Kinetics Integrated Rates - Putting it All Together 2014...

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2016 EDITIONPRESENTER

AP CHEMISTRY

Kinetics—Integrated

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2016 AP Chemistry - Kinetics: Integrated Rates

ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

Throughout the test the following symbols have the definitions specified unless otherwise noted.

L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury g = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s) atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hν

c = λν

E = energy ν = frequency

λ = wavelength

Planck’s constant, h = 6.626 × 10−34 J s Speed of light, c = 2.998 × 108 m s−1

Avogadro’s number = 6.022 × 1023 mol−1 Electron charge, e = −1.602 × 10−19 coulomb

EQUILIBRIUM

Kc = [C] [D]

[A] [B]

c d

a b, where a A + b B c C + d D

Kp = C

A B

( ) ( )

( ) ( )

c dD

a b

P P

P P

Ka = [H ][A ][HA]

+ -

Kb = [OH ][HB ][B]

- +

Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

= Ka × Kb

pH = − log[H+] , pOH = − log[OH−]

14 = pH + pOH

pH = pKa + log [A ][HA]

-

pKa = − logKa , pKb = − logKb

Equilibrium Constants

Kc (molar concentrations)

Kp (gas pressures)

Ka (weak acid)

Kb (weak base)

Kw (water)

KINETICS

ln[A] t − ln[A]0 = − kt

[ ] [ ]0A A1 1

t

- = kt

t½ = 0.693k

k = rate constant

t = time t½ = half-life

ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

Throughout the test the following symbols have the definitions specified unless otherwise noted.

L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury g = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s) atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hν

c = λν

E = energy ν = frequency

λ = wavelength

Planck’s constant, h = 6.626 × 10−34 J s Speed of light, c = 2.998 × 108 m s−1

Avogadro’s number = 6.022 × 1023 mol−1 Electron charge, e = −1.602 × 10−19 coulomb

EQUILIBRIUM

Kc = [C] [D]

[A] [B]

c d

a b, where a A + b B c C + d D

Kp = C

A B

( ) ( )

( ) ( )

c dD

a b

P P

P P

Ka = [H ][A ][HA]

+ -

Kb = [OH ][HB ][B]

- +

Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

= Ka × Kb

pH = − log[H+] , pOH = − log[OH−]

14 = pH + pOH

pH = pKa + log [A ][HA]

-

pKa = − logKa , pKb = − logKb

Equilibrium Constants

Kc (molar concentrations)

Kp (gas pressures)

Ka (weak acid)

Kb (weak base)

Kw (water)

KINETICS

ln[A] t − ln[A]0 = − kt

[ ] [ ]0A A1 1

t

- = kt

t½ = 0.693k

k = rate constant

t = time t½ = half-life


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Typewritten Text

AP Chemistry Equations & Constants

GASES, LIQUIDS, AND SOLUTIONS

PV = nRT

PA = Ptotal × XA, where XA = moles A

total moles

Ptotal = PA + PB + PC + . . .

n = mM

K = °C + 273

D = m

V

KE per molecule = 12

mv2

Molarity, M = moles of solute per liter of solution

A = abc

1 1

pressurevolumetemperaturenumber of molesmassmolar massdensitykinetic energyvelocityabsorbancemolarabsorptivitypath lengthconcentration

Gas constant, 8.314 J mol K

0.08206

PVTnm

DKE

Aabc

R

Ã

- -

=============

==

M

1 1

1 1

L atm mol K

62.36 L torr mol K 1 atm 760 mm Hg

760 torr

STP 0.00 C and 1.000 atm

- -

- -====

THERMOCHEMISTRY/ ELECTROCHEMISTRY

products reactants

products reactants

products reactants

ln

ff

ff

q mc T

S S S

H H H

G G G

G H T S

RT K

n F E

qI

t

D

D

D D D

D D D

D D D

=

= -Â Â

= -Â Â

= -Â Â

= -

= -

= -

�

heatmassspecific heat capacitytemperature

standard entropy

standard enthalpy

standard free energynumber of moles

standard reduction potentialcurrent (amperes)charge (coulombs)t

qmcT

S

H

Gn

EIqt

============ ime (seconds)

Faraday’s constant , 96,485 coulombs per moleof electrons1 joule

1volt1coulomb

F =

=


KINETICS Integrated Rates — Putting it All Together

What I Absolutely Have to Know to Survive the AP Exam

The following might indicate the question deals with kinetics: rate; reactant concentration; order; rate constant; mechanisms; rate determining step; intermediate; catalyst; half-life; instantaneous rate; relative rate; activation energy;

integrated rate law; rate expression; rate law

Instantaneous Rate Instantaneous rate is the rate at any one point in time during the experiment. To find instantaneous reaction rate you find the slope of the curve at the time in question (for those of you in calculus aka…derivative) i.e. the slope of the tangent line to that point in time.


Copyright © 2016 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org

Kinetics – 2

Integrated Rate – Concentration and Time

Where the differential rate law expresses rate as a function of reactant concentration(s) at an instant in time (hence instantaneous rate), integrated rates express the reactant concentrations as a function of time. To solve integrated rate problems, construct a graph with time on the x-axis and then make 3 plots where the y-axis is

§ Concentration of A [A] vs. t § Natural log of the concentration of A ln [A] vs. t

§ Reciprocal of [A]

1[A]

vs. t

LINEAR IS THE WINNER § Zero Order reaction are linear for [A] vs. t § First Order reactions are linear for ln [A] vs. t

§ Second Order reactions are linear for

1[A]

vs. t

Think y = mx + b It is imperative that you can determine reaction order simply by analyzing a graph. What is important?

§ What is plotted on each axis? § What does the slope of the line indicate?

If you know this, the order and rate constant can easily be determined.

Zero Order Reactions For the reaction…

A(g) → products

rate = –tA

ΔΔ ][

= k[A]0

When this relationship is integrated from t to tt then… [A]t = −kt + [A]0

y = mx + b

Where… § [A]t is the concentration at time t – [A]t represents what is plotted on the y-axis § [A]0 is the initial concentration § t is the time – represent what is plotted on the x-axis § k is the rate constant - which is the SLOPE of the graph! § Notice that in this equation k has a negative sign; thus the SLOPE of the graph is negative not the rate constant

If the reaction is zero order then the graph will be linear with a negative slope.

slope = −k

time

[A]



Kinetics – 2

First Order Reactions For the reaction…

A(g) → products

rate = –tA

ΔΔ ][

= k[A]1

When this relationship is integrated from t to tt then… ln [A]t = −kt + ln [A]0

y = mx + b Where…

§ [A]t is the concentration at time t – ln [A]t represents what is plotted on the y-axis § [A]0 is the initial concentration § t is the time – represent what is plotted on the x-axis § k is the rate constant – which is the SLOPE of the graph! § Notice that in this equation k has a negative sign; thus the SLOPE of the graph is negative not the rate constant

If the reaction is first order then the graph will be linear with a negative slope.

slope = −k

time

ln[A

]

Half life… § the time required for the initial concentration, [A]0, to decrease to ½ of its value.

ln [A]t = −kt + ln [A]0 ln 1 = −kt + ln 2

ln 12= −kt

−0.693= −kt0.693

k= t



Kinetics – 2

Second Order Reactions For the reaction…

A(g) → products

rate = –tA

ΔΔ ][

= k[A]2

When this relationship is integrated from t to tt then…

0

1 1[ ] [ ]

ktA A

= +

y = mx + b

Where…

§ [A]t is the concentration at time t –

1[A]

represents what is plotted on the y-axis

§ [A]0 is the initial concentration § t is the time – represent what is plotted on the x-axis § k is the rate constant – which is the SLOPE of the graph! § Notice that in this equation k is positive; thus the SLOPE of the graph is positive

If the reaction is 2nd order then the graph will be linear with a positive slope.

slope = +k

time

1[A]



Kinetics – 2

Kinetics Cheat Sheet

Relationships Differential Rate Law (concentration vs. rate data): Rate = k [A]x[B]y Integrated Rate Laws (concentration vs. time data): Zero order [A] = (−)kt + [A]0 First order ln [A] = (−)kt + ln[A]0

Second order 0

1 1 = t + [A] [A]

k

1/20.693tk

= for first order reactions and all nuclear decay

Connections Stoichiometry − “using up” one component of the system might indicate a limiting reactant in effect

Electrochemistry − if reaction is redox in nature, rate problems could come into play

Thermochemistry − Ea and ΔH°rxn and reaction diagrams Potential Pitfalls

Units on k ! Make sure you can solve for units for k



Kinetics – 2

NMSI SUPER PROBLEM

The decomposition of substance X was experimentally observed at 25°C and shown to be first order with respect to X. Data from the experiment are shown below.

[X] M Time (min)

0.100 0 0.088 2 0.069 6 0.054 10 0.043 14 0.030 ???

time

[X]

time

ln

[X]

(a) For each of the graphs above

i. Sketch the expected curve based on the labeled axes. You do not need to plot the exact data.

Since the reaction is first order, the plot of ln [X] v. t is linear while the plot of [X] vs. t should show the non-linear decrease in [X].

time

[X]

time

ln [X

]

1 point is earned for sketching the plot of ln[X] as linear with a negative slope. 1 point is earned for sketching the plot of [X] as non-linear, decreasing over time.

ii. Write the rate law for the decomposition of substance X.

Since the reaction is decomposition and first order overall the rate law is rate = k [X]

1 point is earned for the correct rate law.



Kinetics – 2

iii. Explain how one of the two graphs above can be used to determine the rate constant, k. Be sure tospecify which graph.

The slope of the plot of ln [X] v. t is equal to the absolute value of therate constant, k.

time

ln [X

] slope = −k

1 point is earned for identify the slope of the plot of ln[X] vs. t will be equal to the rate constant, k.

(b) Based on the above data

i. Calculate the rate constant for this reaction. Be sure to include units.

ln[A][A]o

= −kt

ln[0.043][0.100]o

= −(k)14

0.0603min−1 = k

1 point is earned for the value and the sign of the rate constant.

1 point is earned for the correct units (any time−1 earns the point)

The points can be earned with a description of how the slope of the line was determined with a graphing calculator.

ii. How many minutes will it take for [X] to become 0.030 M?

ln[A][A]o

= −kt

ln[0.030][0.100]o

= −(0.0603)t

20min = t

1 point is earned for the substitution.

1 point is earned for the correct answer.

These points can be earned with a description of how the data was extrapolated using a graphing calculator.



Kinetics – 2

In a different experiment, the decomposition of substance Y at 50°C was determined to have the following rate law.

Rate = k [Y]2

time

__________

(c) On the axes above

i. Sketch the graph that is expected to provide a linear relationship when plotted against time. Be sure to label the y-axis.

time

1[A]

1 point is earned for the reciprocal of [A] on the y-axis 1 point is earned for correctly sketching a straight line with a positive slope.

ii. What does the slope of this line represent?

Slope = k 1 point is earned for stating that slope = k

(d) The temperature of this reaction was increased from 50°C to 100°C. Predict the effect this would have on

each of the following. i. Rate of the reaction

The 50°C increase in temperature will increase the rate of the reaction

1 point is earned for stating that the rate increases

ii. Rate constant, k

The rate constant, k will increase

1 point is earned for stating that rate constant, k increases



Kinetics – 2

(e) Sketch the graph of the reaction at 100°C on the plot in part (c)

time

1[A]

1 point is earned for a linear line that starts at the same place as the original sketch 1 point is earned for a line with a greater positive slope, representing the increase in k.



AP® CHEMISTRY 2004 SCORING GUIDELINES (Form B)

Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

6

Question 3

2 H2O2(aq) → 2 H2O(l) + O2(g)

3. Hydrogen peroxide decomposes according to the equation above.

(a) An aqueous solution of H2O2 that is 6.00 percent H2O2 by mass has a density of 1.03 g mL–1. Calculate each of the following.

(i) The original number of moles of H2O2 in a 125 mL sample of the 6.00 percent H2O2 solution

2 2H On = 125 mL H2O2(aq) × 2 2

2 2

1.03 g H O ( )1.00 mL H O ( )

aqaq

×

2 2

2 2

6.00 g H O100 g H O ( )aq

× 2 2

2 2

1 mol H O34.0 g H O

= 0.227 mol H2O2

1 point for determining mass of H2O2(aq)

1 point for mass of H2O2

1 point for moles of H2O2

(ii) The number of moles of O2(g) that are produced when all of the H2O2 in the 125 mL sample decomposes

2On = 0.227 mol H2O2 × 2

2 2

1 mol O ( )2 mol H O ( )

gaq

= 0.114 mol O2(g) 1 point for moles of O2(g)

(b) The graphs below show results from a study of the decomposition of H2O2 .





7

Question 3 (cont’d.)

(i) Write the rate law for the reaction. Justify your answer.

rate = k[H2O2]1

A plot of ln[H2O2] versus time is a straight line, so the reaction follows simple first-order kinetics.

1 point for correct rate law

1 point for explanation

(ii) Determine the half-life of the reaction.

Using the graph showing [H2O2] versus time, the half-life is about 650 minutes.

OR

Calculate from 1 2t = 0.693k

after determining k

from the slope in part (b)(iii)

1 point for a half-life between 600 and 700 minutes

(iii) Calculate the value of the rate constant, k . Include appropriate units in your answer.

1 2t = 0.693k

k = 1 2

0.693t

= 0.693650 min

= 1.1 × 10−3 min−1

OR k can be obtained from the determination of the slope of the line in the ln k versus time plot

1 point for the magnitude of the rate constant 1 point for the units

(iv) Determine [H2O2] after 2,000 minutes elapse from the time the reaction began.

From the graph of [H2O2] versus time, [H2O2] is approximately 0.12 M.

OR

From the graph of ln[H2O2] versus time, ln[H2O2] is approximately –2.2, so [H2O2] = e− 2.2 = 0.11 M

1 point for 0.09 < [H2O2] < 0.13 M


AP® CHEMISTRY 2012 SCORING GUIDELINES

© 2012 The College Board. Visit the College Board on the Web: www.collegeboard.org.

Question 3 (9 points)

A sample of CH3CH2NH2 is placed in an insulated container, where it decomposes into ethene and ammonia according to the reaction represented above.

Substance Absolute Entropy, S°,

in J/(mol!K) at 298 K

CH3CH2NH2(g) 284.9

CH2CH2(g) 219.3

NH3(g) 192.8

(a) Using the data in the table above, calculate the value, in J/(molrxn!K), of the standard entropy change, "S°, for the reaction at 298 K.

rxn products reactantsS S SD S S �D D D

rxnSD D = [(219.3 + 192.8) – 284.9] J/(molrxn !K)

= 127.2 J/(molrxn !K)

1 point is earned for the correct ǻS°.

(b) Using the data in the table below, calculate the value, in kJ/molrxn , of the standard enthalpy change, "H° , for the reaction at 298 K.

Bond C–C C = C C–H C–N N–H

Average Bond Enthalpy (kJ/mol)

348 614 413 293 391

ǻH° = enthalpy of bonds broken – enthalpy of bonds formed

ǻH° = [5(ǻHC-H) + (ǻHC-N) + (ǻHC-C) + 2(ǻHN-H)] – [4(ǻHC-H) + (ǻHC=C) + 3(ǻHN-H)] = [5(413) + 293 + 348 + 2(391)] – [4(413) + 614 + 3(391)] = 49 kJ/molrxn

OR

ǻH° = [(ǻHC-H) + (ǻHC-N) + (ǻHC- C)] – [(ǻHC=C) + (ǻHN-H)]

= [413 + 293 + 348] kJ/mol – [614 + 391] kJ/mol = 49 kJ/molrxn

1 point is earned for the correct bond count and use

of values from table.

1 point is earned for the correct setup in terms of

bonds broken minus bonds formed and calculated ǻH°.




Question 3 (continued)

(c) Based on your answer to part (b), predict whether the temperature of the contents of the insulated container will increase, decrease, or remain the same as the reaction proceeds. Justify your prediction.

The temperature of the contents should decrease because the reaction is endothermic, as indicated by the positive ǻH°.

1 point is earned for the correct choice with explanation.

An experiment is carried out to measure the rate of the reaction, which is first order. A 4.70 ! 10#3 mol sample of CH3CH2NH2 is placed in a previously evacuated 2.00 L container at 773 K. After 20.0 minutes, the

concentration of the CH3CH2NH2 is found to be 3.60 ! 10# 4 mol/L.

(d) Calculate the rate constant for the reaction at 773 K. Include units with your answer.

� � � �

� � � �

34

2 1

ln[ ] ln[ ]

4.70 10 molln 3.60 10 mol/L ln 20.0min2.00 L

7.929 6.053 20.0min

9.38 10 min

t oA A kt

k

k

k

��

� �

� �È Ø�� É ÙÊ Ú

� � � �

�

1 point is earned for the initial concentration of CH3CH2NH2.

1 point is earned for the correct setup of the first order integrated rate law

equation.

1 point is earned for the calculated result with unit.

(e) Calculate the initial rate, in M min#1, of the reaction at 773 K.

> @ � �3

2 13 2 2

4 1

4.70 10 mol CH CH NH = 9.38 10 min2.00 L

2.20 10 min

initial rate k

M

��

� �

È Ø� � É ÙÊ Ú

�

1 point is earned for the

calculated result.

(f) If 3 2 2

1[CH CH NH ]

is plotted versus time for this reaction, would the plot result in a straight line or would it

result in a curve? Explain your reasoning.

The plot would produce a curve; had the reaction been second order the plot would have been a straight line. A plot of ln[CH3CH2NH2] vs. t would have yielded a straight line.

1 point is earned for the correct choice with explanation.


2012 AP® CHEMISTRY FREE-RESPONSE QUESTIONS


GO ON TO THE NEXT PAGE. -8-

3. A sample of CH3CH2NH2 is placed in an insulated container, where it decomposes into ethene and ammonia according to the reaction represented above.

Substance Absolute Entropy, S°,

in J/(mol⋅K) at 298 K

CH3CH2NH2(g) 284.9

CH2CH2(g) 219.3

NH3(g) 192.8

(a) Using the data in the table above, calculate the value, in J/(molrxn⋅K), of the standard entropy change, ΔS°, for the reaction at 298 K.

(b) Using the data in the table below, calculate the value, in kJ/molrxn , of the standard enthalpy change, ΔH° , for the reaction at 298 K.

Bond C–C C = C C–H C–N N–H

Average Bond Enthalpy (kJ/mol)

348 614 413 293 391

(c) Based on your answer to part (b), predict whether the temperature of the contents of the insulated container will increase, decrease, or remain the same as the reaction proceeds. Justify your prediction.

An experiment is carried out to measure the rate of the reaction, which is first order. A 4.70 × 10−3 mol sample of CH3CH2NH2 is placed in a previously evacuated 2.00 L container at 773 K. After 20.0 minutes, the

concentration of the CH3CH2NH2 is found to be 3.60 × 10− 4 mol/L.

(d) Calculate the rate constant for the reaction at 773 K. Include units with your answer.

(e) Calculate the initial rate, in M min−1, of the reaction at 773 K.

(f) If 3 2 2

1[CH CH NH ]

is plotted versus time for this reaction, would the plot result in a straight line or would it

result in a curve? Explain your reasoning.

S T O P If you finish before time is called, you may check your work on this part only.

Do not turn to the other part of the test until you are told to do so.



© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

Question 2 (8 points)

S2O32–(aq) H+

æææÆ SO32– (aq) + S(s)

A student performed an experiment to investigate the decomposition of sodium thiosulfate, Na2S2O3, in acidic solution, as represented by the equation above. In each trial the student mixed a different concentration of sodium thiosulfate with hydrochloric acid at constant temperature and determined the rate of disappearance of

S2O32−(aq). Data from five trials are given below in the table on the left and are plotted in the graph on the right.

(a) Identify the independent variable in the experiment.

The initial concentration of S2O32−(aq) One point is earned for the correct answer.

(b) Determine the order of the reaction with respect to S2O32−. Justify your answer by using the information

above.

Using trials 1 and 2:

2

1

raterate

= 2

1

22 2 3

21 2 3

[S O ]

[S O ]

m

m

k

k

-

-

1

1

0.030

0.020

M s

M s

-

- = [0.075]

[0.050]

m

m

1.5 = (1.5)m, so m = 1 and the reaction is first order with respect to S2O3

2−.

Note: Other correct justifications are acceptable.

One point is earned for the correct order.

One point is earned for a correct justification.



© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.

Question 2 (continued)

(c) Determine the value of the rate constant, k , for the reaction. Include units in your answer. Show how you

arrived at your answer.

rate = k [S2O32−] ⇒ k =

22 3

rate

[S O ]-

Using the data from trial 1, k = 10.020 s

0.050 M

M

- = 0.40 s−1

OR

the rate constant is equal to the slope of the line

k = 1 1(0.052 0.020) 0.032

(0.13 0.05) 0.08M s M sM M

- -- =-

= 0.40 s−1

One point is earned for the correct value.

One point is earned for the correct units.

(d) In another trial the student mixed 0.10 M Na2S2O3 with hydrochloric acid. Calculate the amount of time it

would take for the concentration of S2O32− to drop to 0.020 M .

ln[A] t − ln[A]0 = − kt ⇒ ln0

[A][A]

t = − kt

ln2

2 32

2 3 0

[S O ]

[S O ]t

-

- = − kt

ln 0.0200.10

= (−0.40 s−1)( t ) ⇒ t = 1

1.61

0.40 s--

- = 4.0 s

One point is earned for the correct setup.

One point is earned for the correct answer with units.

(e) On the graph above, sketch the line that shows the results that would be expected if the student repeated the five trials at a temperature lower than that during the first set of trials.

The line drawn should start on the y-axis at a lower point than the line already plotted and should have a less steep slope.

One point is earned for an acceptable line.




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Question 3

The first-order decomposition of a colored chemical species, X, into colorless products is monitored with a

spectrophotometer by measuring changes in absorbance over time. Species X has a molar absorptivity constant of 5.00 × 103 cm–1 M –1 and the path length of the cuvette containing the reaction mixture is 1.00 cm. The data from the experiment are given in the table below.

[X] ( M )

Absorbance Time (min)

? 0.600 0.0 4.00 × 10– 5 0.200 35.0 3.00 × 10– 5 0.150 44.2 1.50 × 10– 5 0.075 ?

(a) Calculate the initial concentration of the colored species.

A = abc

c = abA

= ( )( )3 1 1

0.6005.00 10 cm 1.00 cmM− −×

= 1.20 × 10− 4 M OR

A0 = abc0 A1 = abc1

0 1

0 1

A Ac c

= 50

0.600 0.2004.00 10c −=

×

c0 = 1.20 × 10− 4 M

1 point for concentration of X

(b) Calculate the rate constant for the first-order reaction using the values given for concentration and time. Include units with your answer.

Using the first two readings,

ln[X]t – ln[X]0 = −kt OR ln 0

[X][X]

t = −kt

ln 5

44.00 101.20 10

−

−××

= −k (35.0 min)

ln (0.333) = −k (35.0 min)

–1.10 = −k (35.0 min)

k = 3.14 × 10−2 min−1

1 point for magnitude and correct sign of rate constant

1 point for correct units




8


(c) Calculate the number of minutes it takes for the absorbance to drop from 0.600 to 0.075.

ln 0

[X][X]

t = −k t

ln 5

41.50 101.20 10

−

−××

= (−3.14 × 10−2 min−1) t

ln (0.125) = (−3.14 × 10−2 min−1) t

–2.08 = (−3.14 × 10−2 min−1) t t = 66.2 min

1 point for correct substitution

1 point for correct answer

Note: students may use half-lives to answer this question.

(d) Calculate the half-life of the reaction. Include units with your answer.

ln 0

[X][X]

t = − k t

ln 0

0

0.5 [X][X]

= (−3.14 × 10−2 min−1) t1/2

ln (0.5) = (−3.14 × 10−2 min−1) t1/2

−0.693 = (−3.14 × 10−2 min−1) t1/2

22.1 min = t1/2 OR

t1/2 = 0.693k

t1/2 = 2 1

0.6933.14 10 min− −×

= 22.1 min

1 point for correct magnitude

1 point for the correct units

(1 point for the half-life equation if no k is given)

(e) Experiments were performed to determine the value of the rate constant for this reaction at various temperatures. Data from these experiments were used to produce the graph below, where T is temperature. This graph can be used to determine the activation energy, Ea , of the reaction.

(i) Label the vertical axis of the graph.

The vertical axis should be labeled ln k. 1 point




9


(ii) Explain how to calculate the activation energy from this graph.

The slope of the line is related to the activation energy:

slope = – aER

To determine the activation energy for the reaction, multiply the slope by – 8.314 J mol−1 K−1.

1 point for recognizing that the slope must be measured

1 point for the correct explanation of how to obtain the activation energy


Kinetics Integrated Rates - Putting it All Together 2014...

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