Exam 2 C Chem 111, Section 2 (Martin, 1:25pm) Spring 1998

h = 6.626 x 10-34 J s

c = 2.998 x 108 m / s

h c = 1.986 x 10-25 J m

No = 6.022 x 1023 "particles" / mol

In the questions below, "MW" refers to molecular weight (molecular mass)

1c. You are manufacturing expensive, hand-built automobiles. You currently have 8 cars ready for delivery to customers, except that they lack tires (each car requires 4 tires and one "skinny" spare). You have in stock 28 regular tires and 9 skinny spares. How many complete cars can you deliver without ordering more tires?

(a) 5 (b) 6 (c•) 7 (d) 8 (e) 9

Limiting "reagent" - tires. Can only build 28/4=7 cars

2c. Hydrogen and oxygen can be used as a fuel, according to the following reaction:

2 H2 (g) + O2 (g) Æ 2 H2O (l) 食 = -483.6 kJ/mol

This reaction is:

(a) endothermic (b•) exothermic

3c. How many moles of O2 are required to completely react with 6 moles of H2?

(a) 24 moles (b) 12 moles (c) 1 mole (d•) 3 moles (e) 6 moles

Require 1 mol O2 per 2 mol H2

4c. In the laboratory, you mix 10.08 g H2 (MW=2.02 g/mol) with 16.0 g O2 (MW=32.0 g/mol) and provide a spark to ignite the reaction. Assuming that the reaction goes to completion, what mass of water (H2O, MW=18.0 g/mol) is produced?

(a) 10.1 g (b) 16.0 g (c•) 18.0 g (d) 36.0 g (e) 26.1 g

We're mixing 5 mol H2 with 0.5 mol O2. O2 is limiting. So we can make (0.5 x 2) mol H2O.

Mass = 1.0 mol x 18.02 g /mol = 18.0 g water.

 

 

 

5c. In the laboratory, you mix 4.032 g H2 (MW=2.016 g/mol) with an excess of O2 (MW=32.0 g/mol) and provide a spark to ignite the reaction. Assuming that the reaction goes to completion, what amount of heat is produced?

(a) 241.8 kJ (b•) 483.6 kJ (c) 967.2 kJ (d) 1,934 kJ (e) 2.016 kJ

The limiting reagent is 2.0 mol H2. From the above reaction, combustion of 2 mols of H2 releases 483.6 kJ.

 

 

 

6c. A small pool of water is sitting on your arm on a hot, dry day. If the water is considered the system and your arm the surroundings, which best describes the spontaneous process which occurs?

(a) no heat is transferred

(b•) heat is transferred from the surroundings to the system

my arm cools down

(c) heat is transferred from the system to the surroundings

my arm heats up

(d) heat is transferred from the system to the surroundings

my arm cools down

(e) heat is transferred from the surroundings to the system

my arm heats up

7c. A 0.14 g sample of a hydrocarbon is burned in oxygen to give 0.44 g CO2 (MW=44.0 g/mol) and 0.18 g H2O (MW=18.0 g/mol). Which molecular formula is most consistent with this result?

(a) C2H2 (b•) C2H4 (c) C2H6 (d) C2H2O (e) C2H4O

Moles C = 0.44 g x (mol/44.0 g) = 0.01 mol

Moles H = 2 x (0.18 g x (mol/18.0 g)) = 0.02 mol

Ratio H/C = 0.02/0.01 = 2.0. Empirical formula is CH2. C2H4 fits that.

If C2H4 then MW=28.05 g/mol. To get 0.01 mol C, 0.005 mol C2H4 would have reacted.

The mass of C2H4 would then be (28.05 g/mol x 0.005 mol) = 0.14 g <- Observed

If C2H4O then MW=44.05 g/mol. To get 0.01 mol C, 0.005 mol C2H6O would have reacted.

The mass of C2H6O would then be (44.05 g/mol x 0.005 mol) = 0.22 g

 

8c. A 1.0 g piece of aluminum at 20.0°C is dropped into a beaker of water. The temperature of the water drops from 50° to 30° C. What quantity of heat did the piece of aluminum absorb? Note that the specific heat of aluminum is 0.902 J/gK, while that of water is 4.184 J/gK.

(a) 18 J (b) 4.2 J (c•) 9.0 J (d) 8.4 J (e) 16.8 J

The aluminum went from 20° to 30° C. Heat = (0.902 J/gK) x (1.0 g) x (10 K) = 9.0 J

9c. 食vap for ethanol (C2H5OH, MW=46.07 g/mol) at its boiling point is 38.56 kJ/mol.
食vap for water (H2O, MW=18.02 g/mol) at its boiling point is 40.73 kJ/mol.
Which statement below is true?

(a) Condensing 4.6 g ethanol requires more heat than condensing 3.6 g water

(b) Condensing 4.6 g ethanol requires more heat than condensing 3.6 g water

(c) Boiling 4.6 g ethanol requires more heat input than boiling 3.6 g water

(d•) Boiling 4.6 g ethanol requires less heat than boiling 3.6 g water

(e) Condensing 4.6 g ethanol releases more heat than condensing 3.6 g water

The table below lists some standard enthalpies of formation at 298 K

Substance 食f°

(kJ/mol)

CO (g) -110.5

CO2 (g) -393.5

CH3OH (l) -238.7

H2O (l) -285.8

10c. What is 食° for the following reaction at 298 K:

CH3OH (l) + O2 (g) Æ CO (g) + 2 H2O (l)

(a) 157.6 kJ/mol (b•) -443.4 kJ/mol (c) -157.6 kJ/mol

(d) 443.4 kJ/mol (e) insufficent information to calculate

食° = [ -110.5 + 2(-285.8) ] - [ -238.7 + 0 ] = -886.8 kJ/mol

 

11c. Both hydrogen (H2) and ethanol (C2H5OH) have been proposed as alternate fuels for automobiles. 食° for the complete combustion of H2 is -285.8 kJ/mol, while that for ethanol is -1,367 kJ/mol. Remembering that weight contributes to overall fuel efficiency, combustion of which fuel produces more heat per gram?

(a•) hydrogen produces more heat per gram

(b) ethanol produces more heat per gram

H2 : heat per gram = -285.8 kJ/mol x (mol/2.016g) = 141.8 kJ/g

C2H5OH : heat per gram = -1,367 kJ/mol x (mol/46.07g) = 29.7 kJ/g

 

 

 

12c. Iron (III) oxide ("rust") can be produced from iron and oxygen in a sequence of reactions that can be written as

2 Fe (s) + 6 H2O (l) Æ 2 Fe(OH)3 (s) + 3 H2 (g) +68.8 kJ/mol

2 Fe(OH)3 (s) Æ Fe2O3 (s) + 3 H2O (l) -35.6 kJ/mol

3 H2 (g) + 3/2 O2 (g) Æ 3 H2O (l) -857.4 kJ/mol

What is 食° for the reaction

2 Fe (s) + 3/2 O2 (g) Æ Fe2O3 (s)

(a) 1,648 kJ/mol (b•) -824.2 kJ/mol (c) -1,682 kJ/mol

(d) -1,648 kJ/mol (e) 824.2 kJ/mol

Correct balancing requires 1 of first equation, 1 of second equation, and 1 of third equation

食° = 1(+68.6) + 1(-35.6) + 1(-857.4) = -824.2 kJ/mol

 

 

 

13c. A given orbital is labeled by the magnetic quantum number ml = -1. This could not be a

(a) g orbital (b) p orbital (c) d orbital (d) f orbital (e•) s orbital

14c. A given orbital has quantum numbers n =3, l = 1, ml = 0. This orbital must be:

(a•) 3p (b) 4d (c) 3d (d) 3s (e) 4p

15c. For the hydrogen atom, which of the following transitions produces an emitted photon with the highest frequency?

(a) n = 2 Æ n = 4 1/16 - 1/4 = -0.1875 (absorbed)

(b) n = 1 Æ n = 2 1/4 - 1/1 = -0.75 (absorbed)

(c) n = 4 Æ n = 2 1/4 - 1/16 = +0.1875 (emitted)

(d•) n = 2 Æ n = 1 1/1 - 1/4 = +0.75 (emitted)

(e) none emits a photon

16c. Which of the following photons has the highest energy?

(a) radio waves (l = 0.5 m) (b) orange light (l = 625 nm)

(c) microwaves (l = 10-3 m) (d•) near UV (l = 250 nm)

(e) infrared (l = 1000 nm)

Lowest wavelength -> highest energy

17c. Radiation in the UV region of the electromagnetic spectrum is quite energetic (remember your sun screen!). If you are bombarded by 1 mol of photons with wavelength l = 300 nm, how much energy are you being subjected to?

(a) 6.62 x 10-37 J (b) 6.62 x 10-28 J (c) 3.99 x 10-13 J

(d•) 3.99 x 105 J (e) 6.62 x 10-19 J

E = hc/l = (1.986 x 10-25 J m / 320 nm) x (109 nm/m) x (6.02 x 1023 /mol) = 3.99 x 105 J

 

 

18c. How many nodes does a 3s orbital have?

(a) 0 (b) 1 (c•) 2 (d) 3 (e) 4

19c. Which best describes the node in a 2pz orbital?

(a) spherical (b•) in the xy plane (c) in the yz plane

(d) in the xz plane (e) there is no node

20c. How many orbitals can have the principle quantum number 3?.

(a•) 9 (b) 5 (c) 3 (d) 2 (e) 1

For n=3: l=0, ml=0; l=1, ml=-1,0,1; l=2, ml=-2,-1,0,1,2; So 1 + 3 + 5 = 9

 

Remember:

You have exam C. Place a "3" in column "P" under "Special Codes"

As a homework assignment, you may earn up to 10% of the points you missed on this exam (eg., if you scored a 60 on the exam, you can earn an extra 4 points), by doing the following:

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2) Turn in the Op-Scan sheet in class on Wednesday, April 1 (no later!).

Your revised exam will be scored and credit applied proportional to the total number of questions answered correctly (if you do less well than you did on the exam, points will not be taken off). Complete the exam exactly as you did previously (except with all correct answers, of course!).

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