Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                      Exam 1-A-Key

                       

Chemisty 1084:  Section 010                                                                                       Fall 2006

 

Name:________________________________________________________

 

Read all directions and questions carefully!!  This exam consists of two parts.  The first part consists of 10 multiple choice questions worth four points each for a total of 40 points.  The second part consists of five numerical problems worth either 10 or 15 points per question for a total of 60 points.  Show all your work necessary for the numerical problems as partial credit will be given for those problems.

 

Possibly Useful Constants and Equations

 

Raoult's Law:                             Freezing point depression: 

Boiling point elevation:        For a 1st order reaction: 

For a second order reaction:                  Osmotic Pressure: 

 

Gas Constant:  R = 0.08206 L atm/(mol K)                 Quadratic formula:  For an equation of the form:

                                                                                                 0 = ax2 + bx + c

                                                                                                

 

Score

 

                                    Part 1 (40 points):_____________________

 

 

                                    Part 2 (60 points):_____________________

 

 

                                    Total (100 points):_____________________

 

 

Don’t forget to put your name on this test!

 

Good Luck!!

 

 

Part 1

 

Multiple Choice

 

Please indicate the answer to each question by putting your choice in the space provided.  There is only one correct answer for each question.  There will be 10 multiple choice questions worth 4 points each.

 

 

1.     Which one of the following substances is more likely to dissolve in CH3OH?

(a)  CCl4                                                                      (c)  N2

(b)  Kr                                                                         (d)  CH3CH2OH

                                          (e)  NaCl

 

Answer:  D

This question involves the concept of “like dissolves like.”  Two compounds which are similar chemically will dissolve.  Since CH3OH and CH3CH2OH are both alcohols that only differ by one carbon length, they are the most similar to each other.

 

2.  As pressure increases, the solubility of a gas in a liquid ________________.

      (a)  increases                                                (c)  remains the same.

      (b)  decreases                                               (d)  may increase or decrease depending on the gas.

 

Answer:  A

The solubility of a gas in a liquid increases with increasing pressure and decreases with increasing temperature.

 

3.  As a nonvolatile solute is added to a solvent, the boiling point of the solution ___________________ and the freezing point of the solution _______________.

      (a)  increases, increases                               (c)  decreases, increases

      (b)  increases, decreases                              (d)  decreases, decreases

 

Answer:  B

Colligative properties:  as a nonvolatile solute concentration increases:  vapor pressure decreases, boiling point increases, freezing point decreases, and osmotic pressure increases.

 

4.  Given the following reaction:

2 A  ¾®   C

      Using the definition of reaction rate, find the relationship between the rate of disappearance of A   and the rate of appearance of C .  In other words,

      (a)  +2                                                          (d)  –1

      (b)  1                                                            (e)  –2

      (c) +½                                                          (f)  –½

 

Answer:  E

The general definition of rate is .  Using this definition in this equation:  .  Multiply both sides of this equation by -2 and you will obtain the correct answer.

5.  Shown above is a plot generated from data obtained from a reaction in which NO2 is a reactant.  From the plot above, it can be concluded that the rate law is _______________ order with respect to NO2.

      (a)  0th                                                          (c)  2nd

      (b)  1st                                                          (d)  impossible to determine order from this plot.

 

Answer:  C

When a plot of 1/[A] versus time gives a straight line, you have a second order reaction.

 

6.  The mechanism for a reaction of the stoichiometry

A + 2 B ¾¾®  C + X

      is found to be:

A +  B  ¾®  C + D     (slow)

B + D  ¾¾®  X     (fast)

      What is the rate law that is consistent with this mechanism?

      (a)  rate = k[B][D]                                      (c)  rate = k[A]

      (b)  rate = k[A][B]2                                     (d)  rate = k[A][B]

 

Answer:  D

Since one step is slow, this mechanism is analyzed using the rate determining step approximation.  The rate law that is consistent with this mechanism is the rate of the slow step.

 

7.  At equilibrium,

(a)  all chemical reactions have ceased.

(b)  the concentrations of reactants and products are equal.

(c)  the rate of the forward reaction is equal to the rate of the reverse reaction.

(d)  the limiting reagent has been totally consumed.

 

Answer:  C

This is the definition of equilibrium.

 

8.  Kp = 0.112 for the reaction:

SO2(g)  +  ½ O2(g)    SO3(g)

      Calculate the value of Kp for the following reaction:

2 SO3(g)     2 SO2(g)  +  O2(g)

      (a)  79.7                                                       (c)  17.86

      (b)  2.99                                                       (d)  4.46

                                                (e)  8.93

 

Answer:  A

The second reaction is the reverse of the two times the first reaction.  Therefore, the equilibrium constant must be squared and then take its reciprocal to get the correct answer (=1/Kp2)

 

9.  Write the letter that represents the enthalpy change (DH) for the reaction in the diagram shown below:

Answer:  E

 

You will need to know what is represented by all the letters on this diagram.

 

 

10.  A reaction with a large value for the equilibrium constant will have _______________________ present at equilibrium.

       (a)  mostly products                                                    (c)  equal amounts of reactants and products

       (b)  mostly reactants                                                   (d)  only products

 

Answer:  A

 

 

 

 

Part 2

 

Numerical Problems

 

Solve the following problems, keeping track of significant figures where applicable.  Please show all the work necessary to obtain your answer in order to receive partial credit for possibly wrong answers.  Generally, full credit will not be given for the correct answer without any of the work performed to obtain the answer being shown on the paper.  Each question is worth either 10 or 15 points.

 

 

11.  (10 points)  A solution is prepared by mixing 10.6 g of Na2SO4 in 483 g of water.  What is the mass percentage, mole fraction, and molality of Na2SO4 in this solution?

 

Using the definition of mass percentage: 

The definition of mole fraction is: 

The number of moles of Na2SO4 and water need to be calculated first:

The mole fraction is equal to: 

The definition of molality is: 

The number of moles of Na2SO4 was calculated previously.  Calculate kg of water:

The molality is equal to: 

 

 

12.  (15 points)  A solution prepared by dissolving 2.154 g of a solute in 22.09 g of carbon tetrachloride.  The freezing point of the solution was measured at –35.1°C.  Given the freezing point of pure carbon tetrachloride at –22.3°C and its freezing point depression constant of 29.8°C/m, calculate the molar mass of the solute.

The formula for freezing point depression is:  DTf = Kfmsolute.

The data given in the problem is:  Tf,solution = –35.1°C           Tf,solvent = –22.3°C      Kf = 29.8°C/m

                                                    mass of solvent (carbon tetrachloride) = 22.09 g

                                                    mass of solute = 2.154 g

Calculate the freezing point depression:  DTf = Tf,solvent – Tf,solution = –22.3°C – (–35.1°C) = 12.8°C

Solve the freezing point depression formula for msolute and calculate the molality:

Multiply the molality of the solute by the number of kilograms of solvent to calculate the number of moles of solute:

Divide the mass of the solute by the number of moles of solute to calculate its molar mass:

 

 

13.  (10 points)  A reaction of the stoichiometry, A ¾® B, has a 1st order rate law with a half-life equal to 25.0 s.  What is the value of the rate constant, k, for this reaction?  What is the rate of the reaction when [A] = 0.10 M?

 

For a 1st order reaction:  .

Solving this equation for the rate constant, k: 

The rate law for this reaction is:  rate = k[A] = (0.0277 s-1)(0.10 M) = 0.00277 M s-1  (3 sig figs)

 

 

14.  (10 points)  At a certain temperature, Kp = 5.24 for the reaction:

Br2(g)  +  Cl2(g)  2 BrCl(g)

       Suppose initially that a flask is filled with 0.25 atm of Br2 and 0.25 atm of Cl2 and allowed to equilibrate.  What are the equilibrium concentrations of Br2, Cl2, and BrCl?

Given the initial partial pressures of 0.25 atm for both Br­2 and Cl2, set up an equilibrium table for this problem:

 

Br2(g)

+

Cl2(g)

2 BrCl(g)

Initial pressure

0.25

 

0.25

 

0

Change in pressure

–x

 

+x

 

+2x

Equilibrium presssure

0.25 – x

 

0.25 + x

 

2x

Let “2x” be equal to the change in the partial pressure of BrCl.  Then, the partial pressures of Br2 and Cl2 both decrease by an amount “x”.  Add the initial and change in pressures to obtain the equilibrium partial pressures.  Place the equilibrium partial pressures into the equilibrium expression and solve for x:

Taking the square root of both sides of this expression:

Multiply both sides by 0.25 – x:

Solving for x:

Calculating the equilibrium partial pressures: 

 

 

15.  (15 points)  Nitrogen monoxide is readily oxidized to nitrogen dioxide by the reaction:

2 NO(g)  + O2(g)    2 NO2(g)

       The following data were collected in a kinetics study of this reaction:

 

Experiment

[O2] (M)

[NO] (M)

Rate (M s-1)

 

 

1

0.0010

0.0010

7.10

 

 

2

0.0040

0.0010

28.4

 

 

3

0.0040

0.0030

255.6

 

 

       Determine the rate law for this reaction and the value of the rate constant, k, with units.

 

The rate law has the form:  rate = k[O2]m[NO]n

Compare experiments 1 and 2:

[O2] quadruples                                          [NO] is constant

                                             The rate quadruples 

From this data:  4m = 4, and therefore m = 1

Compare experiments 2 and 3:

[O2] is constant                                                                  [NO] triples 

                                             The rate increases nine-fold 

From this data:  3n = 9, and therefore n = 2

The rate law is:  rate = k[O2][NO]2

To obtain the value of k, solve the rate law for k and plug in the data from experiment 1:

 

 (10 points extra credit)  Define 2 out the 3 following terms.  Credit will be given only for the complete definition.  No partial credit will be given.  Choose only 2 out of the three terms.  If you give definitions for all three terms without clearly noting which ones you want graded, only the first two terms will be graded.

 

Equilibrium:  the state of a reaction in which the forward reaction rate equals the reverse reaction rate.

 

 

 

 

 

Heterogeneous catalyst:  a catalyst that is in a different phase than the reactants.

 

 

 

 

 

Activation Energy: the minimum energy of a collision required for two molecules to react.