To find how the rate varies with respect to [H2], find two experiments in which [H2] changes but [I2] remains constant.. c When the order of each reactant is known, any one experimental
Trang 1CHAPTER 16 KINETICS: RATES AND
MECHANISMS OF CHEMICAL REACTIONS
a) The balanced equation is 4NO(g) + O2(g) 2N 2 O 3
Choose O2 as the reference because its coefficient is 1 Four molecules of NO (nitrogen monoxide) are consumed for every one O2 molecule, so the rate of O2 disappearance is 1/4 the rate of NO decrease By similar reasoning, the rate of O2 disappearance is 1/2 the rate of N2O3 (dinitrogen trioxide) increase
substance The number in the denominator of each fraction is the coefficient for the corresponding substance in the balanced equation The terms that are negative in the rate equation represent reactants in the balanced
equation, while the terms that are positive represent products in the balanced equation For example, the following term, 1 A
t
a , describes a product (the term is positive) that has a coefficient of a in the balanced equation In
part b), use the rate equation to compare the rate of appearance of H2O with the rate of disappearance of O2 Solution:
a) The balanced equation is 4NH 3 + 5O 2 4NO + 6H 2 O
b) Plan: Because H2O is increasing; its rate of concentration change is positive Substitute its rate into the
expression and solve for [O2]/t
Solution:
Rate = 1
6(2.52x10–2 mol/L•s) = –15 ∆[O2 ]
∆t –56(2.52x10–2 mol/L•s) = ∆[O2]
∆t = –2.10x10–2 mol/L·s
The negative value indicates that [O2] is decreasing as the reaction progresses The rate of reaction is always expressed as a positive number, so [O2] is decreasing at a rate of 2.10x10 –2
mol/L·s 16.2A Plan: The reaction orders of the reactants are the exponents in the rate law Add the individual reaction orders
to obtain the overall reaction order Use the rate law to determine how the changes listed in the problem will affect the rate
Solution:
a) The exponent of [I– ] is 1, so the reaction is first order with respect to I – Similarly, the reaction is first order
with respect to BrO 3 , and second order with respect to H + The overall reaction order is (1 + 1 + 2) = 4, or
fourth order overall
Trang 2b) Rate = k[I–][BrO3–][H+]2 If [BrO3] and [I–] are tripled and [H+] is doubled, rate = k[3 x I–][3 x BrO3–][2 x H+]2, then rate increases to 3 x 3 x 22 or 36 times its original value The rate increases by a factor of 36
16.2B Plan: The reaction orders of the reactants are the exponents in the rate law Add the individual reaction orders
to obtain the overall reaction order Use the rate law to determine how the changes listed in the problem will affect the rate
Solution:
a) The exponent of [ClO2] is 2, so the reaction is second order with respect to ClO 2 Similarly, the reaction is
first order with respect to OH – The overall reaction order is (1 + 2) = 3, or third order overall
b) Rate = k[ClO2]2[OH–] If [ClO2] is halved and [OH–] is doubled, rate = k[1/2 x ClO2]2[2 x OH–], then rate increases to (1/2)2 x 2 or 1/2 its original value The rate decreases by a factor of 1/2
16.3A Plan: Assume that the rate law takes the general form rate = k[H2]m[I2]n To find how the rate varies with respect
to [H2], find two experiments in which [H2] changes but [I2] remains constant Take the ratio of rate laws for those
two experiments to find m To find how the rate varies with respect to [I2], find two experiments in which [I2] changes but [H2] remains constant Take the ratio of rate laws for those two experiments to find n Add m and n
to obtain the overall reaction order Use the rate law to solve for the value of k
Solution:
For the reaction order with respect to [H2], compare Experiments 1 and 3:
rate 3rate 1 =
HH
m m
23 23
9.3 x 101.9 x 10
0.05500.0113
m m
4.8947 = (4.8672566)m
If the reaction order was more complex, an alternate method of solving for m is:
log (4.8947) = m log (4.8672566); m = log (4.8947)/log (4.8672566) = 1
For the reaction order with respect to [I2] , compare Experiments 2 and 4:
rate 4rate 2 =
II
n n
22 22
1.9 x 101.1 x 10
0.00560.0033
n n
k1 = (1.9x10–23 mol/L•s)/[(0.0113 mol/L)(0.0011 mol/L)] = 1.5 x 10–18 L/mol•s
k2 = (1.1x10–22 mol/L•s)/[(0.0220 mol/L)(0.0033 mol/L)] = 1.5 x 10–18 L/mol•s
k3 = (9.3x10–23 mol/L•s)/[(0.0550 mol/L)(0.0011 mol/L)] = 1.5 x 10–18 L/mol•s
k4 = (1.9x10–22 mol/L•s)/[(0.0220 mol/L)(0.0056 mol/L)] = 1.5 x 10–18 L/mol•s
Average k = 1.5x10–18 L/mol•s
16.3B Plan: Assume that the rate law takes the general form rate = k[H2SeO3]m[I–]n[H+]p To find how the rate varies
with respect to [H2SeO3], find two experiments in which [H2SeO3] changes but [I–] and [H+] remain constant
Take the ratio of rate laws for those two experiments to find m To find how the rate varies with respect to [I–], find two experiments in which [I–] changes but [H2SeO3] and [H+] remain constant Take the ratio of rate laws for
those two experiments to find n To find how the rate varies with respect to [H+], find two experiments in which
Trang 3[H+] changes but [H2SeO3] and [I–] remain constant Take the ratio of rate laws for those two experiments to find
p Add m, n, and p to obtain the overall reaction order Use the rate law to solve for the value of k
Solution:
For the reaction order with respect to [H2SeO3], compare Experiments 1 and 3:
rate 3rate 1 =
[H2SeO3]3m
[H 2 SeO 3 ]1m 3.94x10–6 mol/L•s 9.85x10–7 mol/L•s= [1.0x10
–2 mol/L]3m[2.5x10–3 mol/L]1m
4 = (4)m
For the reaction order with respect to [I–], compare Experiments 1 and 2:
rate 2 rate 1 = [I
–]2n[I–]1n 7.88x10–6 mol/L•s 9.85x10–7 mol/L•s= [3.0x10
–2 mol/L]2n[1.5x10–2 mol/L]1n
8 = (2)n
For the reaction order with respect to [H+], compare Experiments 2 and 4:
rate 4 rate 2 = [H
+]4p[H+]2p 3.15x 10–5 mol/L•s 7.88x10–6 mol/L•s= [3.0x10
–2 mol/L]4p[1.5x10–2 mol/L]2p
4 = (2)p
The rate law is rate = k[H2 SeO 3 ][I – ] 3 [H + ] 2 and is sixth order overall
b) Calculation of k:
k = Rate/([H2SeO3][I–]3[H+]2)
k1 = (9.85x10–7 mol/L•s)/[(2.5x10–3 mol/L)(1.5x10–2 mol/L)3(1.5x10–2 mol/L)2] = 5.2x10 5
L 5 /mol 5 •s
16.4A Plan: The reaction is second order in X and zero order in Y For part a), compare the two amounts of reactant X
For part b), compare the two rate values
concentration of X must have doubled to cause a four-fold increase in rate There should be 6 black spheres and 3 green spheres in Experiment 3
5 5
1.0 x 100.25 x 10
=
2 2
x3
4 = 2
x9
x = 6 black spheres
Trang 416.4B 16.4B Plan: Examine how a change in the concentration of the different reactants affects the rate in order to
determine the rate law Then use the rate law to determine the number of particles in the scene for Experiment 4 Solution:
a) Comparing Experiments 1 and 2, we can see that the number of blue A spheres does not change while the number of yellow B spheres changes from 4 to 2 Although the number of yellow spheres changes, the rate does not change Therefore, B has no effect on the rate, which suggests that the reaction is zero order with respect to B Now that we know that the yellow B spheres do not affect the rate, we can look at the influence of the blue A spheres Comparing Experiments 2 and 3, we can see that the number of blue spheres changes from 4 to 2
(experiment 3 concentration of A is half of that of experiment 2) while the rate changes from 1.6x10–3 mol/L•s to 8.0x10–4 mol/L•s (the rate of experiment 3 is half of the rate of experiment 2) The fact that the concentration of blue spheres changes in the same way the rate changes suggests that the reaction is first order with respect to the
blue A spheres Therefore, the rate law is: rate = k[A]
b) The rate of Experiment 4 is twice the rate in Experiment 1 Since the reaction is first order in A, the
concentration of A must have doubled to cause a two-fold increase in rate There should be 2 x 4 particles =
8 particles of A in the scene for Experiment 4
16.5A Plan: The rate expression indicates that the reaction order is two (exponent of [HI] = 2), so use the integrated
second-order law Substitute the given concentrations and the rate constant into the expression and solve for time Solution:
t = 4.6296x1021 s = 4.6x10 21
s
16.5B Plan: The problem states that the decomposition of hydrogen peroxide is a order, reaction, so use the
first-order integrated rate law Substitute the given concentrations and the time into the expression and solve for the rate constant
b) If you start with an initial concentration of hydrogen peroxide of 1.0 M ([A]0 = 1.0 M) and 25% of the sample
decomposes, 75% of the sample remains ([A]t = 0.75 M)
ln 1.00 M 0.75M = (0.041 min–1)(t)
t = 0.2877 0.041 min–1 = 7.0171 = 7.0 min
Trang 516.6A Plan: The initial scene contains 12 particles of Substance X In the second scene, which occurs after 2.5 minutes
have elapsed, half of the particles of Substance X (6 particles) remain Therefore, 2.5 min is the half-life The half-life is used to find the number of particles present at 5.0 min and 10.0 min To find the molarity of X, moles
of X is divided by the given volume
Solution:
a) Since 2.5 minutes is the half-life, 5.0 minutes represents two half-lives:
12 particles of X 2.5 min 6 particles of X 2.5 min 3 particles of X
After 5.0 minutes, 3 particles of X remain; 9 particles of X have reacted to produce 9 particles of
Y Draw a scene in which there are 3 black X particles and 9 red Y particles
b) 10.0 minutes represents 4 half-lives:
12 particles of X 2.5 min 6 particles of X 2.5 min 3 particles of X 2.5 min
1.5 particles of X 2.5 min 0.75 particle of X
Moles of X after 10.0 min = 0.75 particle 0.20 mol
1 particle
= 0.15 mol Molarity = mol X = 0.15 mol
volume 0.50 L = 0.30 M
16.6B Plan: The initial scene contains 16 particles of Substance A In the second scene, which occurs after 24 minutes
have elapsed, half of the particles of Substance A (8 particles) remain Therefore, 24 min is the half-life The half-life is used to find the amount of time that has passed when only one particle of Substance A remains and to find the number of particles of A present at 72 minutes To find the molarity of A, moles of A is divided by the given volume
Solution:
a) The half-life is 24 minutes We can use that information to determine the amount of time that has passed when one particle of Substance A remains:
16 A particles 24 min 8 A particles 24 min 4 A particles 24 min 2 A particles 24 min 1 A particle
After 4 half-lives, only one particle of Substance A remains
4 half-lives = 4 x (24 min/half-life) = 96 min
b)
Number of half-lives at 72 min = 72 min
24 min/half-life = 3 half-lives According to the scheme above in part a), there are 2 A particles left after three half-lives
Amount (mol) of A after 72 minutes = (2 particles A) 0.10 mol A
t = 0.0529119985 = 0.0529 h –1
Trang 616.7B Plan: Rearrange the first-order half-life equation to solve for half-life Determine the number of half-lives that
pass in the 40 day period described in the problem and use this number to determine the amount of pesticide remaining
8 days/half-life = 5 half-lives After each half-life, ½ of the sample remains
After five half-lives, ½ x ½ x ½ x ½ x ½ = (½)5 = 1
32 of the sample remains
16.8A Plan: The activation energy, rate constant at T1, and a second temperature, T2, are given Substitute these values
into the Arrhenius equation and solve for k2, the rate constant at T2
0.286 L/mol • s
k
= –0.49093
20.286 L/mol • s
k
= 0.612057
k2 = (0.612057)(0.286 L/mol•s) = 0.175048 = 0.175 L/mol•s
16.8B Plan: The activation energy and rate constant at T1 are given We are asked to find the temperature at which the
rate will be twice as fast (i.e., the temperature at which k2 = 2 x k1) Substitute the given values into the Arrhenius
equation and solve for T2
Trang 7Solution:
The reaction energy diagram indicates that O(g) + H2O(g) 2OH(g) is an endothermic process, because the
energy of the product is higher than the energy of the reactants The highest point on the curve indicates the transition state In the transition state, an oxygen atom forms a bond with one of the hydrogen atoms on the H2O molecule (hashed line) and the O-H bond (dashed line) in H2O weakens Ea(fwd) is the sum of Hrxn and Ea(rev)
H
OHO
16.9B Plan: Ea(fwd) is the sum of Hrxn and Ea(rev), so Hrxn can be calculated by subtracting Ea(fwd) – Ea(rev) Use Sample
Problem 16.9 as a guide for drawing the diagram
between the hydrogen and the bromine weakens (another hashed line):
16.10A Plan: The overall reaction can be obtained by adding the three steps together The molecularity of each step is the
total number of reactant particles; the molecularities are used as the orders in the rate law for each step
Trang 8(3) HO2(aq) + OH(aq) H2O(l) + O2(g)
Total: 2H2O2(aq) + 2OH(aq) + HO2(aq) 2OH(aq) + 2H2O(l) + HO2(aq) + O2(g)
(overall) 2H2O2(aq) 2H2O(l) + O2(g)
(3) Rate3 = k3 [HO2][OH]
16.10B Plan: The overall reaction can be obtained by adding the three steps together The molecularity of each step is the
total number of reactant particles; the molecularities are used as the orders in the rate law for each step
16.11A Plan: The overall reaction can be obtained by adding the three steps together An intermediate is a substance that
is formed in one step and consumed in a subsequent step The overall rate law for the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the experimental rate law
Solution:
a) (1) H2O2(aq) 2OH(aq)
(2) H2O2(aq) + OH(aq) H2O(l) + HO2(aq)
(3) HO2(aq) + OH(aq) H2O(l) + O2(g)
Total: 2H2O2(aq) + 2OH(aq) + HO2(aq) 2OH(aq) + 2H2O(l) + HO2(aq) + O2(g)
(overall) 2H2O2(aq) 2H2O(l) + O2(g)
2OH(aq) and HO2(aq) are intermediates in the given mechanism 2OH( aq) are produced in the first step and
consumed in the second and third steps; HO2(aq) is produced in the second step and consumed in the third step
Notice that the intermediates were not included in the overall reaction
b) The observed rate law is: rate = k[H2O2] In order for the mechanism to be consistent with the rate law, the
first step must be the slow step The rate law for step one is the same as the observed rate law
16.11B Plan: The overall reaction can be obtained by adding the three steps together An intermediate is a substance that
is formed in one step and consumed in a subsequent step The overall rate law for the mechanism is determined from the slowest step (the rate-determining step) and can be compared to the experimental rate law
N 2 O 2(g) and N2O(g) are intermediates in the given mechanism N2O2(g) is produced in the first step and
consumed in the second step; N2O(g) is produced in the second step and consumed in the third step Notice that
the intermediates were not included in the overall reaction
b) The observed rate law is: rate = k[NO]2[H2], and the second step is the slow, or rate-determining, step The rate
law for step two is: rate = k2[N2O2][H2] This rate law is NOT the same as the observed rate law However, since
Trang 9N2O2 is an intermediate, it must be replaced by using the first step For an equilibrium, rateforward rxn = ratereverse rxn
For step 1 then, k1[NO]2 = k–1[N2O2] Rearranging to solve for [N2O2] gives [N2O2] = (k1/k–1)[NO]2 Substituting this value for [N2O2] into the rate law for the second step gives the overall rate law as rate = (k2k1/k–1)[NO]2[H2] or
rate = k[NO]2[H2], which is consistent with the observed rate law
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B16.1 Plan: Add the two equations, canceling substances that appear on both sides of the arrow The
rate law for each step follows the format of rate = k[reactants] An initial reactant that appears as a
product in a subsequent step is a catalyst; a product that appears as a reactant in a subsequent step
is an intermediate (produced in one step and consumed in a subsequent step)
Step 1 Rate1 = k1[X][O3]
Step 2 Rate2 = k2[XO][O]
b) X acts as a catalyst; it is a reactant in step 1 and a product in step 2 XO acts as an intermediate; it was
produced in step 1 and consumed in step 2
B16.2 Plan: Replace X in the mechanism in B16.1 with NO, the catalyst To find the rate of ozone
depletion at the given concentrations, use step 1) since it is the rate-determining (slow) step
B16.3 Plan: The p factor is the orientation probability factor and is related to the structural complexity of
the colliding particles The more complex the particles, the smaller the probability that collisions
will occur with the correct orientation and the smaller the p factor
Solution:
a) Step 1 with Cl will have the higher value for the p factor Since Cl is a single atom, no matter
` how it collides with the ozone molecule, the two particles should react, if the collision has enough
energy NO is a molecule If the O3 molecule collides with the N atom in the NO molecule,
reaction can occur as a bond can form between N and O; if the O3 molecule collides with the O atom in the NO molecule, reaction will not occur as the bond between N and O cannot form The probability of a successful collision is smaller with NO
b) The transition state would have weak bonds between the chlorine atom and an oxygen atom in ozone, and between that oxygen atom and a second oxygen atom in ozone
OOO
Trang 10unit volume In other words, the gas concentration increases due to increased pressure, so the reaction rate
increases Increased pressure also causes more collisions between gas molecules
16.3 The addition of more water will dilute the concentrations of all solutes dissolved in the reaction vessel If any of
these solutes are reactants, the rate of the reaction will decrease
16.4 An increase in solid surface area would allow more gaseous components to react per unit time and thus would
increase the reaction rate
16.5 An increase in temperature affects the rate of a reaction by increasing the number of collisions, but more
importantly, the energy of collisions increases As the energy of collisions increases, more collisions result in reaction (i.e., reactants products), so the rate of reaction increases
16.6 The second experiment proceeds at the higher rate I2 in the gaseous state would experience more collisions with gaseous H2
16.7 The reaction rate is the change in the concentration of reactants or products per unit time Reaction rates change
with time because reactant concentrations decrease, while product concentrations increase with time
16.8 a) For most reactions, the rate of the reaction changes as a reaction progresses The instantaneous rate is the rate
at one point, or instant, during the reaction The average rate is the average of the instantaneous rates over a period
of time On a graph of reactant concentration vs time of reaction, the instantaneous rate is the slope of the
tangent to the curve at any one point The average rate is the slope of the line connecting two points on the curve The closer together the two points (shorter the time interval), the more closely the average rate agrees with the
b) The initial rate is the instantaneous rate at the point on the graph where time = 0, that is when reactants are mixed
16.9 The calculation of the overall rate is the difference between the forward and reverse rates This complication may
be avoided by measuring the initial rate, where product concentrations are negligible, so the reverse rate is negligible Additionally, the calculations are simplified as the reactant concentrations can easily be determined from the volumes and concentrations of the solutions mixed
16.10 At time t = 0, no product has formed, so the B(g) curve must start at the origin Reactant concentration (A(g))
decreases with time; product concentration (B(g)) increases with time Many correct graphs can be drawn Two
examples are shown below The graph on the left shows a reaction that proceeds nearly to completion, i.e., [products] >> [reactants] at the end of the reaction The graph on the right shows a reaction that does not proceed
to completion, i.e., [reactants] > [products] at reaction end
16.11 a) Calculate the slope of the line connecting (0, [C]o) and (tf, [C]f) (final time and concentration) The negative of
this slope is the average rate
b) Calculate the negative of the slope of the line tangent to the curve at t = x
c) Calculate the negative of the slope of the line tangent to the curve at t = 0
d) If you plotted [D] vs time, you would not need to take the negative of the slopes in a)-c) since [D]
would increase over time
Trang 1116.12 Plan: The average rate is the total change in concentration divided by the total change in time
= 0.00103 = 0.0010 mol/L•s The negative of the slope is used because rate is defined as the change in product concentration with time If a reactant is used, the rate is the negative of the change in reactant concentration The 1/2 factor is included to account for the stoichiometric coefficient of 2 for AX2 in the reaction
b)
The slope of the tangent to the curve (dashed line) at t = 0 is approximately –0.004 mol/L•s This initial rate is
greater than the average rate as calculated in part a) The initial rate is greater than the average rate because
rate decreases as reactant concentration decreases
16.13 Plan: The average rate is the total change in concentration divided by the total change in time
b) The rate at exactly 5.0 s will be higher than the rate in part a)
The slope of the tangent to the curve at t = 5.0 s (the rate at 5.0 s) is approximately –2.8x10–3
mol/L•s
00.010.020.030.040.050.06
Trang 1216.14 Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance A negative sign is used for the rate in terms of reactant A since A is reacting and [A] is decreasing over time Positive signs are used for the rate in terms of products B and C since B and C are being formed and [B] and [C] increase over time Reactant A decreases twice as fast as product C increases because two molecules
of A disappear for every molecule of C that appears
The negative value indicates that [A] is decreasing as the reaction progresses The rate of reaction is always
expressed as a positive number, so [A] is decreasing at a rate of 4 mol/L•s
16.15 Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance A negative sign is used for the rate in terms of reactant D since D is reacting and [D] is decreasing over time Positive signs are used for the rate in terms of products E and F since E and F are being formed and [E] and [F] increase over time For every 3/2 mole of product E that is formed, 5/2 mole of F is produced
Calculating the rate of change of [F]:
0.25 mol E/L•s 5/2 mol F/L•s
3/2 mol E/L•s
= 0.416667 = 0.42 mol/L•s
16.16 Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance A negative sign is used for the rate in terms of reactants A and B since A and B are reacting and [A] and [B] are decreasing over time A positive sign is used for the rate in terms of product C since C is being formed and [C] increases over time The 1/2 factor is included for reactant B to account for the stoichiometric
0 0.01 0.02 0.03 0.04 0.05 0.06
Trang 13coefficient of 2 for B in the reaction Reactant A decreases half as fast as reactant B decreases because one molecule of A disappears for every two molecules of B that disappear
Calculating the rate of change of [A]:
0.5 mol B/L•s 1 mol A/L•s
2 mol B/L•s
= – 0.25 mol/L•s = – 0.2 mol/L•s
The negative value indicates that [A] is decreasing as the reaction progresses The rate of reaction is always
expressed as a positive number, so [A] is decreasing at a rate of 0.2 mol/L•s
16.17 Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance A negative sign is used for the rate in terms of reactants D, E, and F since these substances are reacting and [D], [E], and [F] are decreasing over time Positive signs are used for the rate in terms of products G and H since these substances are being formed and [G] and [H] increase over time Product H increases half as fast as reactant D decreases because one molecule of H is formed for every two molecules of D that disappear
16.18 Plan: A term with a negative sign is a reactant; a term with a positive sign is a product The inverse of the fraction
becomes the coefficient of the molecule
Solution:
N2O5 is the reactant; NO2 and O2 are products
2N 2 O 5(g) 4NO 2(g) + O2(g)
16.19 Plan: A term with a negative sign is a reactant; a term with a positive sign is a product The inverse of the fraction
becomes the coefficient of the molecule
Solution:
CH4 and O2 are the reactants; H2O and CO2 are products
CH 4 + 2O 2 2H 2 O + CO 2
16.20 Plan: The average rate is the total change in concentration divided by the total change in time The initial rate is
the slope of the tangent to the curve at t = 0.0 s and the rate at 7.00 s is the slope of the tangent to the curve at
Trang 14e) Average between t = 3 s and t = 5 s is:
Rate = – [(0.0050 – 0.0063) mol/L]/[5.00 – 3.00) s] = 6.5x10–4 mol/L•s Rate at 4 s 6.7x10–4 mol/L•s, thus the rates are equal at about 4 seconds
16.21 Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance A negative sign is used for the rate in terms of reactants N2 and H2 since these substances are reacting and [N2] and [H2] are decreasing over time A positive sign is used for the rate in terms of the product NH3 since
it is being formed and [NH3] increases over time
[NH ]1
2
t
16.22 Plan: Use Equation 16.2 to describe the rate of this reaction in terms of reactant disappearance and product
appearance A negative sign is used for the rate in terms of the reactant O2 since it is reacting and [O2] is
decreasing over time A positive sign is used for the rate in terms of the product O3 since it is being formed and [O3] increases over time O3 increases 2/3 as fast as O2 decreases because two molecules of O3 are formed for every three molecules of O2 that disappear
16.23 a) k is the rate constant, the proportionality constant in the rate law k represents the fraction of successful
collisions which includes the fraction of collisions with sufficient energy and the fraction of collisions with
correct orientation k is a constant that varies with temperature
b) m represents the order of the reaction with respect to [A] and n represents the order of the reaction with respect
to [B] The order is the exponent in the relationship between rate and reactant concentration and defines how reactant concentration influences rate
Trang 15The order of a reactant does not necessarily equal its stoichiometric coefficient in the balanced equation If a reaction is an elementary reaction, meaning the reaction occurs in only one step, then the orders and
stoichiometric coefficients are equal However, if a reaction occurs in a series of elementary reactions, called a mechanism, then the rate law is based on the slowest elementary reaction in the mechanism The orders of the reactants will equal the stoichiometric coefficients of the reactants in the slowest elementary reaction but may not equal the stoichiometric coefficients in the overall reaction
c) For the rate law rate = k[A] [B]2 substitute in the units:
Rate (mol/L·min) = k[A]1[B]2
k = rate1 2[A] [B] = mol/L•min1 2
k =
3 3
16.24 a) Plot either [A2] or [B2] vs time and determine the negative of the slope of the line tangent to the curve at t = 0
b) A series of experiments at constant temperature but with different initial concentrations are run to determine different initial rates By comparing results in which only the initial concentration of a single reactant is changed, the order of the reaction with respect to that reactant can be determined
c) When the order of each reactant is known, any one experimental set of data (reactant concentration and reaction rate) can be used to determine the reaction rate constant at that temperature
16.25 a) The rate doubles If rate = k[A]1 and [A] is doubled, then the rate law becomes rate = k[2 x A]1 The rate increases by 21 or 2
b) The rate decreases by a factor of four If rate = k[B]2 and [B] is halved, then the rate law becomes
rate = k[1/2 x B]2 The rate decreases to (1/2)2 or 1/4 of its original value
c) The rate increases by a factor of nine If rate = k[C]2 and [C] is tripled, then the rate law becomes
rate = k[3 x C]2 The rate increases to 32 or 9 times its original value
16.26 Plan: The order for each reactant is the exponent on the reactant concentration in the rate law The individual
orders are added to find the overall reaction order
16.27 Plan: The order for each reactant is the exponent on the reactant concentration in the rate law The individual
orders are added to find the overall reaction order
Solution:
The rate law may be rewritten as rate = k[O3]2[O2] –1 The order with respect to [O3] is 2 since it has an
exponent of 2 The order with respect to [O2] is –1 since it has an exponent of –1 The overall reaction order
is 2 + (–1) = 1
second order with respect to O 3 , (–1) order with respect to O 2 , first order overall
16.28 a) The rate is first order with respect to [BrO3] If [BrO3] is doubled, rate = k[2 x BrO3], then rate increases to 21
or 2 times its original value The rate doubles
b) The rate is first order with respect to [Br–] If [Br–] is halved, rate = k[1/2 x Br–], then rate decreases by a factor
of (1/2)1 or 1/2 times its original value The rate is halved
c) The rate is second order with respect to [H+] If [H+] is quadrupled, rate = k[4 x H+]2, then rate increases to 42 or
16 times its original value
Trang 1616.29 a) The rate is second order with respect to [O3] If [O3] is doubled, rate = k[2 x O3]2, then rate increases to 22 or 4
times its original value The rate increases by a factor of 4
b) [O2] has an order of –1 If [O2] is doubled, rate = k[2 x O2] –1, then rate decreases to 2–1 or 1/2 times its original
value The rate decreases by a factor of 2
c) [O2] has an order of –1 If [O2] is halved, rate = k[1/2 x O2] –1, then rate decreases by a factor of (1/2)–1 or 2
times its original value The rate increases by a factor of 2
16.30 Plan: The order for each reactant is the exponent on the reactant concentration in the rate law The individual
orders are added to find the overall reaction order
Solution:
The order with respect to [NO 2 ] is 2, and the order with respect to [Cl 2 ] is 1 The overall order is: 2 + 1 = 3 for the overall order
16.31 Plan: The order for each reactant is the exponent on the reactant concentration in the rate law The individual
orders are added to find the overall reaction order
Solution:
The rate law may be rewritten as rate = k[HNO2]4[NO] –2
The order with respect to [HNO 2 ] is 4, and the order with respect to [NO] is –2 The overall order is:
4 + (– 2) = 2 for the overall order
16.32 a) The rate is second order with respect to [NO2] If [NO2] is tripled, rate = k[3 x NO2]2, then rate increases to 32
or 9 times its original value The rate increases by a factor of 9
b) The rate is second order with respect to [NO2] and first order with respect to [Cl2] If [NO2] and [Cl2] are
doubled, rate = k[2 x NO2]2[2 x Cl2]1, then the rate increases by a factor of 2 2
x 2 1 = 8
c) The rate is first order with respect to [Cl2] If Cl2 is halved, rate = k[1/2 x Cl2]1, then rate decreases to 1/2 times
its original value The rate is halved
16.33 a) The rate is fourth order with respect to [HNO2] If [HNO2] is doubled, rate = k[2 x HNO2]4, then rate increases
to 24 or 16 times its original value The rate increases by a factor of 16
b) [NO] has an order of –2 If [NO] is doubled, rate = k[2 x NO]–2, then rate increases to 2–2 or 1/(2)2 = 1/4 times
its original value The rate decreases by a factor of 4
c) The rate is fourth order with respect to [HNO2] If [HNO2] is halved, rate = k[1/2 x HNO2]4, then rate decreases
to (1/2)4 or 1/16 times its original value The rate decreases by a factor of 16
16.34 Plan: The rate law is rate = [A]m[B]n where m and n are the orders of the reactants To find the order of each
reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes Once
the rate law is known, any experiment can be used to find the rate constant k
Solution:
a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B]
is constant Use experiments 1 and 2 (or 3 and 4 would work) to find the order with respect to [A]
Set up a ratio of the rate laws for experiments 1 and 2 and fill in the values given for rates and concentrations and
solve for m, the order with respect to [A]
m = 2
Using experiments 3 and 4 also gives second order with respect to [A]
To find the order for reactant B, first identify the reaction experiments in which [B] changes but [A]
is constant Use experiments 1 and 3 (or 2 and 4 would work) to find the order with respect to [B]
Set up a ratio of the rate laws for experiments 1 and 3 and fill in the values given for rates and concentrations and
solve for n, the order with respect to [B]
Trang 17The reaction is first order with respect to [B]
b) The rate law, without a value for k, is rate = k[A]2
L 2 /mol 2 •min
16.35 Plan: The rate law is rate = k [A] m[B]n[C]p where m, n, and p are the orders of the reactants To find the order of
each reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes
Once the rate law is known, any experiment can be used to find the rate constant k
1.25x10 mol/L•min 0.1000 mol/L
= 0.0500 mol/L6.25x10 mol/L•min
m = 1 The order is first order with respect to A
To find the order for reactant B, first identify the reaction experiments in which [B] changes but [A] and [C] are constant Use experiments 2 and 3 to find the order with respect to [B] Set up a ratio of the rate laws for
experiments 2 and 3 and fill in the values given for rates and concentrations and solve for n, the order with respect
5.00 x 10 mol/L•min 0.1000 mol/L
= 0.0500 mol/L1.25 x 10 mol/L•min
n = 2 The reaction is second order with respect to B
To find the order for reactant C, first identify the reaction experiments in which [C] changes but [A] and [B] are constant Use experiments 1 and 4 to find the order with respect to [C] Set up a ratio of the rate laws for
experiments 1 and 4 and fill in the values given for rates and concentrations and solve for p, the order with respect
Trang 183 3
6.25x10 mol/L•min 0.0200 mol/L
= 0.0100 mol/L6.25x10 mol/L•min
/mol 2 •s
16.36 Plan: Write the appropriate rate law and enter the units for rate and concentrations to find the units of k The units
of k are dependent on the reaction orders and the unit of time
Solution:
a) A first-order rate law follows the general expression, rate = k[A] The reaction rate is expressed as a change in
concentration per unit time with units of mol/L·time Since [A] has units of mol/L, k has units of time–1:
Rate = k[A]
mol = molL•time k L
k =
molL•timemolL
= mol x LL•time mol =
1time = time –1
b) A second-order rate law follows the general expression, rate = k[A]2 The reaction rate is expressed as a change
in concentration per unit time with units of mol/L·time Since [A] has units of mol2/L2, k has units of
=
2 2
x L•time mol = •
L mol time
c) A third-order rate law follows the general expression, rate = k[A]3 The reaction rate is expressed as a change in concentration per unit time with units of mol/L•time Since [A] has units of mol3/L3, k has units of L2
/mol 2 •time:
Rate = k[A]3
3
mol = molL•time k L
=
3 3
x
2 2
L mol time
d) A 5/2-order rate law follows the general expression, rate = k[A]5/2 The reaction rate is expressed as a change
in concentration per unit time with units of mol/L·time Since [A] has units of mol5/2/L5/2, k has units of
Trang 19k = 5/2
5/2
molL•timemolL
=
5/2 5/2
x
3/2 3/2 L
16.37 Plan: Write the appropriate rate law and enter the units for rate and the rate constant to find the units of
concentration The units of concentration will give the reaction order
1L
x
1/2 1/2
mol = mol mol
molmol = L• min
Trang 207/2 7/2
m must be 7/2 The reaction is 7/2 order
16.38 Plan: The rate law is rate = k [CO] m[Cl2]n where m and n are the orders of the reactants To find the order of each
reactant, take the ratio of the rate laws for two experiments in which only the reactant in question changes Once
the rate law is known, the data in each experiment can be used to find the rate constant k
1.29x10 mol/L•min 1.00 mol/L
= 0.100 mol/L1.33x10 mol/L•min
The reaction is first order with respect to [CO]
To find the order for Cl2, first identify the reaction experiments in which [Cl2] changes but [CO] is constant Use experiments 2 and 3 to find the order with respect to [Cl2] Set up a ratio of the rate laws for experiments 2 and 3
and fill in the values given for rates and concentrations and solve for n, the order with respect to [Cl2]
1.30x10 mol/L•min = 1.00 mol/L
0.100 mol/L1.33x10 mol/L•min
Exp 1: k1 = (1.29x10–29 mol/L•s)/[1.00 mol/L][0.100 mol/L] = 1.29x10–28 L/mol•s
Exp 2: k2 = (1.33x10–30 mol/L•s)/[0.100 mol/L][0.100 mol/L] = 1.33x10–28 L/mol•s
Exp 3: k3 = (1.30x10–29 mol/L•s)/[0.100 mol/L][1.00 mol/L] = 1.30x10–28 L/mol•s
Exp 4: k4 = (1.32x10–31 mol/L•s)/[0.100 mol/L][0.0100 mol/L] = 1.32x10–28 L/mol•s
kavg = (1.29x10–28 + 1.33x10–28 + 1.30x10–28 + 1.32x10–28) L/mol•s/4 = 1.31x10 –28
L/mol•s
16.39 The integrated rate law can be used to plot a graph If the plot of [reactant] vs time is linear, the order is zero If
the plot of ln[reactant] vs time is linear, the order is first If the plot of inverse concentration (1/[reactant]) vs time is linear, the order is second
a) The reaction is first order since ln[reactant] vs time is linear
b) The reaction is second order since 1/[reactant] vs time is linear
c) The reaction is zero order since [reactant] vs time is linear
16.40 The half-life (t1/2) of a reaction is the time required to reach half the initial reactant concentration For a first-order
process, no molecular collisions are necessary, and the rate depends onlyon the fraction of the molecules having sufficient energy to initiate the reaction
16.41 Plan: The rate expression indicates that this reaction is second order overall (the order of [AB] is 2), so use the
second-order integrated rate law to find time We know k (0.2 L/mol•s), [AB]0 (1.50 M), and
Trang 21[AB]t (1/3[AB]0 = 1/3(1.50 M) = 0.500 M), so we can solve for t
16.42 Plan: The rate expression indicates that this reaction is second order overall (the order of [AB] is 2), so use the
second-order integrated rate law We know k (0.2 L/mol•s), [AB]0 (1.50 M), and t (10.0 s), so we can solve for
16.43 Plan: This is a first-order reaction so use the first-order integrated rate law In part a), we know t (10.5 min) Let
[A]0 = 1 M and then [A] t = 50% of 1 M = 0.5 M Solve for k In part b), use the value of k to find the time
necessary for 75.0% of the compound to react If 75.0% of the compound has reacted, 100–75 = 25% remains at
time t Let [A]0 = 1 M and then [A] t = 25% of 1 M = 0.25 M
t
k
1/2
ln 2 =
k
10.5 min = 0.066014 = 0.0660 min
–1 b) ln [A]t = ln [A]0 – kt
Trang 22
16.44 Plan: This is a first-order reaction so use the first-order integrated rate law (the units of k, yr–1, indicates first
order) In part a), the first-order half-life equation may be used to solve for half-life since k is known In part b), use the value of k to find the time necessary for the reactant concentration to drop to 12.5% of the initial
concentration Let [A]0 = 1.00 M and then [A] t = 12.5% of 1 M = 0.125 M
Solution:
a) t1/2 = ln 2
k = ln 2 10.0012 yr = 577.62 = 5.8x10
2
yr b) ln [A]t = ln [A]0 – kt
16.45 Plan: In a first-order reaction, ln [NH3] vs time is a straight line with slope equal to k The half-life can be
determined using the first-order half-life equation
Solution:
a) A new data table is constructed: (Note that additional significant figures are retained in the calculations.)
x-axis (time, s) [NH3] y-axis (ln [NH3])
k = ln 23 13.260x10 s = 212.62 = 2x10
2
s
16.46 The central idea of collision theory is that reactants must collide with each other in order to react If reactants must
collide to react, the rate depends on the product of the reactant concentrations
1.379 1.38 1.381 1.382 1.383 1.384 1.385 1.386 1.387
Trang 2316.47 No, collision frequency is not the only factor affecting reaction rate The collision frequency is a count of the total
number of collisions between reactant molecules Only a small number of these collisions lead to a reaction Other factors that influence the fraction of collisions that lead to reaction are the energy and orientation of the collision A collision must occur with a minimum energy (activation energy) to be successful In a collision, the orientation, that
is, which ends of the reactant molecules collide, must bring the reacting atoms in the molecules together in order for the collision to lead to a reaction
16.48 At any particular temperature, molecules have a distribution of kinetic energies, as will their collisions have a
range of energies As temperature increases, the fraction of these collisions which exceed the threshold energy, increases; thus, the reaction rate increases
16.49 k = e Ea/RT
A
The Arrhenius equation indicates a negative exponential relationship between temperatures and the rate constant,
k In other words, the rate constant increases exponentially with temperature
16.50 The Arrhenius equation, k = e Ea/RT
A , can be used directly to solve for activation energy at a specified
temperature if the rate constant, k, and the frequency factor, A, are known However, the frequency factor is
usually not known To find Ea without knowing A, rearrange the Arrhenius equation to put it in the form of a linear plot: ln k = ln A – Ea/RT where the y value is ln k and the x value is 1/T Measure the rate constant at a series of temperatures and plot ln k vs 1/T The slope equals –Ea/R
16.51 a) The value of k increases exponentially with temperature
b) A plot of ln k vs 1/T is a straight line whose slope is –Ea/R
a)
b)
Trang 24The activation energy is determined from the slope of the line in the ln k vs 1/T graph The slope equals –Ea/R
16.52 a) As temperature increases, the fraction of collisions which exceed the activation energy increases; thus, the
reaction rate increases
b) A decrease in activation energy lowers the energy threshold with which collisions must take place to be
effective At a given temperature, more collisions occur with the lower energy so rate increases
16.53 No For 4x10 –5 moles of EF to form, every collision must result in a reaction and no EF molecule can decompose
back to AB and CD Neither condition is likely All collisions will not result in product as some collisions will occur with an energy that is lower than the activation energy In principle, all reactions are reversible, so some EF molecules decompose Even if all AB and CD molecules did combine, the reverse decomposition rate would result in an amount of EF that is less than 4x10–5 moles
16.54 Collision frequency is proportional to the velocity of the reactant molecules At the same temperature, both
reaction mixtures have the same average kinetic energy, but not the same velocity Kinetic energy equals 1/2 mv2,
where m is mass and v velocity The methylamine (N(CH3)3) molecule has a greater mass than the ammonia molecule, so methylamine molecules will collide less often than ammonia molecules, because of their slower velocities Collision energy thus is less for the N(CH3)3(g) + HCl(g) reaction than for the NH3(g) + HCl(g)
reaction Therefore, the rate of the reaction between ammonia and hydrogen chloride is greater than the
rate of the reaction between methylamine and hydrogen chloride
The fraction of successful collisions also differs between the two reactions In both reactions the hydrogen from HCl is bonding to the nitrogen in NH3 or N(CH3)3 The difference between the reactions is in how easily the H can collide with the N, the correct orientation for a successful reaction The groups (H) bonded to nitrogen in
ammonia are less bulky than the groups bonded to nitrogen in trimethylamine (CH3) So, collisions with correct orientation between HCl and NH3 occur more frequently than between HCl and N(CH3)3 and the reaction
NH3(g) + HCl(g) NH4Cl(s) occurs at a higher rate than N(CH3)3(g) + HCl(g) (CH3)3NHCl(s) Therefore, the
rate of the reaction between ammonia and hydrogen chloride is greater than the rate of the reaction between methylamine and hydrogen chloride
16.55 Each A particle can collide with three B particles, so (4 x 3) = 12 unique collisions are possible
16.56 Plan: Use Avogadro’s number to convert moles of particles to number of particles The number of unique
collisions is the product of the number of A particles and the number of B particles
= 7.76495x1047 = 7.76x10 47
unique collisions 16.57 Plan: The fraction of collisions with a specified energy is equal to the e Ea/RT term in the Arrhenius equation Solution:
Trang 25Fraction =e Ea/RT = e–37.238095 = 6.725131x10–17
The fraction increased by (6.725131x10–17)/(2.9577689x10–18) = 22.737175 = 22.7
16.59 Plan: You are given one rate constant k1 at one temperature T1 and the activation energy Ea Substitute these
values into the Arrhenius equation and solve for k2 at the second temperature
Solution:
k1 = 4.7x10–3 s–1 T1 = 25°C + 273 = 298 K
k2 = ? T2 = 75°C + 273 = 348 K
Ea = 33.6 kJ/mol = 33,600 J/mol 2
16.60 Plan: You are given the rate constants, k1 and k2, at two temperatures, T1 and T2 Substitute these values into the
Arrhenius equation and solve for Ea
Solution:
k1 = 4.50x10–5 L/mol•s T1 = 195°C + 273 = 468 K
k2 = 3.20x10–3 L/mol•s T2 = 258°C + 273 = 531 K
Ea = ? 2
ln
k R
16.61 Plan: The reaction is exothermic (ΔH is negative), so the energy of the products must be lower than
that of the reactants Use the relationship Hrxn = Ea(fwd) – Ea(rev) to solve for Ea(rev) To draw the transition state, note that the bond between B and C will be breaking while a bond between C and D will be forming Solution:
a)
(fwd)
Trang 2616.62 Plan: The forward activation energy Ea(fwd) is larger than the reverse activation energy Ea(rev) which indicates that
the energy of the products must be higher than that of the reactants Use the relationship
Hrxn = Ea(fwd) – Ea(rev) to solve for Hrxn To draw the transition state, note that the bonds in the A2 and B2
molecules will be breaking while bonds between A and B will be forming