temperature increases, the rate increases.
•How is the relationship between temperature and rate reflected in the rate expression?
•The rate law has no temperature term in it, so the rate constant must depend on temperature.
•Consider the first-order reaction CH3NC  CH3CN.
•As temperature increases from 190°C to 250°C the rate constant increases.
•The temperature effect is quite dramatic.
•We see an approximate doubling of the rate of the reaction with each 10°C increase in temperature.
The Collision Model
•Rates of reactions are affected by concentration and temperature.
•We need to develop a model that explains this observation.
•An explanation is provided by the collision model, based on the kinetic-molecular theory.
•In order for molecules to react they must collide.
•The greater the number of collisions the faster the rate.
•The more molecules present, the greater the probability of collision and the faster the rate.
•Thus, reaction rate should increase with an increase in the concentration of reactant molecules.

5. •The higher the temperature, the more energy available to the molecules and the more frequently the
molecules collide.
•Thus, reaction rate should increase with an increase in temperature.
•However, not all collisions lead to products.
•In fact, only a small fraction of collisions lead to products.
•In order for a reaction to occur the reactant molecules must collide in the correct orientation and
with enough energy to form products.
The Orientation Factor
•The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs.
•Consider the reaction between Cl and NOCl:
Cl + NOCl  NO + Cl2
•If the Cl collides with the Cl of NOCl, the products are Cl2 and NO.
•If the Cl collides with the O of NOCl, no products are formed.
Activation Energy
•Arrhenius: Molecules must posses a minimum amount of energy to react. Why?
•In order to form products, bonds must be broken in the reactants.
•Bond breakage requires energy.
•Molecules moving too slowly, with too little kinetic energy, do not react when they collide.
•Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.
•Ea will vary with the reaction.
•Consider the rearrangement of methyl isonitrile to form acetonitrile:
•Energy is required to stretch the bond between the CH3 group and the N≡C group to allow the N≡C to
rotate.
•The C–C bond begins to form.
•The energy associated with the molecule drops.
•The energy barrier between the starting molecule and the highest energy state found along the reaction
pathway is the activation energy.
•The species at the top of the barrier is called the activated complex, or transition state.
•The change in energy for the reaction is the difference in energy between CH3NC and CH3CN.
•ΔErxn has no effect on reaction rate.
•The activation energy is the difference in energy between reactants, (CH3NC) and the transition state.
•The rate depends on the magnitude of the Ea.
•In general, the lower the Ea, the faster the rate.
•Notice that if a forward reaction is exothermic (CH3NC  CH3CN), then the reverse reaction is endothermic
(CH3CN  CH3NC).
•How does this relate to temperature?
•At any particular temperature, the molecules present have an average kinetic energy associated with the
population.
•In the same distribution, some molecules have less energy than the average while others have more than
the average value.
•The fraction of molecules with an energy equal to or greater than Ea is given by:
! Ea
f = e RT
•R is the gas constant (8.314 J/molK) and T is the absolute temperature.
•Molecules that have an energy equal to or greater than ΔEa have sufficient energy to react.
•As we increase the temperature, the fraction of the population that has an energy equal to or greater than
Ea increases. •Thus, more molecules can react.
The Arrhenius Equation
•Arrhenius discovered that most reaction-rate data obeyed an equation based on three factors:
•the number of collisions per unit time,
•the fraction of collisions that occur with the correct orientation, and

6. •the fraction of the colliding molecules that have an energy equal to or greater than ΔEa.
! Ea
k = Ae RT
•From these observations Arrhenius developed the Arrhenius equation.
•Where k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/Kmol), and T is the
temperature in K.
•A is called the frequency factor.
•It is related to the frequency of collisions and the probability that a collision will have a favorable
orientation.
•Both A and Ea are specific to a given reaction. 1. B
2. A
Determining the Activation Energy 3. B
•Ea may be determined experimentally. 4. D
•We need to take the natural log of both sides of the Arrhenius equation: 5. B
6. E
E
ln k = ! a + ln A 7. A
RT 8. C
•A graph of ln k vs 1/T will have a slope of –Ea/R and a y-intercept of ln A. 9. B
•Alternatively we can use: 10. C
11. B
k1 Ea & 1 1# 12. A
ln = $ ' !
k2 R % 2 T1 !
$T
" 13. B
14. E
15. B
14.6 Reaction Mechanisms 16. A
•The balanced chemical equation provides information about substances present at the beginning 17. B
and end of the reaction. 18. A
•The reaction mechanism is the process by which the reaction occurs. 19. C
•Mechanisms provide a picture of which bonds are broken and formed during the course of a reaction. 20. A
21. B
Elementary Reactions 22. A
•Elementary reactions are any processes that occur in a single step. 23. E
24. B
•The number of molecules present in an elementary step is the molecularity of that elementary step. 25. E
•Unimolecular reactions involve one molecule. 26. A
•Bimolecular elementary reactions involve the collision of two molecules. 27. C
•Termolecular elementary reactions involve the simultaneous collision of three molecules. 28. C
•It is not common to see termolecular processes (statistically improbable). 29. B
30. B
31. B
32. B
Multistep Mechanisms 33. C
•A multistep mechanism consists of a sequence of elementary steps. 34. A
•The elementary steps must add to give the balanced chemical equation. 35. D
•Some multistep mechanisms will include intermediates. 36. A
37. C
•These are species that appear in an elementary step but are neither a reactant nor product. 38. C
•Intermediates are formed in one elementary step and consumed in another. 39. D
•They are not found in the balanced equation for the overall reaction. 40. A
41. B
Rate Laws of Elementary Reactions 42. A
•The rate laws of the elementary steps determine the overall rate law of the reaction.
•The rate law of an elementary step is determined by its molecularity.
•Unimolecular processes are first order.
•Bimolecular processes are second order.
•Termolecular processes are third order.
The Rate Determining Step for a Multistep Mechanism
•Most reactions occur by mechanisms with more than one elementary step.
•Often one step is much slower than the others.

7. •The slow step limits the overall reaction rate.
•This is called the rate-determining step (rate-limiting step) of the reaction.
•This step governs the overall rate law for the overall reaction.
Mechanisms with a Slow Initial Step
•Consider the reaction:
NO2(g) + CO(g)  NO(g) + CO2(g)
•The experimentally derived rate law is: Rate = k[NO2]2
•We propose a mechanism for the reaction:
k1
• Step 1: NO2(g) + NO2(g) !!" NO3(g) + NO(g) slow step
• k2
Step 2: NO3(g) + CO(g) !!" NO2(g) + CO2(g) fast step
•Note that NO3 is an intermediate.
•If k2 >> k1, then the overall reaction rate will depend on the first step (the rate-determining step).
•Rate = k1[NO2]2
•This theoretical rate law is in agreement with the experimental rate law.
•This supports (but does not prove) our mechanism.
Mechanisms with a Fast Initial Step
•Consider the reaction:
2NO(g) + Br2(g)  2NOBr(g)
•The experimentally determined rate law is:
Rate = k[NO]2[Br2]
•Consider the following proposed mechanism:
!!#k1
• Step 1: NO(g) + Br2(g) NOBr2(g) fast step
"! !
!
k $1
• Step 2: NOBr2(g) + NO(g) !k 2 2NOBr(g)
!" slow step
•The theoretical rate law for this mechanism is based on the rate-determining step, step 2:
Rate = k2[NOBr2][NO]
•Problem: This rate law depends on the concentration of an intermediate species.
•Intermediates are usually unstable and have low/unknown concentrations.
•We need to find a way to remove this term from our rate law.
•We can express the concentration of [NOBr2] in terms of NO and Br2 by assuming that there is an
equilibrium in step 1.
•In a dynamic equilibrium, the forward rate equals the reverse rate.
•Therefore, by definition of equilibrium we get:
k1[NO][Br2] = k–1[NOBr2]
•Rearranging we get:
k1
[ NOBr2 ] = [ NO][Br2 ]
k !1
•Therefore, the overall rate law becomes
k1
Rate = k 2 [ NO][Br2 ][NO] = k[NO] 2 [Br2 ]
k !1
•Note that the final rate law is consistent with the experimentally observed rate law.
14.7 Catalysis
•A catalyst is a substance that changes the rate of a chemical reaction without itself undergoing a permanent
chemical change in the process.

8. •There are two types of catalysts:
•homogeneous and
•heterogeneous.
•Catalysts are common in the body, in the environment, and in the chemistry lab!
Homogeneous Catalysis
•A homogeneous catalyst is one that is present in the same phase as the reacting molecules.
•For example, hydrogen peroxide decomposes very slowly in the absence of a catalyst:
2H2O2(aq)  2H2O(l) + O2(g)
•In the presence of bromide ion, the decomposition occurs rapidly in acidic solution:
2Br–(aq) + H2O2(aq) + 2H+(aq)  Br2(aq) + 2H2O(l)
Br2(aq) + H2O2(aq)  2Br–(aq) + 2H+(aq) + O2(g)
•Br– is a catalyst because it is regenerated at the end of the reaction.
•The net reaction is still:
2H2O2(aq)  2H2O(l) + O2(g)
•How do catalysts increase reaction rates?
•In general, catalysts operate by lowering the overall activation energy for a reaction.
•However, catalysts can operate by increasing the number of effective collisions.
•That is, from the Arrhenius equation catalysts increase k by increasing A or decreasing Ea.
•A catalyst usually provides a completely different mechanism for the reaction.
•In the preceding peroxide decomposition example, in the absence of a catalyst, H2O2 decomposes
directly to water and oxygen.
•In the presence of Br–, Br2(aq) is generated as an intermediate.
•When a catalyst adds an intermediate, the activation energies for both steps must be lower than the
activation energy for the uncatalyzed reaction.
Heterogeneous Catalysis
•A heterogeneous catalyst exists in a different phase than the reactants.
•Often we encounter a situation involving a solid catalyst in contact with gaseous reactants and gaseous products
(example: catalytic converters in cars) or with reactants in a liquid.
•Many industrial catalysts are heterogeneous.
•How do they do their job?
•The first step is adsorption (the binding of reactant molecules to the catalyst surface).
•Adsorption occurs due to the high reactivity of atoms or ions on the surface of the solid.
•Absorption refers to the uptake of molecules into the interior of another substance.
•Molecules are adsorbed onto active sites on the catalyst surface.
•The number of active sites on a given amount of catalyst depends on several factors such as:
•the nature of the catalyst,
•how the catalyst was prepared, and
•how the catalyst was treated prior to use.
•For example, consider the hydrogenation of ethylene to form ethane:
C2H4(g) + H2(g)  C2H6(g) ΔH° = −137 kJ/mol
•The reaction is slow in the absence of a catalyst.
•In the presence of a finely divided metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at
room temperature.
•First, the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface.
•The H–H bond breaks and the H atoms migrate about the metal surface.
•When an H atom collides with an ethylene molecule on the surface, the C–C π bond breaks and a C–H σ
bond forms.
•An ethyl group, C2H5, is weakly bonded to the metal surface with a metal-carbon σ bond.
•When C2H6 forms it desorbs from the surface.
•When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break bonds.
•The activation energy for the reaction is lowered.
•Thus the reaction rate is increased.

9. Enzymes
•Enzymes are biological catalysts.
•Most enzymes are large protein molecules.
•Molar masses are in the range of 104 to 106 amu.
•Enzymes are capable of catalyzing very specific reactions.
•For example, catalase is an enzyme found in blood and liver cells.
•It catalyzes the decomposition of hydrogen peroxide:
2H2O2(aq)  2H2O(l) + O2(g)
•This reaction is important in removing peroxide, a potentially harmful oxidizing agent.
•The enzyme catalyzes the reaction at its active site.
•The substances that undergo reaction at the active site on enzymes are called substrates.
•A simple view of enzyme specificity is the lock-and-key model.
•Here, a substrate is pictured as fitting into the active site of an enzyme in a manner similar to a specific
key fitting into a lock. This forms an enzyme-substrate (ES) complex.
•Only substrates that fit into the enzyme lock can be involved in the reaction.
•The enzyme’s active site and the substrate thus have complementary shapes.
•However, there may be a significant amount of flexibility at the active site.
•It may change shape as it binds to the substrate.
•A reaction occurs very quickly once the substrate is bound.
•Products depart the active site at the end of the reaction.
•This allows new substrate molecules to bind to the enzyme.
•If a molecule binds so tightly to an enzyme that substrate molecules cannot displace it, then the active site is blocked and
the catalyst is inhibited.
•Such molecules are called enzyme inhibitors.
•Many poisons act by binding to the active site, blocking the binding of substrates.
•Some poisons bind to other locations on the enzyme.
•Binding ultimately causes a change in the enzyme that interferes with enzyme activity.
•Enzymes are extremely efficient catalysts.
•The number of catalytic events occurring at an active site per unit time is called the turnover number.
•Large turnover numbers correspond to very low Ea values.
•For enzymes, turnover numbers are very large (typically 103–107 per second).
1. Read pgs 575-582 HOMEWORK
Pg 617 #1, 2, 11, 13, 18, 19
2. Read pgs 582-587
#3, 21, 23, 27, 29, 30a&b, 31, 32a
AP Chemistry Chapter 14 Practice Test 3. Read pgs 587-593
Kinetics #4, 5, 33, 34, 35, 37, 41, 43, 44
4. Read pgs 593-610
1) Which substance in the reaction below either appears or #7, 8, 45, 49, 51, 53, 57, 59, 61, 65, 67, 69, 71, 79
disappears the fastest? 4NH3 + 7O 2 ! 4NO 2 + 6H 2 O
A) NH3 B) O 2 C) NO 2 D) H 2 O
E) The rates of appearance/disappearance are the same for all
of these.

10. 2) Consider the reaction: A → 2C 9) The overall order of a reaction is 2. The units of the rate
The average rate of appearance of C is given by Δ[C]/Δt. constant for the reaction are __________.
Comparing the rate of appearance of C and the rate of A) M/s B) M -1s-1 C) 1/s
disappearance of A, we get ![C]/!t = _____ " (![A]/!t)
D) 1/M E) s/M 2
A) +2 B) -1 C) +1
D) +1/2 E) -1/2 10) The kinetics of the reaction below were studied and it was
determined that the reaction rate increased by a factor of 9
A flask is charged with 0.124 mol of A and allowed to react to form B when the concentration of B was tripled. The reaction is
according to the reaction A(g) →B(g). The following data are
__________ order in B.
obtained for [A] as the reaction proceeds:
A+B→P
Time(s) 1 10 20 30 40
A) zero B) first C) second
Moles of A 0.124 0.110 0.088 0.073 0.054 D) third E) one-half
3) The average rate of disappearance of A between 20 s and 40
s is __________ mol/s. 11) A reaction was found to be third order in A. Increasing the
A) 8.5 × 10-4 B) 1.7 × 10-3 C) 590 concentration of A by a factor of 3 will cause the reaction rate
D) 7.1 × 10-3 E) 1.4 × 10-3 to __________.
A) remain constant B) increase by a factor of 27
C) increase by a factor of 9 D) triple
4) How many moles of B are present at 10 s?
E) decrease by a factor of the cube root of 3
A) 0.011 B) 0.220 C) 0.110
D) 0.014 E) 1.4 × 10-3 12) A reaction was found to be zero order in A. Increasing the
concentration of A by a factor of 3 will cause the reaction rate
The peroxydisulfate ion (S2 O82- ) reacts with the iodide ion in to __________.
aqueous solution via the reaction: A) remain constant B) increase by a factor of 27
C) increase by a factor of 9 D) triple
(S2 O82- ) (aq) + 3I- ! 2SO 4 (aq) + I3- (aq) E) decrease by a factor of the cube root of 3
An aqueous solution containing 0.050 M of S2 O82- ion and
The data in the table below were obtained for the reaction:
0.072 M of I- is prepared, and the progress of the reaction
A+B→P
followed by measuring [I-]. The data obtained is given below. Experiment [A] (M) [B] (M) Initial Rate
Time(s) 0 400 800 1200 1600 Number (M/s)
[I-] (M) 0.072 0.057 0.046 0.037 0.029 1 0.273 0.763 2.83
2 0.273 1.526 2.83
5) The average rate of disappearance of I in the initial 400 s is 3 0.819 0.763 25.47
__________ M/s.
A) 6.00 B) 3.8 × 10-5 C) 1.4 × 10-4 13) The order of the reaction in A is __________.
D) 2.7 × 10 4 E) 3.2 × 10 -4 A) 1 B) 2 C) 3
D) 4 E) 0
6) The concentration of S2O82- remaining at 800 s is ____ M. 14) The order of the reaction in B is __________.
A) 0.046 B) 0.076 C) 4.00 × 10-3 A) 1 B) 2 C) 3
D) 0.015 E) 0.041 D) 4 E) 0
7) At elevated temperatures, methylisonitrile (CH3NC) 15) The overall order of the reaction is __________.
isomerizes to acetonitrile (CH3 CN): A) 1 B) 2 C) 3
D) 4 E) 0
CH3NC (g) → CH3 CN (g)
At the start of the experiment, there are 0.200 mol of reactant 16) The magnitude of the rate constant is __________.
(CH3NC) and 0 mol of product (CH3 CN) in the reaction A) 38.0 B) 0.278 C) 13.2
vessel. After 25 min of reaction, 0.108 mol of reactant D) 42.0 E) 2.21
(CH3NC) remain. The average rate of decomposition of
methyl isonitrile, CH3NC, in this 25 min period is __ mol/min.
A) 3.7 × 10-3 B) 0.092 C) 2.3 17) For a first-order reaction, a plot of __________ versus
__________ is linear.
D) 4.3 × 10-3 E) 0.54
1
A) In [A]t , B) ln [A]t, t
8) If the rate law for the reaction 2A + 3B → products is t
first order in A and second order in B, then the rate law is ____. 1
C) ,t D) [A]t, t
A) rate = k[A][B] B) rate = k[A]2 [B]3 [A]t
C) rate = k[A][B]2 D) rate = k[A]2 [B] 1
E) t,
E) rate = k[A]2[B]2 [A]t

11. 18) The rate constant for a particular second-order reaction is 26) The reaction
0.47 M -1s-1 . If the initial concentration of reactant is 0.25 CH3 -N ! C " CH3 -C ! N
mol/L, it takes __________ s for the concentration to decrease is a first-order reaction. At 230.3°C, k = 6.29 × 10-4 s-1 If
to 0.13 mol/L.
[CH3 -N ! C] is 1.00 ! 10-3 initially, [CH3 -N ! C] is
A) 7.9 B) 1.4 C) 3.7
D) 1.7 E) 0.13 __________ after 1.000 ! 103s .
A) 5.33 × 10-4 B) 2.34 × 10-4
19) The initial concentration of reactant in a first-order -3
C) 1.88 × 10 D) 4.27 × 10-3
reaction is 0.27 M. The rate constant for the reaction is 0.75 s-1
What is the concentration (mol/L) of reactant after 1.5 s? E) 1.00 × 10-6
A) 3.8 B) 1.7 C) 8.8 × 10-2
The reaction A → B is first order in [A]. Consider the
D) 2.0 × 10-2 E) 0.135 following data.
Time(s) [A] (M)
0.0 1.60
20) The half-life of a first-order reaction is 13 min. If the initial
concentration of reactant is 0.085 M, it takes ________min for 10.0 0.40
it to decrease to 0.055 M. 20.0 0.10
A) 8.2 B) 11 C) 3.6
D) 0.048 E) 8.4 27) The rate constant for this reaction is __________ s-1 .
A) 0.013 B) 0.030 C) 0.14
-3
D) 3.0 E) 3.1 × 10
21) The reaction below is first order in [H 2 O 2 ] :
2H 2 O 2 (l) ! 2H 2 O (l) + O 2 (g) 28) The half-life of this reaction is __________ s.
A) 0.97 B) 7.1 C) 4.9
A solution originally at 0.600 M H 2 O 2 is found to be 0.075 M
D) 3.0 E) 0.14
after 54 min. The half-life for this reaction is __________ min.
A) 6.8 B) 18 C) 14 29) At elevated temperatures, methylisonitrile (CH3NC)
D) 28 E) 54
isomerizes to acetonitrile (CH3 CN):
CH3NC (g) → CH33CN (g)
22) Of the following, all are valid units for a reaction rate
except __________. The reaction is first order in methylisonitrile. The attached
A) mol/L B) M/s C) mol/hr graph shows data for the reaction obtained at 198.9°C.
D) g/s E) mol/L-hr
23) Which one of the following is not a valid expression for
the rate of the reaction below?
4NH3 + 7O 2 ! 4NO 2 + 6H 2 O
1 ![O 2 ] 1 ![NO 2 ]
A) " B)
7 !t 4 !t
1 ![H 2 O] 1 ![NH3 ]
C) D) "
6 !t 4 !t
E) All of the above are valid expressions of the reaction rate.
24) The rate law of a reaction is rate = k[D][X]. The units of
the rate constant are __________. The rate constant for the reaction is __________ s-1.
A) mol L -1s-1 B) L mol-1s-1 A) -1.9 × 104 B) +1.9 × 104
C) -5.2 × 10 -5 D) +5.2 × 10-5
C) mol2 L -2s-1 D) mol L -1s-2
E) +6.2
E) L2 mol-2 s-1
30) As the temperature of a reaction is increased, the rate of
the reaction increases because the __________.
25) The half-life of a first-order reaction __________. A) reactant molecules collide less frequently
A) is the time necessary for the reactant concentration to drop B) reactant molecules collide with greater energy per collision
to half its original value C) activation energy is lowered
B) is constant D) reactant molecules collide less frequently and with greater
C) can be calculated from the reaction rate constant energy per collision
D) does not depend on the initial reactant concentration E) reactant molecules collide more frequently with less energy
E) All of the above are correct. per collision

12. 31) In the energy profile of a reaction, the species that exists at 37) A possible mechanism for the overall reaction
the maximum on the curve is called the __________. Br2 (g) + 2NO (g) ! 2NOBr (g)
A) product B) activated complex is k1
C) activation energy D) enthalpy of reaction
NO (g) + Br2 (g) NO Br2 (g) (fast)
E) atomic state k2
k2
-Ea/RT
NO Br2 (g) + NO (g) 2NOBr (slow)
32) In the Arrhenius equation, k = Ae
__________ is the frequency factor. The rate law for formation of NOBr based on this mechanism is
A) k B) A C) e D) E a E) R rate = ________.
A) k1[NO]1/2 B) k1[Br2 ]1/2
C) (k 2 k1/k -1 )[NO]2 [Br2 ]
33) In general, as temperature goes up, reaction rate _______.
A) goes up if the reaction is exothermic D) (k1/k -1 ) 2 [NO]2 E) (k 2 k1/k -1 )[NO][Br2 ]2
B) goes up if the reaction is endothermic
C) goes up regardless of whether the reaction is exothermic or
endothermic 38) Of the following, __________ will lower the activation
D) stays the same regardless of whether the reaction is energy for a reaction.
exothermic or endothermic A) increasing the concentrations of reactants
E) stays the same if the reaction is first order B) raising the temperature of the reaction
C) adding a catalyst for the reaction
D) removing products as the reaction proceeds
E) increasing the pressure
34) At elevated temperatures, methylisonitrile (CH3NC)
isomerizes to acetonitrile (CH3 CN):
CH3NC (g) → CH3 CN (g) 39) A catalyst can increase the rate of a reaction __________.
The dependence of the rate constant on temperature is studied A) by changing the value of the frequency factor (A)
and the graph below is prepared from the results. B) by lowering the overall activation energy (E a ) of the
reaction
C) by lowering the activation energy of the reverse reaction
D) by providing an alternative pathway with a lower activation
energy
E) All of these are ways that a catalyst might act to increase the
rate of reaction.
40) The combustion of ethylene proceeds
by the reaction C2 H 4 (g) + 3O 2 (g) ! 2CO 2 (g) + 2H 2 O (g)
When the rate of disappearance of O 2 is 0.28 M s-1 , the rate of
The energy of activation of this reaction is __________ kJ/mol. appearance of CO 2 is __________ M s-1 .
A) 160 B) 1.6 × 105 C) 4.4 × 10-7 A) 0.19 B) 0.093 C) 0.84
D) 4.4 × 10 -4 E) 1.9 × 10 4 D) 0.42 E) 0.56
41) SO 2 Cl2 decomposes in the gas phase
by the reaction SO 2 Cl2 (g) ! SO 2 (g) + Cl2 (g)
35) The mechanism for formation of the product X is:
A+B→C+D (slow) The reaction is first order in SO 2 Cl2 and the rate constant is
B+D→X (fast) 3.0 × 10-6 s-1 at 600 K.
The intermediate reactant in the reaction is __________. A vessel is charged with 3.3 atm of SO 2 Cl2 at 600 K. The
A) A B) B C) C
D) D E) X partial pressure of SO 2 at 3.0 ! 105s is __________ atm.
A) 2.1 B) 2.0 C) 1.3
D) 3.0 E) 3.7
36) For the elementary reaction NO3 + CO ! NO 2 + CO 2 42) A particular first-order reaction has a
the molecularity of the reaction is __________, and the rate law rate constant of 1.35 × 102 s-1 at 25°C.
is rate = __________. What is the magnitude of k at 95°C if E a = 55.5 kJ/mol?
A) 2, k[NO3 ][CO] B) 4, k[NO3 ][CO][NO 2 ][CO 2 ]
A) 9.60 × 103 B) 2.85 × 104 C) 576
C) 2, k[NO 2 ][CO 2 ] D) 2, k[NO3 ][CO]/[NO 2 ][CO 2 ] D) 4.33 × 10 87 E) 1.36 × 102
E) 4, k[NO 2 ][CO 2 ]/[NO3 ][CO]