Oct. 24 2003                                       NAME____________________________

 

 

Chemistry 101 Section 1

HOUR EXAM #2

12:10 – 1:30

 

IMPORTANT NOTE--For partial credit be sure to show your reasoning.  Please attempt all parts of all problems--no partial credit can be given for blank spaces.   If you need more paper, write on the back of the page and indicate clearly that you have done so. 

 

 

 

There will be no partial credit on parts of  Problem 1.

 

 

(Possibly Useful )Physical Constants and Equations

NA = 6.022 x 1023                                         Mass of an electron 9.1 x 10-28 g

Mass of a proton      1.67 x 10-24 g                h = 6.63 x 10 -34 J.s/photon

Mass of a neutron    1.67 x 10-24 g                c = 3.0 x 10 8 m/s

 

E = hn             E= hc/l           273 K = 0 °C.  P V = n R T

Visible light 400 nm to 900 nm          UV light 200 to 400 nm

Density of water 1 g/cm3       R = 0.082 L atm / K mol = 8.314 J / K mol

 

Specific heat of water = 4.18 J / g °C  1.00 atm = 760 mm Hg

 

Approximate grades:

> 90 4.0

  75-89           3.0, 3.3, 3.7

50-74             2.0, 2.3,  2.7

 

Congratulations if you improved your score!

 

If your score stayed the same or declined please consider:

 

1)This material may be more unfamiliar to you than the first few chapters

2)Are you studying actively for at least 2 hours for every hour in class?

3)Are you doing HW problems until you master the topic?

4)Are you making good use of office hours and PLI sessions?

5)In addition to solo study consider having a “study buddy” or a “study group”.  Explaining your reasoning to a classmate is excellent practice whether you are a strong or weak student.

 

Problem 1 (20 points)

 

Indicate whether the quantity in Column A is greater than (>), equal to (=), or less than (<) the quantity in column B.  You may write the words, the symbols or both.  No partial credit will be awarded on the parts of this Problem.  Part (a) is done for you as an example.

 

Column A                                                    Column B

(a) Planck’s constant

 

is less than

        <

Avogadro’s number

 

 

 

(b) The pressure of 1 mole of “acetic acid vapor” at STP in 10 L if the gas behaves ideally.

 

     >

The dimers mean that there are  fewer gas molecules and therefore less pressure.

The molecules interact strongly with each other 

The pressure of 1 mole of “acetic acid vapor” at STP in 10 L if the gas behaves as a real gas and 10% of the acetic acid molecules dimerize. (A + A à A2 is a dimerization reaction)

 

 

 

(c) The energy of a photon whose wavelength is 450 nm.

 

 <

Long wavelengths mean less energy

The energy of a photon whose wavelength is 290 nm.

 

 

 

(d D H for an exothermic reaction

 

 

 <

negative value < positive value

D H for an endothermic reaction

 

 

 

(e) At STP, the rms speed of a molecule of 235 UF6 gas.

 

>  the heavier molecule goes more slowly

At STP, the rms speed of a molecule of 237 UF6 gas.

 

 

 

(f) The absolute value of the work done by a piston in expanding against a vacuum ( 0 kg weight).

 

<  zero vs a positive quantity (absolute value)

 The absolute value of the work done by the same piston in expanding against a 2 kg. weight.

 

Problem 2 (20 points)

 The CO2 that builds up in the air of a submerged submarine can be removed by reacting it with sodium peroxide:

 

2 Na2O2 (s)  + 2 CO2 (g) à 2 Na2CO3 (s) + O2 (g)

 

If a sailor exhales 110 mL of CO2 per minute at 25 °C and 1.02 atm, how much sodium peroxide (in grams) is needed per sailor in a 24-hr period?

 

This was a problem from recitation with the numbers  changed.

 

Step 1:  Calculate the number of moles using PV = nRT

 

So n CO2 = P V / R T =  (1.02 atm ) ( 0.110 L) / (.082 L atm/K mol (273+25)

 = 4.59 x 10-3 moles

 

Step 2: There are  4.59 x 10-3  moles of Na2O2

 

Step 3: find the mass of of Na2O2 is needed in 24 hours

 4.59 x 10-3  x 78 g/mol x 60 min/hr x 24 h = 512 g

 

Homework level of difficulty

 

Problem 3 (20 points)

The standard heat of formation of NO2 is + 33.18 kJ/mole. 10.62 grams N2 and 32.1 grams of O2 were mixed in the reaction chamber of a supersensitive, well-insulated calorimeter whose total heat capacity is 9.545 kJ/°C.  After establishing an initial baseline temperature reading of 23.52°C, the NO2-forming reaction was started using an electronic ignition system.  Assuming that the only reaction which took place was the formation of NO2 from N2 and O2 , find the final temperature of the calorimeter.

 

This was hard!  But it’s made of little parts that you should be able to do.

 

Step 1: find the number of moles of each reactant and decide which one is limiting.

 

1/2N2 (g) + O2 (g) à NO2 (g) DH° = + 33.18 kj /mol

 

10.62 g x 1 mol/28.01 g = 0.379 moles N2  Limiting reagent

 

32.01 g x 1 mol/ 32 g = 1 mol O2

 

so 2 x 0.379 = 0.758 moles of NO2 are produced

 

Step 2

D H for the reaction = + 33.18 kJ/ mol x 0.758 moles = 25.18 kJ so q for surroundings is –25.18 kJ

Step 3

The calorimeter’s capacity is + 9.545 kJ/ °C so q = C DT

 

DT = q / C = -25.18 kJ/ 9.545  kJ/ °C = -2.64 °C

so the final temperature is 23.52- 2.64 = 20.88 °C

 

Draw an energy level diagram for the system and the surroundings and clearly identify each part.

 

The reaction is endothermic so the calorimeters temperature will drop as the reaction absorbs heat from the surroundings.

 

 


system                                     surroundings lose energy

 

Energy

 

 

Short Answers: (5 points each)—Basic to HW level.   One step problems that require one formula.

 

1)         What is a State Function?

 

Function whose value depends only on the initial and final values and not on the path take.  Examples are V, P, T, E, H

 

 

2)         If 5.00 mL of a 2.00 M solution of NaCl is mixed with 95.0 mL of water the final concentration of diluted NaCl is ____0.100 M_______.

Show your work.

 

5.00 mL / 100.0 mL x 2.0 M = .100 M

 

3)         One mole of blue photons having a wavelength of 450 nm has an energy of____266______kJ.

 

Show your work.

 

E = (h x c / l) x Na

= 6.63 x 10 -34 J s / photon x 3 x 10 8 m/s x 6.02 x 10 23 photon/mol / 450 x 10 –9 m

 

= 266, 000 J or 266 kJ

 

4)         The amount of heat required to heat 26.3 grams of water from 20.0 °C to 85.7 °C is _____7.22____ kJ.

 

 

Show your work.  Q = m s DT = 26.3 g x 4.18 J/g °C (85.7-20 °C) = 7222 J

Problem 5 (20 points)

 

Some baseball fans from Florida may be heartened to learn that baseballs travel further in the hot, humid south Florida air as opposed to the cooler, dryer air of the Bronx, New York.  Therefore, if the baseball stadiums were identical, we might expect more home runs in Florida.

 

a)          Calculate the density of Bronx air ( in grams per Liter) at 17 ° C assuming that the molar mass of an average “air” molecule is 28.8 g /mol and the pressure is 760 mm Hg.

 

 

Homework to slightly hard—many of you correctly derived the needed  information

 

 

D = grams / liter  We already know that one mole is 28.8 g so we need to find the volume of one mole:  V = n RT / P = 1 mole x 0.802 L atm/K mol x (17+273 K) / 1 atm

= 23.78 L

 

D = 28.8 g / 23.78 L = 1.21 g /L

 

 

 

 

b)         Calculate the density of South Florida air (in grams per Liter) at 30 °C assuming that the molar mass of an average “90 % relative humidity wet air molecule” is 28. 4 g / mol and the pressure in 760 mm Hg.

 

Same method: V = 1 mol (.082 L atm/K mol) (273 + 30 K) / 1 atm = 24.85 L

 

D= 28.4 g / 24.85 L = 1.14 g /L

 

 

c)          If you wanted to hit more home runs where should you go to find the lowest density air?  (Hint: in the Ideal Gas Equation n / V is directly proportional to the density and is equal to what?)

 

Density a n /V = P /RT  so for low densities we can increase the temperature (Florida) or reduce the pressure (Denver).  The pressure is actually the more important factor and more home runs are hit in Denver due to the high altitude and low pressure.