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This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points Which of the following statements about sub- limation is true?
- none of these
- ∆S < 0
- ∆H < 0
- ∆S > 0 correct
Explanation: For all phase changes, the signs of both ∆H and ∆S are the same. Melting, vaporization and sublimation are endothermic, while freez- ing, deposition, and condensation are exother- mic.
002 10.0 points Rank the following in order of increasing va- por pressure: BeO, CH 3 CH 2 NH 2 , C 6 H 6 , CsF, C 3 H 6 O, N 2.
- CsF < BeO < C 3 H 6 O < CH 3 CH 2 NH 2 < C 6 H 6 < N 2
- BeO < CsF < C 3 H 6 O < CH 3 CH 2 NH 2 < C 6 H 6 < N 2
- BeO < CsF < CH 3 CH 2 NH 2 < C 3 H 6 O < C 6 H 6 < N 2 correct
- CsF < BeO < CH 3 CH 2 NH 2 < C 3 H 6 O < N 2 < C 6 H 6
Explanation: Ranking in terms of increasing vapor pres- sure requires ranking in terms of decreasing IMF.
003 10.0 points When attempting to dissolve a salt in water, if the lattice energy is larger in magnitude than the hydration energy, which of the following
statements is true?
- The salt cannot dissolve because the heat of solution is endothermic.
- The salt will dissolve because the heat of solution is exothermic.
- The salt can only dissolve if the tempera- ture of solution is increased. correct
- The salt can only dissolve if the tempera- ture of the solution is decreased.
Explanation: ∆Hsolutuon = ∆Hlattice + ∆Hhydration
004 10.0 points The phase diagram for a pure compound is given below.
Solid
Liquid
50 Vapor
Temperature, K
Pressure, atm
The triple point occurs at
- 200 atm and 400 K.
- greater than 50 atm and greater than 200 K.
- 50 atm and 200 K. correct
- 0 atm and 200 K.
- 320 atm and 250 K.
Explanation: The triple point is where three phases co- exist: here it is where the three lines on the diagram intersect.
005 10.0 points The phase diagram for a pure substance is given below.
Temperature, K
Pressure, atm
The substance is stored in a container at 150 atm at 25◦C. Describe what happens if the container is opened at 25◦C.
- The liquid in the container freezes.
- The solid in the container melts.
- The vapor in the container escapes.
- The liquid in the container vaporizes. correct
- The solid in the container sublimes.
Explanation:
006 10.0 points Given the constants below, which of the fol- lowing statements are true about the heating curve that would be produced when a sample of ice is heated to turn it into steam?
cice = 2.05 J · g−^1 · K−^1 cwater = 4.184 J · g−^1 · K−^1 csteam = 2.08 J · g−^1 · K−^1 ∆Hvap = 2256 J · g−^1 ∆Hf us = 334 J · g−^1
- The slope for heating ice is steeper than the slope for heating liquid. correct
- The slope for heating water is steeper than the slope for heating vapor.
- More heat must be supplied to melt ice than to vaporize liquid.
- none of these
Explanation:
007 10.0 points Which of the following would increase the solubility of nitrogen gas, N 2 (g), in the solvent water? I) decreasing the pressure of N 2 (g) II) increasing the temperature of the sample III) changing to a more polar solvent
- III only
- II only
- I and II
- I only
- I and III
- II and III
- none of these correct
Explanation: The solubility of a gas is directly propor- tional to the gas’ pressure, inversely propor- tional to the temperature of the solvent. Con- sidering the principle of like dissolve like, a non-polar species such as N 2 (g) will be more soluble in non-polar solvents. The volume of the solvent does not impact the solubility, which is an intensive property; this is evident when one considers the unity of solubility such as grams per liter, etc.
008 10.0 points Octane (C 8 H 18 ) is a major component of gaso- line. Which of the following would you expect to be the LEAST soluble in gasoline?
- candle wax
Explanation: There are 15.625 moles of particles in the solution, which corresponds to 7.8125 moles of NaCl, or 456.6 grams NaCl.
013 10.0 points Referring to the graph
Molar Gibbs energy,
G
m
Temperature, T
vapor
pure liquid
solvent
and using your knowledge of ∆Hsolution, com- plete the following statement concerning boil- ing point elevation: Adding solute decreases the molar Gibbs energy of the liquid at all temperatures by (increasing/decreasing) the molar (en- thalpy/entropy) of the liquid. This causes the vapor and solvent lines to intersect at a higher temperature.
- decreasing; enthalpy
- increasing; entropy correct
- decreasing; entropy
- increasing; enthalpy
Explanation: G = H − T S, so in order to decrease G at a given temperature, we have to either decrease H or increase S. We know the ∆Hsolution can be positive or negative depending on the so- lute, but colligative properties don’t depend on the identity of the solute, so the effect on the enthalpy is not a good explanation. We also know that mixtures always have more en- tropy than pure substances, so that must be
the best explanation.
014 10.0 points Consider the reaction
2 HgO(s) ⇀↽ 2 Hg(ℓ) + O 2 (g).
What is the form of the equilibrium constant Kc for the reaction?
- Kc = [Hg]^2 [O 2 ]
- Kc = [O 2 ] correct
- None of the other answers is correct.
- Kc =
[Hg]^2 [O 2 ] [HgO]^2
- Kc =
[O 2 ]
[HgO]^2 Explanation: Solids and liquids are not included in the K expression.
015 10.0 points Consider the reaction: 2 HI(g) ↔ H 2 (g) + I 2 (g) If we start out with pure HI and the equi- librium hydrogen gas concentration is 0. M at 730 K and at this temperature Kc = 0.12, what is the correct expression for the equilibrium concentration of HI(g)?
1. [HI] =
correct
- [HI] = (0. 233 · 0. 233 · 0 .12)^1 /^2
3. [HI] =
4. [HI] = (0. 233 · 0. 233 · 0 .12)
5. [HI] =
Explanation: Since we started from pure HI, the equilib- rium concentrations of hydrogen and iodine gas must be equal.
016 10.0 points Consider the reaction:
NO(aq) + CO 2 (aq) ↔ NO 2 (aq) + CO(aq)
If at some temperature Kc = 4, and the initial concentrations of NO, CO 2 , NO 2 and CO are 5, 10, 2 and 0 M respectively, what is the equilibrium concentration of CO?
- 6
- 9
- 1
- 16
- 4 correct
- 14
Explanation:
K =
[NO 2 ] · [CO]
[NO] · [CO 2 ]
(2 + x) · x (5 − x) · (10 − x)
x = 4
017 10.0 points The equilibrium constant Kc for the reaction
2 SO 2 (g) + O 2 (g) → 2 SO 3 (g)
is 11.7 at 1100 K. A mixture of SO 2 , O 2 , and SO 3 , each with a concentration of 0.015 M, was introduced into a container at 1100 K. Which of the following is true?
- [SO 3 ] = 0.045 M at equilibrium.
- SO 2 (g) and O 2 (g) will be formed until equilibrium is reached. correct
- [SO 3 ] = [SO 2 ] = [O 2 ] at equilibrium.
- SO 3 (g) will be formed until equilibrium is reached.
- [SO 3 ] = 0.015 M at equilibrium.
Explanation:
018 10.0 points Consider the reaction
2 H 2 O(ℓ) + O 2 (g) ←→ 2 H 2 O 2 (aq)
at equilibrium. If you remove some O 2 from the system, which direction will the reaction shift?
- the reaction will shift to the right
- nothing will occur
- not enough information
- the reaction will shift to the left correct
Explanation: The Le Chatelier principle states that when a stress is applied to a system that is at equi- librium, the system will respond in such a way to restore that equilibrium and mitigate the effects of the applied stress. In this case, our system is:
2 H 2 O(ℓ) + O 2 (g) ←→ 2 H 2 O 2 (aq)
and it is at equilibrium. The act of removing O 2 will force the system to shift left in order to make more O 2 and restore equilibrium.
019 10.0 points Consider a reaction with ∆Hrxn = 63.8 kJ. Which pair of K values and temperatures is possible for this reaction?
- K 1 = 10.5, T 1 = 130 K; K 2 = 9.33, T 2 = 327 K
- K 1 = 21, T 1 = 111 K; K 2 = 21, T 2 = 316 K
- None of these
- K 1 = 14.6, T 1 = 122 K; K 2 = 16.6, T 2 = 315 K correct Explanation:
- 1 × 10 −^6 mol
- 1 × 10 −^3 mol
- 8 × 10 −^3 mol correct
- 2 × 10 −^6 mol
Explanation:
025 10.0 points Rank the following compounds from most sol- uble to least soluble. Assume that all bonds except the OH are ionic. (You can estimate this ranking without using a calculator.)
Compound Ksp Bi 2 S 3 1. 0 × 10 −^97 Fe(OH) 2 1. 6 × 10 −^14 PbI 2 2. 6 × 10 −^13 HgS 1. 6 × 10 −^52
- Fe(OH) 2 > PbI 2 > HgS > Bi 2 S 3
- HgS > PbI 2 > Fe(OH) 2 > Bi 2 S 3
- PbI 2 > Fe(OH) 2 > Bi 2 S 3 > HgS correct
- PbI 2 > Fe(OH) 2 > HgS > Bi 2 S 3
- Bi 2 S 3 > Fe(OH) 2 > HgS > PbI 2
Explanation:
026 10.0 points Consider five generic acids (HA, HB, HC, HD, and HE) that have the following ionization constants.
Ionization Acid Constant Ka value HA 4. 6 × 10 −^3 HB 1. 3 × 10 −^3 HC 7. 6 × 10 −^4 HD 6. 3 × 10 −^3 HE 9. 3 × 10 −^5
Which of the following anions will be the WEAKEST base?
- B−
- C−
- D−^ correct
- A−
- E− Explanation: The weakest base anion will pair up with the strongest acid in the list which is HD.
027 10.0 points What is the approximate pH of a solution labeled 2. 0 × 10 −^4 M KOH?
- 10.3 correct Explanation: [KOH] = 2. 0 × 10 −^4 M KOH is a strong base which completely dissociates in aqueous solution:
KOH −→ K+^ + OH−
[OH−] = [KOH] = 0.0002 M
[H 3 O+] =
Kw [OH−]
1 × 10 −^14
M
= 5 × 10 −^11 M
pH = − log[H 3 O−] = 10. 301
028 10.0 points A weak acid HX has Ka = 1. 00 × 10 −^6. What would be the pH of a 0.10 M solution of HX in water?
- 3.50 correct
Explanation:
029 10.0 points Calculate the pH of a 0.30 M solution of NaF. Ka for HF is 7. 2 × 10 −^4. (NaF completely dissociates in water.)
- 8.3 correct
Explanation:
030 10.0 points Calculate the [OH−] in an aqueous solution that is 0.125 M NH 3 and 0.300 M NH 4 Cl. The value of Kb for NH 3 is 1.8 × 10 −^5.
- 0.125 M
- 4.3 × 10 −^5 M
- 7.5 × 10 −^6 M correct
4. 1.8 × 10 −^5 M
5. 0.425 M
Explanation:
