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Chemistry 1A, Fall 2008 KEY
Midterm Exam III-Version A
November 12, 2008
(90 min, closed book) Name: TA:
SID: Section:
Please read this first: Write your name and that of your TA on all 9 pages of the exam
Test-taking Strategy
In order to maximize your score on the exam:
- Do the questions you know how to do first.
- Go back and spend more time on the questions you find more challenging.
- Budget your time carefully -- don't spend too much time on one problem.
- Show all work for which you want credit and don't forget to include units.
Question Page Points Score
A 2 10
Bi-ii 3 10
Biii-vii 4 17
C,D 5 21
E 6 14
F (MC) 7-9 28
Total 100
Useful Equations and constants: ∆G° = ∆H° - T∆S° ∆H° = Σ n∆H°f (products) - Σ n∆H°f (reactants) ∆S° = Σ nS° (products) - Σ nS° (reactants) ∆G° = Σ n∆G°f (products) - Σ n∆G°f (reactants) PV=nRT 1L·atm = 0.1kJ w = - Pext∆V ∆G° = - RTln K ∆G =∆G° + RTln Q ∆G° = - nF∆Єº ∆Є = ∆Єº - (RT/nF) lnQ or ∆Є = ∆Єº - (0.05916V/n) logQ oxidation occurs at the anode reduction occurs at the cathode ∆Ecell = Ecathode-Eanode
R
S
R T
H
K
ln
N 0 = 6.02214 x 10^23 mol - T (K) = T (C) + 273. F = 96,485 C / mol 1 V = 1 J / C R = 8.31451 J K-1^ mol - R = 8.20578 x 10-2^ L atm K-1^ mol
A.) Bond Energy and Potential Energy The following figure is a plot of the potential energy as a function of separation (r) between atoms. The black line corresponds to a hydrogen atom (H·) and a chlorine atom (Cl·) approaching each other. H·(g) + Cl·(g) → HCl (g)
Potential Energy
i. Which curve would correspond to hydrogen and fluorine (H· + F·) atoms?
Curve A same as H+Cl Curve B
ii. Explain your choice in terms of the separation distance.
The minimum in the curve corresponds to the bond length of the molecule. HF has a shorter bond length than HCl because fluorine is smaller and more electronegative than chlorine. Therefore the separation distance between atoms should be shorter for HF than for HCl which corresponds to Curve A.
iii. Explain your choice in terms of the minimum potential energy.
The depth of the well in the curve corresponds to the bond strength of the molecule. HF has a stronger bond than HCl because fluorine is smaller and more electronegative than chlorine. Therefore the plot for HF would have a deeper well than that of HCl. Curve A has a deeper well.
Note: Test version B used HBr instead of HF. The arguments are similar.
r, separation distance
H + Cl
Curve A
Curve B
Blood Alcohol Levels and Electrochemistry (continued) Ethanol fuel cell
CH 3 COOH (aq) + 4 H+^ (aq) + 4 e−^ → CH 3 CH 2 OH (aq) + H 2 O (l) E° = +0.23 V
O 2 (g) + 4 H+^ (aq) + 4 e−^ → 2 H 2 O (l) E° = +1.23 V
CH 3 CH 2 OH (aq) + O 2 (g) → CH 3 COOH (aq) + H 2 O (l)
iii. What is ∆Ecell°?
∆ Ecell = Ecathode-Eanode = 1.23-(0.23) -1.00V
iv. How many electrons are transferred in the net reaction?_____ 4 ________
v. What is being oxidized? CH 3 CH 2 OH, ethanol
vi. In the illustration to the right,
- Label the reactants that flow in and the products that flow out on each side (boxes provided).
- Show the direction the electrons are transferred. (box provided)
- What species travels through the membrane? ____ H+ ______
vii. What is used to determine if the individual being tested had been drinking a lot of alcohol or only a little? Use both words and equations to explain.
The measured cell voltage is correlated to the concentration of the species present. The Nernst equation (shown below) will determine the observed Ecell. The Nernst shows the correction to the observed voltage when you are not at standard conditions. ∆Ecell = ∆E°cell - (0.05916V/n) logQ Q = [CH 3 COOH] /[CH 3 CH 2 OH] PO 2
There can be a variety of conclusions to draw depending on your assumption about the starting (Q) conditions.
H 2 O
O 2
CH 3 COOH
CH3CH2OH, H 2 O
? V
MEMBRANE
e- flow
C.) Huts and Heat Capacity
Imagine two small huts in Antartica. One is made of snow (an igloo) and the other is made of concrete. Both have similar mass.
For most of the day, the temperature is 265 K (−8°C). As the temperature drops overnight, would you be warmer in the igloo or the hut made of concrete? Explain your thinking. Cp (snow) = 2.1 J/g K Cp (concrete) = 0.88 J/g K
Heat capacity relates temperature change to energy flow. The snow has a higher heat capacity than the concrete, meaning that it would have to lose more energy to drop the temperature by 1 degree. Both huts will lose heat to the surroundings at the same rate but will not undergo the same temperature change. Assuming that both huts have reached 265K, the hut made of snow (igloo) will lose heat less quickly than the hut made of concrete.
D.) Entropy of Vaporization
The enthalpy of vaporization at the boiling point and the boiling temperature of four liquids are given below.
liquid
carbon disulfide CS 2
chloroform CHCl (^3)
benzene C 6 H 6
formic acid HCOOH
∆Hvap ° (^) 28.4 kJ/mol 29.6 kJ/mol 30.7 kJ/mol 23.1 kJ/mol
boiling point 319 K 334 K 353 K 374 K
∆Svap ° 89.0 J/molK 88.6 J/molK 87.0 J/molK 61.8 J/molK
i. Calculate the entropy of vaporization, ∆Svap °, for the compounds above. Show an example calculation below and place your values in the table above.
∆ G ° = ∆ H ° - T ∆ S ° for a phase change, ∆G° = 0 because the system is at equilibrium
∆ H ° = T ∆ S ° ∆ S ° = ∆ H ° /T for CS 2 28.4 kJ/mol÷319 K = 0.0890 kJ/molK = 89.0 J/molK
ii. What evidence is there from the data that two molecules of formic acid, HCOOH, are attracted to each other in the gas phase?
The entropy for the liquid → gas is similar for CS 2 , CHCl 3 and C 6 H 6. The ∆S is lower for formic acid than for the other molecules. This indicates the gas phase formic acid is more ordered than the other gas phase molecules.
iii. The structure of formic acid is shown below. Draw a possible structure showing how two formic acid molecules are attracted to each other.
Multiple Choice (2 points each) For questions 1-7, there is only one correct answer.
Answer the questions below for the galvanic cell (battery) shown.
Sn2+^ (aq) + 2e- → Sn (s) E° = -0.14 V
Au+^ (aq) + 1e- → Au (s) E° = +1.69 V
1) Which electrode will dissolve?
A)
Sn
B)
Au
C)
both
D)
neither
2) What is that maximum amount of work from the cell if all species are in their standard states? ( either A or B answers accepted)
A)
+353 kJ
B)
-353 kJ
C)
-0.054 J
D)
+0.054 J
3) How would ∆Ecell change if the volumes of both solutions were doubled but the concentrations remain at 1.0M?
A)
increase
B)
decrease
C)
no change
4) How would ∆Ecell change if the concentrations of both solutions were doubled?
A)
increase
B)
decrease
C)
no change
Sn Sn2+ Au+^ Au
? V
The heat of combustion of CH 3 OH (l) is −24 kJ/g. You burn 1 gram of methanol, CH 3 OH (l), to heat 3 different substances. The results are given below. Experiment 1: The temperature of 100 g H 2 O (l) increases from 300 K to 357 K. Experiment 2: The temperature of 100 g NaCl (s) increases from 300 K to 579 K. Experiment 3: The temperature of 100 g N 2 (g) increases from 300 K to 540 K
5) The 100 g of water at 357 K will transfer____ compared with the 100 g of salt at 579 K to the surroundings as they cool.
A)
more heat
B)
less heat
C)
the same amount of heat
6) The best reason for your choice for #5) is:
A) NaCl is at a higher temperature B) H 2 O has a higher specific heat capacity C) NaCl is a solid. D) Energy is conserved.
7) How much work is done by the N 2 (g) when it is heated? Assume the external pressure is 1 atm.
A) 7 kJ B) 20 kJ C) 24 kJ
