Name_____________KEY____________________
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Your name is worth 1 point, Merry Xmas.
Do not open this exam until told to do so. The exam consists of 16 pages, including this one. Count them to insure that they are all there. The last five pages of the exam are the lists of constants and equations. No additional notes are allowed. You should only have your exam, writing implements and a calculator on your desk.
Do not write in the area below
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For each of the following problems, a right answer is worth 3 points, a wrong answer results in -1 points and no answer is scored as a zero. You do not need to justify your answer.
1) In general, the temperature of a system is most accurately described as:
2) During the isothermal expansion of an ideal gas:
3) At constant pressure, when CsCl is dissolved in water, the beaker becomes quite cold to the touch. The enthalpy change for dissolving CsCl in water is:
4) For an isothermal expansion of an ideal gas, the change in enthalpy is:
5) For an adiabatic compression of an ideal gas, the change in internal energy is:
6) An ideal gas is compressed isothermally and reversibly from 2 liters to 1 liter, then expanded adiabatically and reversibly back to 2 liters. The overall change in temperature of the gas is:
7) The freezing of liquid water to form ice
8) When a spontaneous, endothermic reaction occurs at constant temperature and pressure, the entropy change of the surroundings is:
9) When a spontaneous, endothermic reaction occurs at constant temperature and pressure, the entropy change of the system is:
10) Consider a mixture of ice and water at equilibrium in an adiabatic container. Heat is transferred into the system until half the ice melts. The resulting change in temperature of the system will be
11) Consider the transfer of heat between a hot block of copper and a cold block of copper. Which of the following is true?
12) For the process of heat transfer between the hot and cold blocks in the last problem, the free energy change is:
13) For a spontaneous chemical reaction, the enthalpy change is:
14) At 20 C and 1 atm, the chemical potential of ice is
15) Upon dissolving salt into a mixture of ice and water at equilibrium, the temperature of the mixture will:
16) You can cause ice to melt by exerting pressure on it. This means that:
17) Carbon tetrachloride and water are immiscible. That is, they do not mix together. This means that:
18) What is the maximum number of phases that can be present in a two component system?
19) When the reaction Aß à B is at equilibrium, the reaction free energy is:
20) If I add salt to water, the chemical potential of the water will:
21) After dissolving 0.01 M NaOH in pure liquid water, the pH is approximately (assuming all activity coefficients are 1.0):
22) If one dissolves an equal amount of the conjugate base and the conjugate acid form of a molecule in a beaker of water, the final pH will be approximately (assuming activity coefficients are all one):
23) According to the Debey-Huckle theory, the activity (not the activity coefficient!) of NaCl in a 0.1 molal aqueous solution is:
24) The anode of an electrochemical cell is always:
25) For a spontaneous electrochemical reaction, the zero current potential (E) is:
26) The rate of a chemical reaction has units of:
27) The reaction A ß à B has an equilibrium constant of 10. This means that:
28) For the irreversible reaction A à B, doubling the concentration of B will cause the rate of change of A (dA/dt) to:
29) For the elementary reaction, A+B+C à D, the units of the forward rate constant are:
30) What molecular properties give rise to the red color of the paint on a stop sign?
31) When you measure the absorbance spectrum of a complex organic dye molecule in solution, you typically find that it is 10 - 30 nanometers wide. A likely explanation for the spectrum being so broad is:
32) Your mirror is probably made of a piece of glass with an aluminum backing. Almost all the light that hits the mirror reflects back. Only about 5% of the light that hits the surface of the glass in your window reflects back. The reason why light reflects so well from the aluminum surface compared to the glass surface is:
33) Phophorescence arises from:
Each of the following problems is worth 25 points. You must show your work or reasoning for credit!
34) 40 grams of ice at 0 C is placed in 100 mls of water at 25 C in an adiabatic container. What is the final temperature when the system comes to equilibrium? How much ice will there be (in grams) and how much water (in mls) at equilibrium? The density of water is 1.00 g/ml under these conditions.
The heat required to bring 25 C water to 0 C: q = nwater CP DT =
(5.56 moles)(75.5 J/(K mole))(25 K) = 10,500 J
How much ice will this much heat melt? q = nice DHfus; 10,500 J = nice (6008 J/mole)
nice = 1.75 moles. This is 1.75 moles x 18 g/mole = 31.4 grams. Since this is less than the 40 grams of ice we started with, ice and water are still present at equilibrium so the temperature must be O C. 40 - 31.4 grams is 8.6 grams which is how much ice remains. The amount of liquid water is just 100 mls + 31.4 grams x 1 ml/g = 131.4 mls.
35) 3.00 moles of an ideal gas is expanded isothermally against a constant pressure of 1 atm from 2.0 liters to 10.0 liters at a temperature of 20.0 C.
Determine each of the following for this process (express all energies in Joules and entropy changes in J/K):
w = -PDV = -(1atm)(8 l) = -(101.3 J/l)(8 l) = -811.2 J
q = -w (since DU = 0) = 811.2 J
DU = 0 (since isothermal ideal gas)
DH = 0 (since isothermal ideal gas)
DSSYS = qrev/T = -wrev/T = nRln(VF/VI) = 40.14 J/K
DSSUR = -qsys/T = (-811.2 J)/(293 K) = -2.77 J/K
Is this process spontaneous (explain why or why not)?
It is spontaneous. The total entropy change for the system is greater than zero (adding the change in energy for the system and for the surroundings together above).
36) For the reaction A + B à C ß à D, the rate constant for the first irreversible reaction forming C is k1, for the reaction from C to D the rate constant is k2f and for the reverse reaction from D to C the rate constant is k2r. These rate constants have the following values: k1 = 2.00 M-1s-1, k2f = 1.00 s-1, k2r = 5.00 s-1. The initial concentrations of each of the components in the reaction (A, B, C and D) is 0.100 M.
d[A]/dt = -k1 [A][B]
d[B]/dt = -k1 [A][B]
d[C]/dt = k1 [A][B] - k2f [C] + k2r [D]
d[D]/dt = k2f [C] - k2r [D]
b) Evaluate the initial rates of [A], [B], [C] and [D] change for this reaction (before any A, B, C or D is used up)
d[A]0/dt = -(2.00 M-1s-1)(0.1M)(0.1M) = -0.02 M/s
d[B]0/dt = -(2.00 M-1s-1)(0.1M)(0.1M) = -0.02 M/s
d[C]0/dt = (2.00 M-1s-1)(0.1M)(0.1M) - (1.00 s-1)(0.1M) + (5.00 s-1)(0.1M)
= 0.42 M/s
d[D]0/dt = (1.00 s-1)(0.1M) - (5.00 s-1)(0.1M) = -0.40 M/s
c) Determine the final (equilibrium) concentrations of each component.
[A]eq = 0.00 M [B]eq = 0.00 M [C]eq = 0.25 M [D]eq = 0.05 M
At equilibrium, [A] = [B] = 0 since the first reaction is irreversible. That means all of the material is in C and D which implies that [C] + [D] = 0.3 M. From the fact that we are at equilibrium we can say that K = [D]/[C] = k2f/k2r = 0.2. Solving the two equations for [C] and [D] gives the values listed above.
37) One thing we work on in my lab is the process of photosynthesis. The light driven reactions of photosynthesis convert light energy into chemical energy using the overall reactions diagramed below. The light energy is stored by generating an oxidant, P+, and a reductant, Q-. The final state, P+ Q- is higher in energy than the initial state, P Q, and therefore some of the light energy is stored.
In the organisms that we study, light at a wavelength of 860 nm is absorbed by the pigment P. During the lifetime of the excited state of P (P*), an oxidation reduction reaction occurs forming oxidized P (P+) and a reduced quinone (Q-). The half reaction P+ + e- à P has a standard potential of about 0.500 V. The half reaction Q + e- à Q- has a standard potential of about 0.00 V.
For one molecule, E = hc/l = 2.31 x 10-19 J. Multiplying by Avagodro's number to give the number of joules for a mole of reactants gives 139,100 J/mole.
P à P+ + e- -0.500 V
Q + e- à Q- 0.000 V
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P + Q à P+ + Q- -0.500 V
converting potential to reaction free energy: DrG0 = -nFE0 = 48,300 J/mole
(48.3 kJ/mole)/(139.1 kJ/mole) = 0.347