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Chem 121, section 03, Midterm 1 Page 1 CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. Cartwright Midterm 1 February 18 th 2015. Time 1 ½ hours Name _____Brief answers________ Student Number ____________________ Attempt as many questions as possible. Write your answers on this examination paper. If you need extra space, continue on the reverse of the paper, stating within your answer to the question that you have done so. The numbers in square brackets indicate the approximate number of marks available for each part of the question. Do not turn over the examination paper until you are instructed to do so.
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CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

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Page 1: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 1

CAMOSUN COLLEGE

Chemistry 121 Section 03, Instructor: H. M. Cartwright

Midterm 1

February 18th 2015.

Time 1 ½ hours

Name _____Brief answers________

Student Number ____________________

Attempt as many questions as possible.

Write your answers on this examination paper. If you need extra space,

continue on the reverse of the paper, stating within your answer to the

question that you have done so.

The numbers in square brackets indicate the approximate number of marks

available for each part of the question.

Do not turn over the examination paper until you are instructed to do so.

Page 2: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 2

Integrated rate laws:

0th order: [𝐴]𝑡 = −𝑘𝑡 + [𝐴]𝑜

1st order: ln [𝐴]𝑡 = −𝑘𝑡 + ln [𝐴]𝑜

2nd order: 1

[𝐴]𝑡= 𝑘𝑡 +

1

[𝐴]𝑜

Half-life of a 1st order reaction: t ½ = ln 2/k

Page 3: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 3

1. Circle the letter that corresponds to the one statement that is INCORRECT. A catalyst: A. Lowers the activation energy for a reaction. B. Can only increase reaction rate at elevated temperature. C. Does not change the position of equilibrium. D. Can be in the same phase or a different phase as the reactants. E. Always changes the rate of a reaction. [1] 2. For each of the following types of compound: (a) give the general formula; (b) draw the full structure of one example of this type of compound; (c) name the compound that you have drawn. [3 x 1.5]

(i) Alkene

CnH2n

Examples include ethene, propene, etc. For structure see text or lecture notes.

(ii) Alcohol

R-OH, or CnH2nOH

Examples include, methanol, ethanol, etc

(iii) Aromatic

Formula varies, but must include an aromatic ring

Examples include benzene and its many substituted forms; also molecules with multiple

rings such as naphthalene, anthracene, ...

3. Using suitable equations, explain the difference between a substitution reaction and

an addition reaction. [4]

In a substitution reaction one atom or group is replaced by a second atom or group. An

example would be the reaction of an alkane with molecular chlorine, in which a single H

atom is replaced by a single Cl atom (the other chlorine atoms joins with the displaced H

atom to form HCl).

In an addition reaction, a double bond is opened during the reaction and atoms are added

across what was a double (or triple) bond. A typical example would be the addition of HCl to

ethene, to form chloroethane.

Page 4: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 4

4. At least one of the following statements is WRONG:

A. Increasing the temperature speeds up most reactions.

B. For a 1st order reaction the units of the rate constant k are time-1.

C. The rate of a chemical reaction in the forward direction is always greater than the

rate in the backward direction.

D. Catalysts provide an alternate pathway between reactants and products.

E. The magnitude of a rate constant depends on the concentration of reactants.

The INCORRECT statement(s) is/are:

(i) A and B (ii) E (iii) E and B (iv) C (v) C and E [1]

5. Of the following changes, which ONE will not affect the rate of most reactions? [1]

A. Increasing the concentration of reactants.

B. Raising the temperature of the reaction.

C. Adding a catalyst.

D. Removing products as they are formed.

E. Increasing the pressure of a gas-phase reaction.

6. (a) Explain the difference between geometric isomers and structural isomers. [3]

The bonding between atoms in two geometric isomers is the same, but the orientation in

space of some bonded groups is different; for example, cis- and trans- isomers in the

alkenes. Structural isomers differ in the bonding between atoms, for example, 1-

chloropropane and 2-chloropropane, not just the orientation of groups in space.

(b) Draw all the isomers of butene, C4H8. [3]

Draw 1-butene, and cis- and trans-2-butene.

7. What is an electrophile? [3]

Literally, an electron-loving species – a chemical that is expected to react with electron-rich

species.

Page 5: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 5

8. (a) What is a chiral molecule? [1]

A molecule that contains at least one carbon atom which is joined to four different entities.

(b) Is the molecule below chiral? If so, indicate any chiral atom(s) with a star. [2]

Three of the four carbons at the "top" of the cyclopentane rings are chiral.

9. Which ONE of the following experimental techniques can be used to distinguish

between two enantiomers? [1]

A. Infrared spectroscopy

B. Nuclear Magnetic Resonance spectroscopy

C. Accurate analytical weighing

D. Optical rotation

E. Free radical reaction with an alkene

10. (a) What is a chain reaction? [2]

A reaction in which at least one of the steps is initiated by a very reactive species, usually a

free radical. The species is consumed during the reaction, but regenerated later, so that,

after bringing about the reaction of one molecule it can do the same trick again later, in

theory continuing the process as long as there is unreacted material available. This free

radical is then the “chain carrier”.

(b) Name the three main types of steps in a chain reaction. [1.5]

Initiation, propagation, termination

(c) Give an example of a chain reaction, showing each of the three steps you have

named in part (b). [3]

Page 6: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 6

See lecture notes, alkanes reacting with chlorine.

11. (a) Sketch the potential energy vs. progress of reaction (the reaction coordinate or

reaction profile diagram) for an endothermic reaction. Label the axes and show on your

diagram

(i) location of the reactants and products

(ii) position of any transition states (activated complexes).

(iii) activation energy for each step. [4]

See lecture notes. During an endothermic reaction there is an overall increase in enthalpy, so the enthalpy of the products must be shown on the diagram to be greater than that of the reactants. (b) Based on your diagram, which of the forward or the reverse reaction would you predict is the slower? Justify your choice. [1] The forward reaction would be the slower, as it has the greater activation energy. 12. The reaction between NO(g) and Br2(g) to form NOBr (g) was examined at 0 oC. The initial rate of formation of product was studied in the experiments tabulated below. Expt. [NO](mol L-1) [Br2] (mol L-1) Initial rate (mol L-1 min-1) 1 0.10 0.20 24 2 0.25 0.20 150 3 0.10 0.50 61 4 0.35 0.50 734 (a) What is the rate law for this reaction? [2] Rate = k[NO]2[Br2] (b) Calculate the average value of the rate constant, k. Include the correct units. [3] k=12000 L2 mol-2 min-1 (c) Suppose an experimenter measured the change in the concentration of Br2 vs time for the reaction. Taking into account the rate law you have just derived, what function of the Br2 concentration should she plot against time to get a straight line? [1]

Page 7: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 7

The rate law shows that the reaction is 1st order w.r.t. bromine, so she should plot ln[Br2] vs t.

13. (a) Draw a circle around the carbocation(s) shown below which is/are the LEAST

stable, based on the position of the positive charge (there may be more than one correct

answer). Note that the charge actually resides on the carbon atom, not on a hydrogen

atom. [1]

The last three are the least stable.

(b) Justify your choice in part (a): [2]

Primary cations ions are less stable than secondary, which in turn are less stable than

tertiary. Explanation in terms of the increased ability of carbon to donate electron density to

the positive carbon atom compared to that of the hydrogen atom.

14. (a) What is the difference between a conformer and an isomer? [1]

Isomers, although they share the same formula, can only be interconverted if the large

energy barrier which prevents one isomer turning into the other is overcome. Typically this

barrier arises because, to turn one isomer into the other, one or more bonds must be

broken; there’s not enough energy available at room temperature to do this. By contrast,

conformers are separated by a low energy barrier, so constantly interconvert at room

temperature.

(b) Why is there "free" rotation around every bond in butane, C4H10, but not around

every bond in 2-butene, C4H8? [2]

Because rotation around the double bond in 2-butene is only possible if the -bond is

broken (since it is formed by the sideways overlap of two p-orbitals and that sideways

overlap diminishes if one end of the molecule is rotated relative to the other end). This

requires more energy than is available thermally at room temperature, so does not happen.

Page 8: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 8

Rotation around the sigma bond in butane however does not alter the amount of overlap

between the atomic orbitals that combine to form the bond, so is easy to accomplish.

15. (i) The value of the rate constant for a reaction depends upon temperature; why? [3]

In most reactions molecules need to acquire energy to surmount an activation barrier

before they can react. Hot molecules have more energy than cold ones, so as T goes up a

greater proportion of molecules have the energy to react. This temperature dependence is

encapsulated in the rate constant k, given by k = Ae-Ea/RT

(ii) What factors other than temperature influence the value of a rate constant? [2]

Particular reaction studied, presence of a catalyst, orientation of collisions (but not the

concentrations of the reactants!).

16. (a) What is meant by the half-life of a chemical reaction? [1]

The time required for the concentration of a reactant to fall by one half.

(b) The rate constant for the first-order decomposition of N2O5 is 2.75 x 10-2 s-1.

Determine the half-life of the decomposition. [1]

ln 2/k = 25 seconds

(c) Sketch how the concentration of N2O5 will vary as a function of time, showing

clearly at least two half-lives on your diagram.

See lecture notes. The concentration will show an exponential decay with all half lives of the

same length.

17. Hydrogen bromide gas, HBr, reacts with oxygen to form water in the gas phase and

bromine, also in the gas phase.

(a) Write down the equation for the reaction and balance it. [2]

4 HBr + O2 = 2 H2O + 2 Br2

Page 9: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 9

(b) The concentration of which material changes most rapidly during the reaction?

[1]

HBr

(c) Using your balanced equation, write an expression for the rate at which the

concentration of bromine changes, expressed in terms of the rate at which the

concentration of HBr changes. (Note that the rate of change of concentration of HBr can be

written as t

HBr

][ ) [1]

d[Br2]/dt = - ½ d[HBr]/dt

18. The following sequence of steps is one of the main routes by which ozone is

destroyed in the Antarctic atmosphere:

Cl(g) + O3(g) → ClO(g) + O2(g)

ClO(g) + light → Cl(g) + O(g)

O(g) + O(g) → O2(g)

(a) Identify any reactive intermediate(s) and/or catalyst(s) in this mechanism. [2]

Intermediates ClO and O

Catalyst Cl

(b) Write down the rate law for each step in the mechanism. [3]

Rate1 = k1[Cl][O3]

Rate2 = k2[ClO

Rate3 = k3[O]2

(c) The rate law for the overall reaction is Rate = k [Cl][O3]. Which of the three steps

given above is probably the slowest? Justify your choice. [2]

1st, since the rate law for the overall reaction matches that of reaction 1.

(d) Identify the probable origin of the chlorine atoms in the first step of the

mechanism. [1]

Page 10: CAMOSUN COLLEGE Chemistry 121 Section 03, Instructor: H. M. …web.uvic.ca/.../Chem_121_section_3_midterm_1_2015_brief_answers.pdf · clearly at least two half-lives on your diagram.

Chem 121, section 03, Midterm 1 Page 10

Chlorofluorocarbons (CFCs).

19. For the reaction A + 3B 2C the rate law is Rate = k[A][B]2. Circle the letter that

corresponds to the correct answer.

If the concentrations of A and B are both halved, then the rate will fall by a factor of:

A. 1/3

B. 1/4

C. 1/8

D. 1/9

E. 1/16 [1]

20. Give an example of a catalyst and describe briefly how it works. [3]

Enzyme, catalytic converter, nickel catalysis of hydrogenation, … Explanation will depend

upon the type of catalyst.

21. The reaction A → products has been found to be 1st order. In a typical experiment,

the concentration of A decreased by 4% in 8.00 minutes.

(a) Calculate the rate constant for this reaction, quoting the appropriate units. [3]

k=5.103 x 10-3 min-1

(b) If the initial concentration of A was 0.61 M, after how many seconds would the

concentration have fallen to 0.1M? [2]

t = 21300 s, or 355 min.