Department of Chemistry, Geosciences and Environmental Sciences
Exam 1AKey
Chemisty 1084: Sections 010 and 030 Spring 2007
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: _{}
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.
1. Pairs of liquids that mix in all proportions are called:
(a) miscible (c) polar
(b) unsaturated (d) saturated
(e) supersaturated
Answer: A
definition of miscible
2. Which one of the following concentration units varies with temperature?
(a) mass percent (c) mole fraction
(b) molarity (d) molality
(e) All of these are temperature dependent
Answer: B
3. A 0.100 m solution of which of the following solutes will have the lowest freezing point?
(a) CaCl_{2} (c) Fe(NO_{3})_{3}
(b) C_{6}H_{12}O_{6} (d) LiCl
(e) All of these solutions have the same freezing point
Answer: C
The solution with the greatest number of particles (ions) will have the lowest freezing point. CaCl_{2} produces 3 ions. Fe(NO_{3})_{3} produces 4 ions. C_{6}H_{12}O_{6} is a nonelectrolyte and produces no ions. LiCl produces 2 ions.
4. If the rate law for the reaction: 2 A + 3 B ¾® product, is first order with respect to A and second order with respect to B, then the rate law is rate = _________________.
(a) k[A][B]^{2} (c) k[A]^{2}[B]^{3}
(b) k[A][B] (d) k[A]^{2}[B]
(e) k[A]^{2}[B]^{2}
Answer: A
5. Which graph is linear for a 2^{nd} order rate law?
(a) ln[A] versus t (c) [A] versus t
(b) 1/[A] versus t (d) lnk versus 1/T
Answer: B
6. The reaction mechanism for the following reaction:
2 NO_{2}(g) + Cl_{2}(g) ¾¾® 2 ClNO_{2}(g)
is believed to be:
NO_{2}(g) + Cl_{2}(g) ¾¾® ClNO_{2}(g) + Cl(g) (slow)
NO_{2}(g) + Cl(g) ¾¾® ClNO_{2}(g) (fast)
What is the rate law that is consistent with this mechanism?
(a) rate = k[NO_{2}]^{2}[Cl_{2}] (c) rate = k[NO_{2}][Cl_{2}]
(b) rate = k[NO_{2}][Cl] (d) rate = k[NO_{2}]^{2}
Answer: C
The rate law consistent with this mechanism is the rate of the slow first step.
7. A certain reaction has an energy profile shown above. What is the value of DE for this reaction?
(a) –15 kJ (c) 45 kJ
(b) 30 kJ (d) 60 kJ
(e) 15 kJ
Answer: E
DE = Energy of products minus energy of reactants = 30 kJ – 15 kJ = 15 kJ
8. If the temperature of a reaction is increased from 0°C to 40°C, then the reaction rate should increase by a factor of:
(a) 2 (c) 6
(b) 4 (d) 8
(e) 16
Answer: E
Temperature rose by 40°C. Since rate doubles for every 10°C rise, the rate increases by a factor of 2^{4} = 16.
9. If the value of the equilibrium constant is very small, then at equilibrium:
(a) there will be mostly reactants present.
(b) there will be equal amounts of reactants and products.
(c) there will be mostly products present.
(d) all of the reactants will have been consumed.
Answer: A
10. If K_{p} = 0.0752 for the reaction:
2 Cl_{2}(g) + 2 H_{2}O(g) _{} 4 HCl(g) + O_{2}(g)
what is the value of K_{p} for the reaction:
2 HCl(g) + ½ O_{2}(g) _{} Cl_{2}(g) + H_{2}O(g)
(a) 13.3 (c) –0.036
(b) 3.65 (d) 0.00566
(e) 0.274
Answer: B
The second reaction is equal to half of the reverse of the 1^{st} reaction. Take the first K_{p} and take the reciprocal of its value for reversing the reaction and then take the square root for halving the reaction.
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) Concentrated phosphoric acid, H_{3}PO_{4}(aq), is 85.0% H_{3}PO_{4} by mass and has a density of 1.70 g/mL. What is the molarity of H_{3}PO_{4} in this solution?
Assume 1 L (1000 mL) of solution. The mass of 1 L of solution is equal to:
_{}
Use the percent by mass to calculate the mass of H_{3}PO_{4} in 1 L of solution:
_{}
Divide by the molar mass of H_{3}PO_{4} to calculate the number of moles of H_{3}PO_{4} in 1 L of solution. This is equal to the molarity of H_{3}PO_{4}:
_{}
12. (15 points) A solution prepared by dissolving 1.278 g of a solute in 15.426 g of cyclohexane has a freezing point equal to –3.9°C. Given the freezing point of pure cyclohexane of 6.6°C and its freezing point depression constant of 20.8°C/m, calculate the molar mass of this solute.
Calculate DT_{f}:
DT_{f} = 6.6°C – (–3.9°C) = 10.5°C
Calculate the molality of solute using DT_{f} = K_{f}m_{solute}:
_{}
Calculate the number of moles of solute by multiplying the molality by the number of kilograms of solvent (cyclohexane):
_{}
Calculate the molar mass of the solute by dividing the mass of solute (1.278 g) by the number of moles:
_{}
13. (10 points) The equilibrium constant, K_{c}, is equal to 2.25 × 10^{4} at 1130 K for the reaction:
2 H_{2}S(g) _{} 2 H_{2}(g) + S_{2}(g)
An equilibrium mixture of all three gases has [H_{2}S] = 0.00484 M and [H_{2}] = 0.00150 M. What is the equilibrium concentration of S_{2} in this mixture?
Using the law of mass action, the K_{c} expression is equal to:
_{}
Place the given concentrations and value of the K_{c} into the expression and solve for [S_{2}]:
_{}
14. (10 points) The decomposition of PH_{3},
4 PH_{3}(g) ¾¾® P_{4}(g) + 6 H_{2}(g)
has a first order rate law with a rate constant, k = 0.0198 s^{1}.
(a) (4 points) What is the value of the halflife, t_{1/2}, when [PH_{3}]_{0} = 0.0500 M?
For a 1^{st} order reaction: _{}
(b) (6 points) How long will it take for [PH_{3}] to fall from 0.0500 M to 0.0100 M?
For a 1^{st} order reaction: _{}
Plugging in the numbers from the problem and solving for t:
_{}
15. (15 points) The following data were collected for the reaction:
2 HgCl_{2}(aq) + C_{2}O_{4}^{2}(aq) ¾¾® 2 Cl¯(aq) + 2 CO_{2}(g) + Hg_{2}Cl_{2}(s)

Experiment 
[HgCl_{2}]_{0} (M) 
[C_{2}O_{4}^{2}]_{0} (M) 
Initial rate (M/s) 


1 
0.164 
0.15 
3.2 × 10^{5} 


2 
0.164 
0.45 
2.88 × 10^{4} 


3 
0.082 
0.45 
1.44 × 10^{4} ^{} 

Deduce the rate law from this data. Calculate the value of the rate constant, with units.
The rate law with have the form: rate = k[HgCl_{2}]^{m}[C_{2}O_{4}^{2}]^{n}
^{ }
To find the value of m, compare experiments 3 and 2 (in that order):
[HgCl_{2}] doubles _{}
[C_{2}O_{4}^{2}] is constant 2^{m} = 2; m = 1
Rate doubles _{}
To find the value of n, compare experiments 1 and 3:
[HgCl_{2}] is constant
[C_{2}O_{4}^{2}] triples _{} 3^{n} = 9; n = 2
Rate increases by a factor of 9 _{}
The rate law is: rate = k[HgCl_{2}][C_{2}O_{4}^{2}]^{2}
To find the value of k, plug in the numbers from experiment 1 into the rate law and solve for k:
_{}
(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.
Homogeneous catalyst: a catalyst that is in the same phase as the reactants.
Equilibrium: when the forward reaction rate equals the reverse reaction rate for a reaction.
Unsaturated solution: a solution whose concentration is less than the solute’s solubility.