Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                     Exam 1-A-Key

                       

Chemisty 1084:  Sections 010 and 030                                                                   Spring 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:                             For a 1^st order reaction: 

Freezing point depression: 

Boiling point elevation: 

For a second order reaction:                  Osmotic Pressure: 

 

Gas Constant:  R = 0.08206 L atm/(mol K)

 

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.

 

Italacized comments are from the instructor.  They would not appear on a regular test.

 

1.  Which of the following solutions is unstable?

       (a)  a supersaturated solution                                      (c)  an unsaturated solution

       (b)  a saturated solution

 

Answer:  C

See definitions of unsaturated, saturated and supersaturated solutions.

 

 

2.  Which of the following compounds would you expect to be the most soluble in CH₃OH?

       (a)  CCl₄                                                                     (c)  HCl

       (b)  Kr                                                                        (d)  CH₃CH₂OH

 

Answer:  D

This question tests the concept of “like dissolves like”.  CH₃CH₂OH is most like CH₃OH.

 

3.  Which one of the following 0.10 m aqueous solutions will have the lowest boiling point?

       (a)  NaCl                                                                    (c)  Al(NO₃)₃

       (b)  NH₃                                                                      (d)  K₂SO₄

       (e)  Since they have the same molality, they have the same boiling point

 

Answer:  B

The solution with the lowest boiling point in this list would be the non-electrolyte NH₃.  The species that creates the most ions per formula unit would have the highest boiling point.  This would be Al(NO₃)₃ which forms 4 ions (Al³⁺ and 3 NO₃¯) per formula unit.

 

4.  Which one of the following plots would be linear for a 1^st order reaction?

       (a)  [A] versus time                                                    (c)  1/[A] versus time

       (b)  ln[A] versus time                                                 (d)  lnk versus 1/T

 

Answer:  B

You need to know this about 1^st order reactions.  You also need to know which plots are linear for 2^nd order reactions and 0^th order reactions.

 

5.  One possible mechanism for a reaction with the

stoichiometry:  2 A + 2 B ¾¾®  C, is:

A  +  B  ¾¾®  D  (slow)

D + B  ¾¾®  E  (fast)

A + E  ¾¾®  C  (fast)

       The rate law that is consistent with this mechanism is:

       (a)  rate = k[A][B]                                                     (c)  rate = k[A]²[B]²

       (b)  rate = k[A][E]                                                     (d)  rate = k[D][B]

 

Answer:  A

The rate law that is consistent is the rate of the slow step.

 

 

 

6.  An energy profile for a particular reaction is shown above.  Using this diagram, give the value for the activation energy for the reverse direction.

       (a)  20 kJ                                                                    (c)  50 kJ

       (b)  70 kJ                                                                    (d)  10 kJ

                                                            (e)  60 kJ

 

Answer:  E

The activation energy for the reverse reaction is the energy difference between the energy at the right-hand side of the reaction pathway (10 kJ) and the top of the hill (70 kJ)

 

7.  Which statement is not correct concerning the function of a catalyst?

       (a)  A catalyst lowers the energy of the products making the reaction more exothermic.

       (b)  A catalyst changes the mechanism of a reaction.

       (c)  A catalyst lowers the activation energy for a reaction.

       (d)  A catalyst increases the rate of the reaction.

 

Answer:  C

This is how a catalyst works.  See the definition of a catalyst.

 

8.  Under which of the following conditions will a reaction definitely be at equilibrium:

       (a)  the concentration of reactants and products do not change with time.

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

       (c)  the rate of the forward reaction equals the rate of the reverse reaction.

       (d)  all of the reactants have been consumed leaving only products.

 

Answer:  C

Choice (c) is the definition of equilibrium.  Choice (a) is a consequence of choice (c).  There will be times where the concentrations of reactants and products do not change with time and the reaction will not be at equilibrium.

 

9.  Given the value of K_p is 0.25 for the reaction:

SO₂(g)  +  NO₂(g)    SO₃(g)  +  NO(g)

       calculate the value of K_p for the reaction:

2 SO₂(g)  +  2 NO₂(g)    2 SO₃(g)  +  2 NO(g)

       (a)  0.50                                                                      (c)  0.12

       (b)  0.063                                                                    (d)  0.25

                                                            (e)  16

 

Answer:  B

Since the second reaction is 2 times the first reaction, the second K_p value is equal to the first K_p value squared.

 

 

 

10.  When the value of the equilibrium constant is very small,

       (a)  products predominate at equilibrium.

       (b)  reactants predominate at equilibrium.

       (c)  roughly equal amounts of reactants and products exist at equilibrium.

       (d)  only products are present at equilibrium.

       (e)  only reactants are present at equilibrium.

 

Answer:  B

Since the equilibrium expression consists of the concentration of products over the concentration of reactants, when the value of the constant is very small, the concentration of products will be much smaller than the concentration of reactants at equilibrium.

 

 

 

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 dissolving 12.74 g of phenol (C₆H₅OH) in 223.5 g of methanol (CH₃OH).  Calculate the mass percent of phenol, the mole fraction of phenol, and the molality of phenol in this solution.

 

Given in the problem:  12.74 g C₆H₅OH; 223.5 g CH₃OH.  

The definition of mass percent is: 

Placing the given masses into the definition: 

The definition of mole fraction is: 

Calculate the number of moles of C₆H₅OH and CH₃OH:

Placing the calculated number of moles into the definition for mole fraction:

The definition of molality is: 

Calculate the number of kg of CH₃OH: 

The molality is equal to: 

 

12.  (15 points)  The reaction

2 NO(g)  +  O₂(g)  ¾¾®  2 NO₂(g)

            was studied with the following results:

 

Experiment	[NO], M	[O2], M	Rate, M/s
1	0.0126	0.0125	0.0141
2	0.0252	0.0250	0.1130
3	0.0252	0.0125	0.0564

 

            (a)  Determine the rate law for this reaction:

From the table of data, the rate law will look like:  rate = k[NO]^m[O₂]ⁿ

Compare experiments 1 and 3 to determine the value of the rate exponent, m:

[NO] doubles   and [O₂] remains constant.

The rate increases by a factor of 4

2^m=4.  Therefore m = 2.

Compare experiments 3 and 2 (in that order) to determine the value of the rate exponent, n:

[NO] is constant and [O₂] doubles

The rate doubles

2ⁿ=2.  Therefore n =1

The rate law is: rate = k[NO]²[O₂]

 

            (b)  Calculate the value of the rate constant, k (with units!!).

Solving the rate law for the rate constant: 

Using the data from experiment 1: 

 

            (c)  Calculate the rate when [NO] = 0.100 M and [O₂] = 0.0200 M

Plug the concentrations given and your calculated value of k into the rate law and calculate the rate of the reaction:

   (3 sig figs)

 

13.  (10 points)  Calculate the boiling point and the melting point of a solution prepared by dissolving 21.73 g of vanillin (C₈H₈O₃) in 173.9 g of chloroform (CHCl₃).  The boiling and freezing points of pure chloroform are equal to 61.2°C and –63.5°C, respectively.  The values of K_b and K_f are 3.63°C/m and 4.68°C/m, respectively, for chloroform.

 

Given:  21.73 g of C₈H₈O₃ (solute) dissolved in 173.9 g of CHCl₃ (solvent)

            K_f = 4.68°C/m; K_b = 3.63°C/m

Equations needed:  DT_f = K_fm_solute; DT_b = K_bm_solute

 

Calculate the molality of C₈H₈O₃:   

Calculate the change in freezing point, DT_f = K_fm_solute= (4.68°C/m)((0.8213 m) = 3.84°C

Since freezing point goes down, the freezing point of solution is equal to:

–63.5°C – 3.84°C = –67.3°C  (answer ends in the tenths digit)

Calculate the change in the boiling point:  DT_b = K_bm_solute = (3.63°C/m)(0.8213 m) = 2.98°C

Since boiling point increases, the boiling point of solution is equal to:

61.2°C + 2.98°C = 64.2°C     (answer ends in the tenths digit)

 

14.  (10 points)  The gas phase decomposition of N₂O₅ is first order in N₂O₅.  The half-life for this reaction is 6.82 × 10^-3 s.  The decomposition reaction is:

N₂O₅(g)  ¾¾®  2 NO₂(g)  +  O₂(g)

 

(a)  What is the value of the rate constant, k?

 

For a first order reaction:  .  Solving this for the rate constant k: 

 

 

(b)  What is the reaction rate when [N₂O₅] = 0.00100 M?

 

As given in the problem, the rate law is: 

rate = k[N₂O₅] = (102 s^-1)(0.00100 M) = 0.102 M s^-1  (3 sig figs)

 

15.  (15 points)  At a particular temperature, the value of K_p is 0.11 for the reaction:

CO₂(g)  +  H₂(g)    CO(g)  +  H₂O(g)

     Suppose that a flask is filled with 0.50 atm of CO₂ and 0.50 atm of H₂ and allowed to equilibrate.  What are the equilibrium partial pressures of CO, H₂O, CO₂, and H₂?

 

Initially, the partial pressures are are 0.50 atm for CO₂ and 0.50 atm for H₂.  Setting up an equilibrium table:

	CO2(g)	+	H2(g)		CO(g)	+	H2O(g)
Initial pressures	0.50		0.50		0		0
Change in pressure	–x		–x		+x		+x
Equilibrium pressure	0.50 – x		0.50 – x		x		x

 

Let an unknown quantity,+x, be equal to the change in pressure of CO.  Using the stoichiometry of the reaction, then the change in pressure of H₂O will also be +x, the change in pressure of CO₂ will be –x and the change in pressure of H₂ will also be –x  (Both CO₂ and H₂ are on the opposite sides of the arrows than CO.  If the pressure of CO goes up, the pressures on the opposite side must go down)

Add the initial pressures and the change in pressures to obtain the equilibrium pressures in terms of x.

Place the equilibrium pressures into the equilibrium expression and solve for x:

Calculate the equilibrium partial pressures:  P_CO = P_H2O = x = 0.12 atm

                                                    P_H2=P_CO2 = 0.50 – x = 0.38 atm

 

(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:  When the rate of the forward reaction equals the rate of the reverse reaction, then the reaction is in equilibrium.

 

 

 

 

 

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

 

 

 

 

 

Activation Energy:  The minimum energy necessary in a collision between two molecules for a reaction to occur.