# Nafshun Worksheet 1 Notes

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```							Chemistry 121S                        Summer 2006            Oregon State University
Worksheet 1 Notes                                            Dr. Richard Nafshun

(13) 2. For each system below, give the signs (positive or negative) for w, q, and E:

E      =       q       +       w
Energy         heat            work

(A)    A system does 10 kJ of work and gives off 220 kJ of heat.

E = (-220 kJ) + (-10 kJ) = -230 kJ

The heat (-220 kJ) is negative because the system is giving off heat (exothermic).
The system now has 220 kJ less energy than it had before.

The work (-10 kJ) is negative because the system did work (it no longer has the
ability to do that work—it give it away).

E is calculated to be negative (although a calculation was not needed—both the
heat and work were negative leading to E being negative).
(B)    A system does 50 kJ of work and absorbs 65 kJ of heat.

E = (+65 kJ) + (-50 kJ) = +15 kJ

The heat (+65 kJ) is positive because the system is received heat (endothermic).
The system now has 65 kJ more energy than it had before.

The work (-50 kJ) is negative because the system did work (it no longer has the
ability to do that work—it give it away).

E is calculated to be positive (the system gained energy overall because it gained
a greater amount of heat than the work it output).

(C)    A system does 10 kJ of work and gives off 220 kJ of heat.
[Accidentally the same question as Part A].

(D)    A system has 120 kJ of work done on it and gives off 50 kJ of heat.

E = (-50 kJ) + (+120 kJ) = +70 kJ

The heat (-50 kJ) is negative because the system is giving off heat (exothermic).
The system now has 50 kJ less energy than it had before.

The work (+120 kJ) is positive because the system received work (it now has the
ability to do that work—it received work and can give it away later).
E is calculated to be positive (the system gained energy).

(E)   A system has 70 kJ of work and gives off 85 kJ of heat.

I took this question to mean that the system has 70 kJ of work done on it (it
should be worded a bit clearer)

E = (-85 kJ) + (+70 kJ) = -15 kJ

The heat (-85 kJ) is negative because the system is giving off heat (exothermic).

The work (+70 kJ) is positive because the system received work (it now has the
ability to do that work—it received work and can give it away later).

E is calculated to be negative (the system lost energy).

(F)   A system does 75 kJ of work and gives off 75 kJ of heat.

E = (-75 kJ) + (-75 kJ) = -150 kJ

(G)   A system cools down while expanding.

E = (-) + (-) = (-)

q is negative because the system cools down—it gives off energy to become
cooler.

w is negative because the system expands (does work).

(H)   A system heats up a little while expanding a lot.

E = (+) + (-) = (-)

q is positive because the system heats up—it receives energy to become hotter.
Endothermic.

w is negative because the system expands (does work).

E is negative because the (+) is small and the (-) is large.
(I)    A endothermic reaction does not expand.

E = (+) + 0 = (+)

q is positive because the system heats up—it receives energy to become hotter.
Endothermic.

w is zero (no work is done—either on the system or by the system).

E is positive because q is (+) and w = 0.

(14) 4. The heat of formation of urea, CO(NH2)2 (s) is –333.19 kJ/mol. Write the
chemical equation associated with this reaction (include phases—s, l, g, or aq).

C (graphite) + 2 H2 (g) + ½ O2 (g) + N2 (g) → CO(NH2)2 (s) ΔHºf = –333.19 kJ

5.     (Table 6.2 will be helpful) How much energy is given off when 8.00 moles of
CO (g) is formed from the elements (at 25 ˚C and 1 atm).

C (graphite) + ½ O2 (g) → CO (g) ΔHºf = –110.5 kJ

110.5 kJ of energy is given off when one mole of CO (g) is formed. So, when 8
moles of CO (g) is formed:

  110 .5kJ 
8.00 moles CO              = -884.0 kJ
 1 mole CO 

884.0 kJ of heat energy is given off when 8 moles of CO is formed.

How much energy is given off when 500.0 g CO (g) is formed from the elements
(at 25 ˚C and 1 atm).

 1moleCO 
 28.01g  = 17.85 moles CO (g)
500.0 g CO          
         

  110 .5kJ 
17.85 moles CO              = -1973 kJ
 1 mole CO 

1973 kJ of heat energy is given off when 500.0 g of CO is formed.

6.     Consider:
2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (l) ΔH˚reaction = -10900 kJ
How much energy is released when one mole of octane, C8H18, is combusted?
10900 kJ of heat energy is released when two moles of octane are combusted (this
is what the balanced equation says).

When one mole of octane is combusted:

     10900kJ    
 2 moles C H (l)  = -5450 kJ
1 mole                  
          8 18   

5450 kJ of heat energy are released.

What is ΔH˚ when 9 moles of water are formed?

  10900kJ 
 18 moles H O (l)  = -5450 kJ
9 moles H2O (l)                   
           2      

7.   (Table 6.2 will be helpful) Calculate ΔH˚ for the following reaction:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) ΔH˚reaction = ?

CH4 (g)       +            2 O2 (g) →   CO2 (g)       +     2 H2O (l)
ΔHºf = -74.85 kJ/mol              0 kJ/mol     -393.5 kJ/mol       -285.9 kJ/mol

ΔH˚reaction = Σ ΔH˚products - Σ ΔH˚reactants

ΔH˚reaction = [(1mol CO2)(-393.5 kJ/mol) + (2 mol H2O)(-285.9 kJ/mol)] –
[(1 mol CH4)(-74.85 kJ/mol) + (2 mol O2)(0 kJ/mol)]

ΔH˚reaction = [-965.3 kJ] – [-74.85 kJ] = -890.45 kJ

(Exothermic—energy is released; of course it is! This is the combustion of
methane—natural gas. We do this reaction to cook and heat our living spaces.]

8.   (Table 6.2 will be helpful) Calculate ΔH˚ for the following reaction:
C2H2 (g) + 5/2 O2 (g) → 2 CO2 (g) + H2O (l) ΔH˚reaction = ?

C2H2 (g)         +        5/2 O2 (g) → 2 CO2 (g)       +   H2O (l)

ΔHºf = 226.75 kJ/mol              0 kJ/mol     -393.5 kJ/mol       -285.9 kJ/mol

ΔH˚reaction = Σ ΔH˚products - Σ ΔH˚reactants

ΔH˚reaction = [(2 mol CO2)(-393.5 kJ/mol) + (1 mol H2O)(-285.9 kJ/mol)] –
[(1 mol C2H4)( 226.75 kJ/mol) + (5/2 mol O2)(0 kJ/mol)]
ΔH˚reaction = -1300 kJ

(Exothermic—energy is released; of course it is! This is the combustion of
another hydrocarbon; this one is acetylene—used in torches. We do this reaction
to wield and cut holes in metal.]

9.    The heat of formation of NH4Cl (s) is –315.4 kJ/mol. Write the chemical
equation associated with this reaction (include phases—s, l, g, or aq).

½ N2 (g) + 2 H2 (g) + ½ Cl2 (g) + → NH4Cl (s) ΔHºf = –315.4 kJ
10.   From the following heats of reaction

2 SO2(g) + O2(g)  2 SO3(g)             H = – 196 kJ                  (1)
2 S(s) + 3 O2(g)  2 SO3(g)             H = – 790 kJ                  (2)
calculate the heat of reaction for
S(s) + O2(g)  SO2(g)                   H = ? kJ                      (3)

[Flipped equation 1 and divided it by 2; we divided equation 2 by 2. H = -297
kJ for (3)]

11.   Calculate the standard reaction enthalpy for the photosynthesis reaction,
o
6 CO2(g) + 6 H2O(l)  C6H12O6(s) + 6 O2(g) Hrx n = ? kJ
The heat of formation of glucose, C6H12O6(s), is – 1274.5 kJ/mol; other values are listed in the
text.

ΔH˚reaction = Σ ΔH˚f (products) – Σ ΔH˚f (reactacnts)

ΔH˚reaction = [(1 mol C6H12O6(s))(-1274.5 kJ/mol) + (6 mol O2 (g))(0 kJ/mol)] –
[(6 mol CO2(g))(-393.5 kJ/mol) + (6 mol H2O(l))(-285.8 kJ/mol)]
12.    Given the following reactions and their enthalpy changes, calculate the enthalpy
change for

2 NO2(g) ----> N2O4(g)

Equations                                   Change in Heat Energy
N2(g) + 2 O2(g) ---> 2 NO2(g)                       67.8 kJ
N2(g) + 2 O2(g) ---> N2O4(g)                        9.67 kJ

Leave the bottom reaction and flip the first:

N2(g) + 2 O2(g) ---> N2O4(g)
2 NO2(g) ---> N2(g) + 2 O2(g)

The nitrogen and oxygen cancel leaving:

2 NO2(g) ----> N2O4(g)          ΔH˚reaction = (-67.8 kJ) + (9.67 kJ) = -58.13 kJ

(16) 1. What is meant by "quanta?"

Packets of energy.

2.     Sketch the entire electromagnetic (EM) spectrum with the highest energy on the
left. Label each region; gamma, x-ray, ultraviolet (UV), visible etc. Label the
highest and lowest energy regions. Label the highest and lowest frequency
regions. Label the longest and shortest wavelength regions. Does it make sense
that UV is to the left of visible? Explain. Which has a longer wavelength, blue or
red light? Which has a higher frequency, blue or red light? Which has a higher
energy, blue or red light?
High Energy                                                  Low Energy
High Frequency                                               Low Frequency
Short Wavelength                                             Long Wavelength

Red light is to the right of blue. Red is lower in energy, lower in frequency, and
has a longer wavelength.

3.   What is meant by "ROY G BIV?"

Red, Orange, Yellow, Green, Blue, Indigo, Violet.
4.   Why is there a demand for blue lasers? What color is the laser in a CD player?
Which one (a blue or a red laser) has a shorter wavelength? Which one (a blue or
a red laser) has a greater energy?

The current lasers in CD players are red or IR. Blue lasers are higher in energy
and shorter wavelength. Blue lasers enable more data access (a CD will hold

5.   What is the color of light that has a wavelength of 532 nm? What is the
frequency of this light? What is the energy of one photon of this frequency?
What is the energy of one mole of photons of this frequency?

532 nm looks to be in the green visible region.

ν = c/λ = (3.00 x 108 m/s)/(532 x 10-9 m) = 5.64 x1014 1/s or Hz

E = hν = (6.626 x 10-34 J·s)(5.64 x1014 1/s) = 3.74 x 10-19 J (per photon)
[Note: When an electron "falls" from one energy level to another electromagnetic
radiation is emitted. One electron "falling" corresponds to one photon.]

 6.02 x10 23 photons 
3.74 x 10-19 J/photon 
                      = 224935 J/mol or 225 kJ/mol

       1mole         

6.   What is meant by the ultraviolet radiation regions UVA (315-400 nm), UVB
(290-315 nm), and UVC (100-290 nm)? Which is most dangerous? Why?

UVC has the shortest wavelength; it is the most dangerous because it is the
highest in energy.

UVA has the longer wavelength of the three, it sits nearest violet light.

7.   What is the wavelength, in nanometers, of light that has an energy content of 508
kJ/mol photons. In what portion of the electromagnetic spectrum will this light be
found? What portion of the electromagnetic spectrum lies to the right? Is this
portion higher or lower in energy?

E = hc/λ (this energy corresponds to the energy of one photon; the energy given in
this problem is for one mole of photons so we will change this after we change the
units from kJ to J).

 1000 J     1mol photons 
508 kJ/mol photons            
 6.02 x10 23 photons  = 8.44 x 10 J/photon
-19
 1kJ 

                     
E = hc/λ

Rearranged: λ = hc/E

           34              8 m 
 (6.626 x10 J  s )(3.00 x10 ) 
λ=                               s  = 236 x 10-9 m or 236 nm
                   -19
8.44 x 10 J           
                                
                                

From the EM figure above, this appears to be in the UV region. Question #6
above confirms this and pinpoints this emission in the UVC region. The region
that lies to the right of UV is visible (it is lower in energy than UV).

8.   What is the wavelength (in meters) of the FM signal broadcast from KBVR radio
at a frequency of 88.7 MHz (88.7 x 106 Hz). Is this long or short compared to the
wavelength of the color you determined in Question #5?

ν = c/λ

Rearranged, λ = c/ν = (3.00 x 10-8 m/s)/(88.7 x 106 1/s) = 3.38 m (Much longer
wavelength than visible light.)

9.   [CONCEPT ON FINAL EXAM, BUT THE
CALCULATION IS NOT ON FINAL EXAM!]
Given that the energy level for the n = 1 level in the hydrogen spectrum is
-1312 kJ/mol, calculate the energy for the n = 2, n = 3, n = 4, n = 5, and n = ∞
levels.
         kJ 
1           1312      
Recall, E is proportional to 2 . E =           mol 
n              n2      
             
             
n     Energy (kJ/mol)

1      -1312.00
2      -328.00
3      -145.78
4      -82.00
5      -52.48
6      -36.44
7      -26.78
8      -20.50
9      -16.20
10     -13.12

        kJ               kJ 
  1312            1312     
E1 =         mol  =           mol  = -1312 kJ/mol
     n2               12     
                             
                             

        kJ               kJ 
  1312            1312     
E2 =         mol  =           mol  = -328 kJ/mol
     n2               22     
                             
                             
        kJ               kJ 
  1312            1312     
E3 =         mol  =           mol  = -146 kJ/mol
     n2               32     
                             
                             

        kJ               kJ 
  1312            1312     
E4 =         mol  =           mol  = -82 kJ/mol
     n2               42     
                             
                             

        kJ               kJ 
  1312            1312     
E5 =         mol  =           mol  = -52 kJ/mol
     n2               52     
                             
                             

        kJ               kJ 
  1312            1312     
E∞ =         mol  =           mol  = 0 kJ/mol
     n2               2     
                             
                             

Do these values match those in the table above? They should!

10.   Which hydrogen atom has absorbed more energy; one in which the electron
moved from the first to the third energy levels or one in which the electron moved
from the second to the fourth energy levels? Explain.

First to the third energy level. Look at the energy diagram above. The gap
between n = 1 and n = 3 is huge compared to n = 2 to n = 4. The energy
relationship is:

[CONCEPT ON FINAL EXAM, BUT THE
CALCULATION IS NOT ON FINAL EXAM!]
        kJ 
  1312     
E=         mol  , so the energy is proportional to 1/n2.
     n2     
            
            
11.   In the hydrogen atom, how much energy is required to remove one mole of
electrons from the n = 1 energy level? How much energy is required to remove
one mole of electrons from the n = 5 energy level? How much energy is required
to excite one mole of electrons from n = 1 to n = 3?

[CONCEPT ON FINAL EXAM, BUT THE
CALCULATION IS NOT ON FINAL EXAM!]
        kJ               kJ 
  1312            1312     
E1 =         mol  =           mol  = -1312 kJ/mol
     n2               12     
                             
                             

so, it takes 1312 kJ to remove one mole of electrons from the n = 1 energy level.

        kJ               kJ 
  1312            1312     
E5 =         mol  =           mol  = -52 kJ/mol
     n2               52     
                             
                             

so, it takes 52 kJ to remove one mole of electrons from the n = 5 energy level.

The energy required to excite one mole of electrons from n = 1 to n = 3 is:

E3 – E1 =

        kJ           kJ 
  1312        1312     
        mol  -       mol  = 1166 kJ/mol
     32           12     
                         
                         

so, it takes 1166 kJ to excite one mole of electrons from n = 1 to n = 3.

12.   How much energy is released from one mole of electrons when they relax from
n = 3 to n = 2? What is the wavelength and color of this emission?
[CONCEPT ON FINAL EXAM, BUT THE
CALCULATION IS NOT ON FINAL EXAM!]

E2 – E3 =

        kJ           kJ 
  1312        1312     
        mol  -       mol  = -182 kJ/mol
     22           32     
                         
                         

so, 182 kJ of energy is released when one mole of electrons "falls" from n = 3 to n
= 2.

[THIS CALCULATION IS GOOD TO KNOW
FOR THE FINAL EXAM]
E = hc/λ (this energy corresponds to the energy of one photon; the energy
calculated in this problem is for one mole of photons so we will change this after
we change the units from kJ to J).

 1000 J     1mol photons 
182 kJ/mol photons            
 6.02 x10 23 photons  = 3.02 x 10 J/photon
-19

 1kJ 

                     

E = hc/λ

Rearranged: λ = hc/E

           34              8 m 
 (6.626 x10 J  s )(3.00 x10 ) 
λ=                               s  = 657 x 10-9 m or 657 nm
          3.02 x 10 -19 J       
                                
                                

From the EM figure above, this appears to be in the red region of the visible
spectrum (this is hydrogen's red line; listed as 656 nm in the Bohr hydrogen atom
figure on the handout.)

```
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