CHEMISTRY 123-01 Midterm #2 – answer key October 23, 2007

CHEMISTRY 123-01
Midterm #2 – answer key
October 23, 2007
Statistics:
• Average: 71 p (71%);
• Highest: 98 p (98%); Lowest: 43 p (43%)
• Number of students performing at or above average: 50 (49%)
PART I: MULTIPLE CHOICE (Each multiple choice question has a 2-point value).
1.
How many moles of NaOH are present in 25.0 mL of a 0.1000 M NaOH solution?
A. 100 mol
B. 2.50 × 10-3 mol
C. 0.100 mol
D. 2.50 mol
E. 25.0 mol
2.
Which of the methods described below will yield 100 mL of a 0.100 M NaOH solution?
A. Add exactly 100 mL of water to 4.00 g of NaOH.
B. Dissolve 0.400 g NaOH in water and dilute to exactly 100 mL.
C. Add exactly 100 mL of water to NaOH.
D. Dissolve 4.00 g of NaOH in water and dilute to exactly 100 mL.
E. Dilute 50.0 mL of 1.00 M NaOH to exactly 100 mL.
3.
Which of the following correctly describes the relationship between the pressure of a gas and the volume of a gas at constant
temperature?
A. as one increases the other increases
B. unrelated
C. directly proportional
D. irreversibly proportional
E. inversely proportional
4.
A sample of N2 gas occupies 2.40 L at 20°C. If the gas is in a container that can contract or expand at constant pressure, at
what temperature will the N2 occupy 4.80 L?
A. 10°C
D. 313°C
B. 40°C
E. 685°C
C. 146°C
5.
If the pressure on a gas sample is tripled and the absolute temperature is quadrupled, by what factor will the volume of the
sample change?
D. 1/3
A. 12
E. 4
B. 4/3
C. 3/4
6.
At what temperature will a fixed amount of gas with a volume of 175 L at 15°C and 760 mmHg occupy a volume of 198 L at
a pressure of 640 mm Hg?
A. 274°C
D. 1°C
B. 214°C
E. -59°C
C. 114°C
7.
How many molecules of N2 gas can be present in a 2.5 L flask at 50°C and 650 mmHg?
A. 2.1 × 10-23 molecules
B. 4.9 × 1022 molecules
C. 3.1 × 1023 molecules
D. 3.6 × 1025 molecules
E. 0.081 molecules
8.
Calculate the volume occupied by 35.2 g of methane gas (CH4) at 25°C and 1.0 atm (R = 0.0821 L.atm/mol.K)
A. 0.0186 L
B. 4.5 L
C. 11.2 L
9.
D. 49.2 L
E. 53.7 L
Calculate the density, in g/L, of CO2 gas at 27°C and 0.50 atm pressure.
D. 46.0 g/L
A. 0.89 g/L
E. 2.17 kg/L
B. 1.12 g/L
C. 9.93 g/L
10. Which of the following gases will have the greatest density at the same specified temperature and pressure?
D. C2H6
A. H2
E. CF4
B. CClF3
C. CO2
11. Two moles of chlorine gas at 20.0°C are heated to 350°C while the volume is kept constant. The density of the gas
A. increases.
B. decreases.
C. remains the same.
D. Not enough information is given to correctly answer the question.
12. A sample of hydrogen gas was collected over water at 21°C and 685 mmHg. The volume of the container was 7.80 L.
Calculate the mass of H2(g) collected. (Vapor pressure of water = 18.6 mmHg at 21°C.)
A. 0.283 g
D. 7.14 g
E. 435 g
B. 0.572 g
C. 0.589 g
13. How many liters of chlorine gas at 200°C and 0.500 atm can be produced by the reaction of 12.0 g of MnO2 with HCl as
follows?
MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)
A. 10.7 L
B. 3.09 L
C. 4.53 L
D. 0.138 L
E. 0.093 L
14. If equal masses of O2(g) and HBr(g) are in separate containers of equal volume and temperature, which one of the following
statements is true?
A. The pressure in the O2 container is greater than that in the HBr container.
B. There are more HBr molecules than O2 molecules.
C. The average velocity of the O2 molecules is less than that of the HBr molecules.
D. The average kinetic energy of HBr molecules is greater than that of O2 molecules.
E. The pressures of both gases are the same.
15. Which of the following IS NOT a postulate of the kinetic-molecular theory of gases?
A. All gases consist of atoms and molecules moving randomly in all directions at various speeds.
B. The average kinetic energy of gas molecules increases as the temperature increases.
C. The volume of a gas molecule is negligible when compared to the space that separates it from other gas molecules.
D. Gas molecules will eventually stop colliding due to the energy lost in each collision.
E. There are no attractive or repulsive forces between gas molecules.
16. Which of the following gas molecules have the highest average kinetic energy at 25°C?
D. Cl2
A. H2
B. O2
E. All the gases have the same average kinetic energy.
C. N2
17. Oxygen gas at 1.000 atm, placed in a container having a pinhole opening in its side, leaks from the container 2.14 times
faster than does a sample of an unknown gas at 1.000 atm placed in this same apparatus. Which of the following species
could be the unknown gas?
C. Kr
A. Cl2
B. SF6
D. UF6
E. Xe
18. Which of the following is not an example of kinetic energy?
A. The motion of a molecule
B. The motion of a golf ball
C. The vibration of an object
D. A loosely held brick on the top of a building
E. The motion of electrons through a wire
19. Determine the incorrect relationship given below.
A. 80.0 cal/g = 312 J/g
B. 1000 cal = 1 kcal
C. 44.0 kJ = 1.05 x 104 cal
D. 1 µJ = 1 x 0-6 J
E. 1000 J = 1 kJ
20. An exothermic reaction causes the surroundings to
A. warm up.
B. become acidic.
C. expand.
D. decrease its temperature.
E. release CO2.
21. Which statement is true?
A. A positive change in enthalpy occurs with endothermic processes.
B. A positive change in enthalpy occurs with exothermic processes.
C. A positive change in enthalpy occurs when work is done on the surroundings.
D. A positive change in enthalpy occurs when work is done on the system.
E. A positive change in enthalpy occurs when a process is endothermic and does work on the surroundings.
22. How many degrees of temperature rise will occur when a 25.0 g block of aluminum absorbs 10.0 kJ of heat? The specific
heat of Al is 0.900 J/g·°C.
D. 360°C
A. 0.44°C
B. 22.5°C
E. 444°C
C. 225°C
23. To which one of these reactions occurring at 25°C does the symbol
A. H(g) + N(g) + O3(g) → HNO3(l)
B. ½ H2(g) + ½ N2(g) + 3/2 O2(g) → HNO3(l)
C. HNO3(l) → ½ H2(g) + ½ N2(g) + 3/2 O2(g)
D. HNO3(l) → H(g) + N(g) + 3O(g)
E. H2(g) + N2(g) + O3(g) → HNO3(l)
[HNO3(l)] refer?
24. When 0.560 g of Na(s) reacts with excess F2(g) to form NaF(s), 13.8 kJ of heat is evolved at standard-state conditions. What
) of NaF(s)?
is the standard enthalpy of formation (
A. 24.8 kJ/mol
D. -7.8 kJ/mol
B. 570 kJ/mol
E. -570 kJ/mol
C. -24.8 kJ/mol
25. Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation:
4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g)
Given that
A.
B.
C.
D.
E.
[CO2(g)] = -393.5 kJ/mol and
= -3857 kJ.
[H2O(l)] = -285.8 kJ/mol, calculate the enthalpy of formation of glycine.
-537.2 kJ/mol
-268.2 kJ/mol
2,149 kJ/mol
-3,178 kJ/mol
-964 kJ/mol
26. Given H2(g) + (1/2)O2(g) → H2O(l), ΔH° = -286 kJ/mol, determine the standard enthalpy change for the reaction
A.
B.
C.
D.
E.
ΔH° = -286 kJ/mol
ΔH° = +286 kJ/mol
ΔH° = -572 kJ/mol
ΔH° = +572 kJ/mol
ΔH° = -143 kJ/mol
2H2O(l) → 2H2(g) + O2(g)
27. Pentaborane B5H9(s) burns vigorously in O2 to give B2O3(s) and H2O(l). Calculate
B5H9.
[B2O3(s)] = -1,273.5 kJ/mol
[B5H9(s)] = 73.2 kJ/mol
A.
B.
C.
D.
E.
for the combustion of 1 mol of
[H2O(l)] = -285.8 kJ/mol
-1273.5 kJ/mol
-4,543 kJ/mol
-18,170 kJ/mol
-9,086 kJ/mol
-8,448 kJ/mol
28. According to the first law of thermodynamics:
A. Energy is neither lost nor gained in any energy transformations.
B. Perpetual motion is possible.
C. Energy is conserved in quality but not in quantity.
D. Energy is being created as time passes. We have more energy in the universe now than when time began.
29. Calculate the amount of work done, in joules, when 2.5 mole of H2O vaporizes at 1.0 atm and 25°C. Assume the volume of
liquid H2O is negligible compared to that of vapor. [1 L·atm = 101.3 J]
A. 6,190 kJ
D. 5.66 kJ
E. 518 J
B. 6.19 kJ
C. 61.1 J
30. A gas is allowed to expand, at constant temperature, from a volume of 1.0 L to 10.1 L against an external pressure of 0.50
atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w, and ΔE (Hint: 1 L.atm = 101.3 J)
A. Row 1
B. Row 2
C. Row 3
D. Row 4
E. Row 5
31. A calorimeter
A. is equal to the molar enthalpy of a reaction.
B. is a dieting aid.
C. is an indicator of a spontaneous reaction.
D. is a device used to measure the transfer of heat energy.
E. is useful in measuring the amount of heat released by endothermic reactions.
32. For which of these reactions will the difference between ΔH° and ΔE° be the greatest?
A. 2H2O2(l) → 2H2O(l) + O2(g)
B. CaCO3(s) → CaO(s) +CO2(g)
C. NO(g) + O3(g) → NO2(g) + O2(g)
D. 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
E. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
PART II: SHORT ANSWER (Each short answer question has a 2-point value!!)
33. Molarity is a unit of solution concentration expressed in moles per liter.
34. According to the First Law of Thermodynamics, the total energy of the universe is constant.
35. The quantity of energy required to increase the temperature of one gram of a sample by 1°C is called the specific heat.
36. In an endothermic reaction, heat is transferred from the surroundings to the system.
37. Enthalpy change is equal to heat transfer at constant pressure.
PART III: CONCEPTS
38. (3 pts) What is T in the table below?
Based on V1/T1 = V2/T2
182 K
39. (4 pts) Below is given a process that results in a change of both volume and temperature.
a.
Has any work been done? If so, is the work positive or negative with respect to the
system?
There is a change in volume so work was done. Since the final volume is greater than
the initial, the system has expanded against the surroundings and wsystem < 0 (i.e. the
system does work on the surroundings)
b. Has there been an enthalpy change? If so, is the reaction exothermic or endothermic?
There is a change of temperature, so heat transfer occurred as well. Since Tfinal >
Tinitial, q > 0. Hence the process is endothermic.
40. (4 pts) A reaction is carried out in a cylinder, fitted with a movable piston (see below). The starting volume is 5.00 L. At
constant pressure of 1 atm the changes in enthalpy and energy are: ΔH = -35.0 kJ and ΔE = -34.8 kJ. Does the volume
increase, decrease or remain the same?
Solution: We know that
ΔE = ΔH + w = ΔH - pΔV
pΔV = ΔH - ΔE
pΔV < 0
But the pressure is constant and p > 0. Therefore ΔV < 0.
ΔV = Vf – Vi < 0
V f < Vi
i.e. the volume decreased
PART V: CALCULATION PROBLEMS (Show your work in its entirety. Do not provide just a single number!).
41. (5 pts) When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00 oC to 14.14 oC. The
heat capacity of the calorimeter is 682 J/oC. Write the complete thermochemical equation for the process:
-
K+ (aq) + NO3 (aq)
KNO3 (s)
Solution: It is based on the fact that qsoln = - qcal
qcal = (heat capacity)calorimeter x ΔT = 682 x (- 10.86 oC) = - 7407 J
Therefore qsoln = + 7407 J
This is the amount of heat absorbed by 21.45 g x 1 mol/101.11 g = 0.2121 mol of KNO3
Per mole: 7407 J/0.2121 mol KNO3 = 34922 J/1 mol KNO3 = 34.92 kJ/1 mol KNO3
-
K+ (aq) + NO3 (aq)
KNO3 (s)
ΔHo = + 34.92 kJ
42. (5 pts) Determine the standard enthalpy change for the process:
C2H4 (g) + 6 HCl (g)
2 CHCl3 (g) + 4 H2 (g)
using the information given below:
2 C (s) + 2 H2 (g)
C2H4 (g)
ΔHo = 52.3 kJ
H2 (g) + Cl2 (g)
2 HCl (g)
ΔHo = -184.6 kJ
C (s) + ½ H2 (g) + 3/2 Cl2 (g)
ΔHo = -103.1 kJ
CHCl3 (g)
Solution: We begin by identifying reactants and products in the equation in question and also where they appear in the set
of equations. The following changes/manipulations are necessary:
Equation 1: Reverse:
C2H4 (g)
2 C (s) + 2 H2 (g)
ΔHo = - 52.3 kJ
3 H2 (g) + 3 Cl2 (g)
ΔHo = +553.8 kJ
Equation 2: Reverse and multiply by 3:
6 HCl (g)
Equation 3: Multiply by 2:
2 C (s) + H2 (g) + 3 Cl2 (g)
2 CHCl3 (g)
ΔHo = -206.2 kJ
Inspection now shows that adding the three new equations gives precisely the process under study. Therefore the summing
of the enthalpy changes must give us the enthalpy change for the above process: ΔHo = -52.3 + 553.8 – 206.2 = + 295.3 kJ
43. (5 pts) There is a power plant in Portland, Oregon that is very concerned about global warming. This plant takes all of its
exhaust gases from its boilers and recycles the CO2 using the Solvay process to make sodium hydrogen carbonate. The
reaction is shown below.
NH3(g) + H2O(l) + CO2(g) + NaCl(aq) → NaHCO3(aq) + NH4Cl(aq)
How many liters each of NH3 and CO2 (both at STP) are needed to make 3.00 kg of sodium bicarbonate?
Solution: At STP the volume occupied by 1 mol of gas is 22.4 L (standard molar volume). All we need therefore is to find
the # mol of NH3 and CO2:
3.00 kg NaHCO3 x 1000 g/1 kg x 1 mol NaHCO3/84.0 g NaHCO3 x 1 mol NH3/1 mol NaHCO3 = 35.7 mol NH3
3.00 kg NaHCO3 x 1000 g/1 kg x 1 mol NaHCO3/84.0 g NaHCO3 x 1 mol CO2/1 mol NaHCO3 = 35.7 mol CO2
Volume of NH3: 35.7 mol x 22.4 L/mol = 800 L NH3
Volume of CO2: 35.7 mol x 22.4 L/mol = 800 L CO2
Note: The volume amounts can also be found by applying the ideal gas law, once we find the # mol of each gas.
44. (3 pts) BONUS PROBLEM (In order to receive credit, you have to solve the problem completely!).
A mixture was made form the following: 25.0 g of water at 15.0 oC, 45.0 g of water at 50.0 oC, and 15.0 g of water at 37.0
o
C. What is the final temperature of the mixture?
Solution: It is based on the fact that: q1 + q2 + q3 = 0, due to the law of conservation of energy
c x m1 x (ΔT)1 + c x m2 x (ΔT)2 + c x m3 x (ΔT)3 = 0
m1 x (ΔT)1 + m2 x (ΔT)2 + m3 x (ΔT)3 = 0
m1 x (Tf – T1i) + m2 x (Tf – T2i) + m3 x (Tf – T3i) = 0
(m1 + m2 + m3) x Tf = m1 x T1i + m2 x T2i + m3 x T3i
Tf = (m1 x T1i + m2 x T2i + m3 x T3i) / (m1 + m2 + m3)
Tf = (25.0 x 15.0 + 45.0 x 50.0 + 15.0 x 37.0) / (25.0 + 45.0 + 15.0) = 37.4 oC
Tf = 37.4 oC