1. The specific heat of iron is 0.451 J/g·K. When a 200.0 g iron slug is heated to 368^oC and added to 60.0 g of water at 20.0^oC, all of the heat lost by the iron is gained by the water. If the heat of vaporization of water is 2260 J/g, how many grams of water will vaporize?

q_{lost iron} = q_{gained water}

= q_{water warming} + q_{water vaporizing}

(200.0 g)(0.451J/g·K)(268K) = (60.0g)(4.184J/g·K)(80K) + (mass water vaporized)(2260J/g)

mass water vaporized = 1.80 g

2. When gasoline burns in a car engine, the heat released causes the products CO₂ and H₂O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (DE) in kcal.

w = -451 J (negative because energy leaves system)

q = -325 J (negative because heat leaves system)

E = q + w = -451 J + -325 J = -776 J

3. The standard enthalpy of combustion of liquid heptane (C₇H₁₆) is -4816.0 kJ/mol. The products of this combustion are liquid water and carbon dioxide gas. Calculate the standard enthalpy of formation of liquid n-heptane.

C₇H₁₆(g) + 11 O₂(g) ¾® 7 CO₂(g) + 8 H₂O(l)

DH^o_rxn = 8(DH_f^o,H₂O,l) + 7(DH_f^o,CO₂,g) - (DH_f^o,C₇H₁₆,l)

-4816 kJ = 8(-285.5 kJ) + 7(-393.5 kJ) - (DH_f^o,C₇H₁₆,l)

-4816 kJ = -2284 kJ + -2754.5 kJ - (DH_f^o,C₇H₁₆,l)

DH_f^o,C₇H₁₆,l = -222.5 kJ

4. Acetylene reacts with bromine according to the following equation:

C₂H₂ + 2 Br₂ ¾® C₂H₂Br₄

From bond energies, calculate the standard enthalpy change for the reaction.

Bonds Broken Bonds Formed

1 CºC 812 kJ 4 C-Br 4(-285 kJ)

2 Br-Br 2(193 kJ) 1 C-C -346 kJ

1198 kJ -1468 kJ

DHrxn = 1198 kJ (energy required to break bonds) - 1468 kJ (energy released forming bonds)

= -288 kJ

5. Calculate the standard enthalpy of formation of solid magnesium hydroxide from the following data:

2 Mg(s) + O₂(g) ¾® 2 MgO(s) DH^o = -1203.7 kJ

MgO(s) + H₂O(l) ¾® Mg(OH)₂(s) DH^o = -36.7 kJ

2 H₂O(l) ¾® 2 H₂(g) + O₂(g) DH^o = +571.6 kJ

Equation for the formation of Mg(OH)₂ from the elements

Mg(s) + O₂(g) + H₂(g) ¾® Mg(OH)₂(s) DH^o_f = ?????

Mg(s) + ¹/₂ O₂(g) ¾® MgO(s) DH^o = -1203.7/2 kJ = -601.8 kJ

¹/₂ O₂(g) + H₂(g) ¾® H₂O(l) DH^o = (-1)(571.6/2) kJ = -285.8 kJ

MgO(s) + H₂O(l) ¾® Mg(OH)₂(s) DH^o = -36.7 kJ

Mg(s) + O₂(g) + H₂(g) ¾® Mg(OH)₂(s) DH^o_f = -924.3 kJ

6. Ethanol is a major ingredient of a camping fuel called “Sterno”. If the energy released in the combustion of 5.00 g of ethanol (C₂H₆O) is transferred without loss to 200.0 g of liquid water at 22.0^oC, what will be the final state and temperature of the water? (Combustion of one mole of ethanol yields 1235 kJ of energy)

1^st find out how much energy produced by Sterno

2^nd find out how much energy required to heat all water up to 100^oC. If not enough energy then go back and determine final temperature.

Plenty of energy to heat all water to 100^oC. 134 – 65.3 kJ = 69 kJ energy left to begin converting water at 100^oC to steam. Next find out how much energy needed to convert all water to steam.

How much water can 69 kJ of energy evaporate?

So we get 30 g of steam at 100^oC and 200.0 – 30 = 170 g water at 100^oC!

7. Aspirin is produced commercially from salicylic acid, C₇O₃H₆. A large shipment of salicylic acid is contaminated with boric oxide, which like salicylic acid is a white powder. The heat of combustion of salicylic acid at constant volume is known to be –3.00 x 10³ kJ/mol. Boric oxide, because it is fully oxidized, does not burn. When a 3.556 g sample of contaminated salicylic acid is burned in a bomb calorimeter, the temperature increases 2.556^oC. From previous measurements, the heat capacity of the calorimeter is known to be 13.62 kJ/K. What is the amount of boric oxide in the sample, in terms of mass percent?

8. A sample of neon gas has a volume of 3.15L and a pressure of 0.951 atm at 21°C. If the pressure increases to 1.564 atm and temperature remains constant, what is the final volume?

PV = nRT, T and n are constant so P₁V₁ = P₂V₂

P₁ = 0.0951 atm V₁ = 3.15 L

P₂ = 1.564 atm V₂ = ?

9. Calculate the volume of O₂ at 2.89 atm and 15^oC required for the complete combustion of 125 g octane (C₈H₁₈) to CO₂ and H₂O.

2 C₈H₁₈ + 25 O₂ --> 16 CO₂ + 18 H₂O

10. Consider the three boxes, all at 27^oC, in the diagram below. Assuming the connecting tubes have negligible volumes, what is the partial pressure of each gas and the density of the final mixture when both of the stopcocks are opened?

P_total = 6.55 atm

Find total moles of gas from total P, T, and V

Total mass = 756 g Total volume = 38.0 L density = 756 g/38.0 L = 19.9g/L

11. A 0.700 g sample of lithium metal is placed in an evacuated 1.00 L flask connected by a stopcock to a 1.00 L flask containing oxygen at a temperature of 25.0^oC and 5.00 atm pressure. The stopcock is opened and the exothermic reaction allowed to proceed to completion. When the temperature of the system returns to 25.0^oC, what will be the pressure? (assume that the volumes of Li(s) and Li₂O(s) are negligible).

	4 Li (s) + O2(g) ¾® 2 Li2O(s)
initial moles	0.101	0.204	0
change	- 4x (-0.101)	-x (-0.025)	+2x
final moles	0.00	0.178	0.050

12. The gaseous reaction below is followed by monitoring the total gas pressure in the reaction vessel. The temperature is held constant by a large water bath at 100^oC. The initial pressure of CS₂ and H₂ was 0.558 torr. The final pressure was 0.480 torr. The limiting reactant, CS₂ was entirely used up. Using Dalton’s Law of Partial Pressure, calculate the partial pressures of the other gases when the reaction was finished and the partial pressures of the two initial gases.

	Total Pressure	CS2(g) + 4 H2(g) ® CH4(g) + 2 H2S(g)
Initial	0.558 torr			0 torr	0 torr
D	-0.078 torr	-X	-4X	+X	+2X
Final	0.480 torr	0 torr

Pressure change = -0.078 torr = (-X) + (-4X) + (+X) + (+2X) ® ® X = 0.039 torr

	Total Pressure	CS2(g) + 4 H2(g) ® CH4(g) + 2 H2S(g)
Initial	0.558 torr	0.039 torr	(0.558-0.039)torr =0.519 torr	0 torr	0 torr
D	-0.078 torr	-0.039 torr	-4(0.039 torr)	+0.039 torr	+2(0.039 torr)
Final	0.480 torr	0 torr	0.480-(.039+.078)torr =0.363 torr	0.039 torr	0.078 torr

Check final pressure of H₂ H₂ initial - D = H₂ final

0.519 torr - 4(0.039 torr) = 0.363 torr it works!!

13. One of the raw materials needed for the manufacture of sulfuric acid is the oxygen in the air, which is used to convert sulfur to sulfur trioxide. The sulfur trioxide can then be reacted with water to produce sulfuric acid. The reactions are

2 S(s) + 3 O₂(g) à 2 SO₃(g)

SO₃(g) + H₂O(l) à H₂SO₄(l)

What volume of air, measured at 1.10 atm and 500^oC, would be needed to produce 1.000 x 10³ kg of 98.9% sulfuric acid? Air is 20.9% oxygen by volume.

14. If at a temperature T, the average speed of a CO₂ molecule is 3.5 x 10⁴ ms^-1.

a. What would be the average speed of a methane (CH₄) molecule at the same temperature?

rate₂ = 5.8 x 10⁴ m/sec

b. A third gas has an average speed of 3.0 x 10⁴ ms^-1 and is composed of 40. %C, 6.7% H, and 53% O. Determine the molecular formula of this gas.

MW = 60 g/mol (by doing la calculation like one above) mol formula C₂H₄O₂

15. Consider that you have three containers of a gas at the following conditions.

Container A 800 mL CO₂ @ 37^oC and 1.00 atm pressure MW = 44

Container B 500 mL Ne @ 47^oC and 1.80 atm pressure MW = 20

Container C 450 mL Ar @ 37^oC and 1.20 atm pressure MW = 40

Identify

a. The container with molecules possessing the highest kinetic energy. Why?

kinetic energy =3/2 kT \ highest temperature corresponds to highest kinetic energy

Ne has highest kinetic energy

b. The container with the largest number of molecules. Why?

calculate sample with most moles (molecules)

for CO₂ n = 0.0314

Ne n = 0.0343 ÜÜ most molecules

Ar n = 0.0212

c. The container with the largest number of collisions per second. Why?

High speed and high density contribute to a high collision frequency. Ne atoms are moving fastest under the most crowded conditions \Ne has the highest collision frequency

d. The container with the fastest RMS (average) velocity. Why?

\ container with highest value has fastest velocity.

CO₂ = 2.6 Ne =4.0 Ar = 2.8

Ne has the highest velocity

e. The container with molecules possessing the highest average momentum (mass x velocity). Why?

CO₂ momentum µ 114 ÜÜ highest momentum

Ne momentum µ 80

Ar momentum µ 112

f. The container with the densest sample of gas. Why?

mass density = particle density =

CO₂ density_mass = 1.73g/L density_particle = 0.0393 particle/L

Ne density_mass = 1.37g/L density_particle = 0.0686 particle/L

Ar density_mass = 1.88g/L density_particle = 0.0471 particle/L

16. What is the frequency of an X ray with a wavelength of 1.5 x 10^-2 nm? What is the energy, in joules, associated with a photon of this frequency? What would be the energy of a mole of such photons?

17. One way to gain information about the origin and functions of stars within the universe is to study the origin and distribution of the hundreds of gamma ray sources in the sky. If one such source were producing high-energy gamma rays of 1.6 x 10^-8J, what wavelength would astronomers have detected? What is the frequency of this radiation?

18. Taking a photograph of a moving object – for example, a person sprinting—will cause some blurring or the actual person’s position when the photograph is developed. Therefore, the more blur, the better you represent movement. If the entire scene were re-shot using a faster shutter speed, what information would you gain and what would you lose in the photo?

You would get a sharper image showing the position clearly, but the speed would not be evident.

19. When an electron is in the fifth energy level, how many sublevels are possible? How many orbitals are possible?

n=5

l = 4,3,2,1,or 0 so 5 possible sublevels

for l = 4, ml = 4,3,2,1,0,-1,-2,-3,-4 so 9 orbitals

for l = 3, ml = 3,2,1,0,-1,-2,-3 so 7 orbitals

for l = 2, ml = 2,1,0,-1,-2 so 5 orbitals

for l = 1, ml = 1,0,-1 so 3 orbitals

for l = 0, ml = 0 so 1 orbital

total 25 orbitals

20. If the four quantum numbers for an electron were (3,2,1,+½) what would be the four quantum numbers for an electron in the same orbital as the first electron?

(3,2,1,−½)

21. The sequence in each line that follows represents values for the quantum numbers for an electron in a hydrogen atom. Select any sequence(s) that are not possible and explain the problem(s)?

n	l	ml
3	-1	0	Possible
2	+2	+1	Not possible, l must be less than n
3	+2	+3	Not possible ml = l to –l
1	+1	+1	Not possible, n must be greater than l
4	+3	-2	Possible

22. A photon with a wavelength of 415 nm strikes a metal surface ejecting an electron with a velocity of 9.0 x 10⁵ m/sec. What is the threshold frequency of the metal? (kinetic energy = ¹/₂mv², J = kg·m²/sec²)

Energy of excitation

Energy of electron

Threshold Energy

E_threshold = E_excitation- E_electron = (4.79 – 3.69) x 10^-19J = 1.10 x 10^-19 J

frequency

n= E/h = 1.10 x 10^-19J/6.626x10^-34 J sec = 1.66 x 10¹⁴ /sec

23. Draw the orbitals associated with the following quantum numbers.

A 4s orbital

a. n = 4, l = 0, m = 0

A 2p orbital

b. n = 2, l = 1, m = 0

A 3d orbital

c. n = 3, l = 2, m = -2

24. Consider the following levels of a hypothetical atom:

E_∞ 0 J

E₄ -1.0 x 10^-19

E₃ -5.0 x 10^-19

E₂ -10 x 10^-19

E₁ -15 x 10^-19

a. An electron in the E₁ energy level will jump to a higher energy level when excited by light with a wavelength of 190 nm. Calculate the energy of the light beam.

b. Predict the energy level to which the electron will jump upon excitation by 190 nm light.

The electron will jump to the E₃ level.

c. How many different emission lines would you predict as electrons fall from the E₄ state back to the ground state by all possible routes. (Draw the possible transitions on the diagram using arrows.)

Six lines see above