Department of Chemistry, Geosciences and Environmental Sciences

 

Exam 1-B

Chemisty 1084: Sections 010 and 040 Spring 2008

 

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)

 

Score

 

Part 1 (40 points):_____________________

 

 

Part 2 (60 points):_____________________

 

 

Total (100 points):_____________________

 

 

Dont 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. The solubility of a gaseous solute in a liquid solvent:

(a) decreases with increasing pressure

(b) increases with increasing pressure

(c) is not affected by pressure changes.

 

Answer: B

 

 

2. In a saturated solution __________________________________________________________.

(a) the rate of solute dissolving is greater than the rate of solute precipitating out of solution.

(b) the rate of solute dissolving is less than the rate of solute precipitating out of solution.

(c) the rate of solute dissolving is equal to the rate of solute precipitating out of solution.

(d) the solute will rapidly precipitate out of solution when a seed crystal is added..

(e) the solute has stopped dissolving and precipitating out of solution.

 

Answer: C

 

3. Which of the following solutions will have the highest boiling point?

(a) 0.10 m Al2(SO4)3 (c) 0.10 m Ca(NO3)2

(b) 0.10 m HCl (d) 0.10 m C6H12O6

(e) All of these solutions have the same freezing point.

 

Answer: A

Al2(SO4)3 dissolved in water to make 5 ions so it would have the effect of a 0.50 m solution.

 

4. Which of the following is not a colloid?

(a) air (c) whipped cream

(b) smoke (d) homogenized milk

(e) Styrofoam

 

Answer: A

Air is a homogeneous gas mixture, a gaseous solution.


5. Given the following reaction:

4 NH3 + 7 O2 4 NO2 + 6 H2O

Using the definition of rate, complete the following equality:

(a) (c)

(b) (d)

(e) 1

 

Answer:

 

Using the definition of rate:

 

 

6. Shown above is a plot obtained from concentration data in the decomposition of nitrogen dioxide (NO2(g)). According to this plot, the rate law will be _____________ order with respect to NO2.

(a) zeroth (c) second

(b) first (d) cannot determine order from this plot.

 

Answer: C

 

A plot of 1/[A] versus time is a straight line for a second order rate law.

 

 

 

7. If a reactions temperature increases from 10C to 40C, the reaction rate should increase by a factor of:

(a) 2 (c) 6

(b) 4 (d) 8

(e) 16

 

Answer: D

 

Reaction rates double for every 10C rise in temperature. For a 30C rise, rates increase by a factor of 23=8.

 

 

 

 

8. Show above is a reaction profile for a chemical reaction. What is the activation energy for the reverse reaction?

(a) 25 kJ (c) 40 kJ

(b) 10 kJ (d) 30 kJ

(e) 55 kJ

 

Answer: C

 

This is equal to 55 kJ minus 15 kJ.

 

9. Which of the following conditions holds only for reactions in equilibrium?

(a) the reactant and product concentrations do not change with time.

(b) the reactant and product concentrations are equal.

(c) all of the reactants have been consumed to form products.

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

(e) the reaction has ceased.

 

Answer: D

This is the definition of equilibrium.

 

10. A reaction with a very large value for the equilibrium constant will:

(a) have mostly reactants present at equilibrium.

(b) have mostly products present at equilibrium.

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

 

Answer: B

 

 


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 25.00 g of ethanol (C2H5OH) in 125.0 g of water (H2O).

 

(a) (3 points) What is the mass percent of ethanol (%C2H5OH) in this solution?

 

(b) (4 points) What is the mole fraction of ethanol () in this solution?

 

(c) (3 points) What is the molality of ethanol in this solution?

 


12. (15 points) A solution prepared by dissolving 0.984 g of a solute in 9.287 g of cyclohexane has a freezing point of 5.0C. Given the freezing point of cyclohexane at 6.6C and its Kf value of 20.8C/m, calculate the molar mass of the solute.

Calculate the freezing point depression:

Calculate the molality of the solute from the freezing point depression and the Kf value:

Calculate the number of moles of solute by multiplying the molality by the number of kilograms of pure solvent (cyclohexane):

Calculate the molar mass by dividing the mass of solute (given in problem) by the number of moles of solute:


13. (10 points) When the reaction shown below is at equilibrium, the partial pressures of SO2, O2, and SO3 are equal to 0.135 atm, 0.455 atm, and 0.053 atm, respectively. Calculate the value of Kp for the reaction shown below.

2 SO2(g) + O2(g) 2 SO3(g)

 

The equilibrium expression, Kp, for this reaction is:

Since the equilibrium partial pressures are given directly, just place the numbers into the expression and evaluate KP:

 

14. (10 points) The decomposition of N2O5 has a 1st order rate law with respect to N2O5. The value of the half-life (t1/2) when [N2O5]0 = 0.100 M is 144 s.

2 N2O5 4 NO2 + O2

 

(a) (5 points) What is the value of the rate constant, k, for the rate law?

For a 1st order rate law:

 

(b) (5 points) What is the rate of the reaction when [N2O5] = 0.100 M?

In the first sentence of this problem, you were told that the rate law is equal to: rate = k[N2O5]. Using the value of k from part (a), the rate is equal to:

 


15. (15 point) The following data was collected for a reaction with the stoichiometry:

2 A + B products

 

 

Experiment

[A]0 (M)

[B]0 (M)

Initial rate (M/s)

 

 

1

0.025

0.025

2.40 10-3

 

 

2

0.025

0.075

2.16 10-2

 

 

3

0.050

0.075

2.16 10-2

 

 

From this data, determine the rate law for this reaction and the value of the rate constant, k, with units.

 

From the data, it is implied that the rate law has the general form: rate = k[A]m[B]n

 

In order to find the value of the rate exponent m, compare experiments 2 and 3:

[A]0 doubles

[B]0 remains constant

Rate remains unchanged.

From this data: 2m = 0 , the value of m is equal to 0.

 

In order to find the value of the rate exponent n, compare experiments 1 and 2:

[A]0 is constant

[B]0 triples

Rate increases nine-times

From this data: 3n = 9, the value of n is equal to 2.

 

The rate law is: rate = k[B]2.

 

To find 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.

 

 

Parts per million (ppm):

 

 

 

 

Equilibrium: when the forward rate of a reaction equals the reverse rate of a reaction.

 

 

 

 

Unimolecular elementary step: a step in a reaction mechanism that has only one reactant.