Chem 101 - Homework 1 - DOC by Y7xT503

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									                                                                                  Dr. Guldan
                                                                               Chem. 108 S09

                   Environmental Chemistry - Homework 1
                          Energy - SOLUTIONS

1. Calculate the energy content of methanol (CH3OH) in kJ/mole and kJ/g. How does it
   compare to the fuels listed in Table 2.2?
         2 CH3OH         +       3 O2                2 CO2        +      4 H2 O
               H
               
       HCOH               +    O=O             O=C=O         + HOH
               
               H
       Energy used to break bonds:
       H  C:      (3)(410 kJ/mol)(2 mols) = 2460 kJ
       C  O:      (1)(360 kJ/mol)(2 mols) = 720 kJ
       O  H:      (1)(460 kJ/mol)(2 mols) = 920 kJ
       O = O:      (1)(494 kJ/mol)(3 mols) = 1482 kJ
       Total                                    5582 kJ
       Energy released when bonds are formed:
       C = O:      (2)(799 kJ/mol)(2 mols) = 3196 kJ
       H  O:      (2)(460 kJ/mol)(4 mols) = 3680 kJ
       Total                                    6876 kJ
       Net energy released = 6876 kJ  5582 kJ = 1294 kJ
       Note that this is the energy released when 2 mols of CH3OH are burned. So the
       energy released per mole of methanol is:
       1294 kJ / 2 mols = 647 kJ/mol          (ENERGY CONTENT)
       To find in kJ per g, first determine the molar mass of methanol:
       12.011 g/mol + 4(1.00794g/mol) + 15.9994 g/mol = 32.042 g/mol CH3OH
           647 kJ     1 mol CH 3OH 
                                    20.2        kJ
        1 mol CH OH   32.042 g                         g
                3                  
       On a per-gram basis, methanol is less efficient than all the other fuels listed except
       cellulose. Methanol has a greater energy content than biomass, but has less
       energy per gram than natural gas, petroleum, coal, and ethanol.
                                                                               Dr. Guldan
                                                                            Chem. 108 S09

2. Calculate the CO2 emission (in moles CO2 per 1000 kJ) expected from the
   combustion of methanol. How does that compare to the CO2 emission for the other
   fuels?
   Find how much CH3OH is needed to produce 1000 kJ:
       647 kJ        1000 kJ   
                                              x = 1.5 mols CH3OH
    1 mol CH OH   x mols CH OH 
            3              3   
   So 1.5 moles CH3OH must be burned to yield 1000 kJ. According to the balanced
   chemical reaction, burning 2 mols CH3OH produces 2 mols CO2, so if 1.5 ols of
   CH3OH is burned, 1.5 mols CO2 will be produced (since the ratio is 1:1).
       The CO2 emission is 1.5 mols CO2 per 1000 kJ
   Compared to other fuels, the CO2 emission from methanol are lower than the
   emission from other fuels except natural gas (CH4) and hydrogen (H2).


3. Spiro, p. 134, problem 5.
   The H to C atomic ratio is higher in oil than in coal because oil is derived from
   hydrocarbon based lipids (see Figure 2.2) in marine environments. In hydrocarbons,
   each carbon bonds to four other atoms, which are either hydrogen or carbon atoms.
   The chemical formula for alkanes is CnH2n+2, where n is the number of carbon atoms.
   Thus, there are about 2 hydrogen atoms for each carbon atom. The chemical formula
   for petroleum molecules do not conform exactly to this rule because it is not made up
   of pure alkanes. Nevertheless, the 2 H to 1 C ratio is approximately correct.
   Coal is derived from terrestrial plants, composed mostly of lignin and cellulose. Over
   time, dead plants are enriched in lignin, which is more resistant to bacterial
   decomposition than is cellulose. Under the right geological conditions, lignin is
   transformed into peat, which is a precursor of coal formation. Lignin is a complex
   three-dimensional polymer based on benzene rings. Because of the presence of
   aromatic C=C double bonds in benzene, each carbon is bonded to 2 other carbons and
   1 hydrogen (the chemical formula for benzene is C6H6). As the coal formation
   process continues, benzene rings fuse together, and the H to C ration may be less than
   1:1. Anthracite, for example, contains about 80 percent fixed carbon.
   Methane (CH4) is formed during the geological process of oil formation. Under high
   pressures and temperatures, organic rearrangement reactions occur in the buried
   biomass. These release large quantities of methane and other light hydrocarbon gases.
   With regard to fuel quality, methane is the best in terms of energy per gram of fuel
   (see Table 2.2). Hydrogen atoms are 12 times lighter than carbon atoms. Hence,
   fuels with higher H to C ratios will produce more energy per gram. Methane, with 4
   C-H bonds per C, produces more energy per unit weight than oil, for which each C
   atom is bonded (approximately) to 2 C-H and 2 C-C bonds.
                                                                               Dr. Guldan
                                                                            Chem. 108 S09

4. Spiro, p.135, problem 7.
   The reaction for combustion of methane is:
       CH4 + 2 O2            CO2   + 2 H2O
       Net energy gained = {[2(EC=O) + 4(EO-H)]  [4(EC-H) + 2(EO=O)]} kJ
       Where EC=O, EO-H, EC-H, and EO=O are the bond energies for the C=O, O-H, C-H,
       and O=O bonds, respectively. These values are provided in Table 2.1 on p.25.
       Thus,
       Energy gained = {[2(799) + 4(460)]  [4(410) + 2(494)]} kJ = 810 kJ / mole fuel
       Since only one mole of CO2 is produced per mole of fuel burned, energy / mol
       CO2 = 810 kJ
       Since 1 mole of methane weighs 16 grams,
       Energy / g fuel = (810 kJ / mol CH4)(1 mol CH4 / 16 g) = 50.6 kJ/g fuel
5. Spiro, p.135, problem 9.
   a) For insolation of 270 W / m2, the energy for one year is:
       270 W  24 hours  365 days 
                                   2.36  103 kWh 2
           2            1 year                    m  yr
        1 m  1 day               
       Since 1 kg coal = 8.14 kWh, the coal equivalent of solar energy/m2 is:
                      1 kg coal 
       2.36  103 kWh             290 kg
                      8.14 kWh 
                                
       Energy derived from the PV cell per m2/yr is:
       2.36  103 kWh (0.15) = 345 kWh
       Energy derived from the equivalent coal combustion is:
       2.36  103 kWh (0.33) = 779 kWh
   b) If coal is 72 percent carbon, there are 720 g C/kg of coal. Also, we know that
      upon combustion, the carbon reacts with oxygen to form carbon dioxide:
       C + O2  CO2
       Which means 12 g of carbon (or one mole) reacts to form 44 g of CO2 (or one
       mole). Therefore, the grams of CO2 from 720 g of C is:
                1 mol C  1 mol CO2  44 g CO2 
       720 g C                                 2640 g CO2 or 2.64 kg CO2
                12 g C  1 mol C  1 mol CO 
                                           2 




       For 290 kg of coal (as determined in part a), CO2 emissions are:
                                                                                      Dr. Guldan
                                                                                   Chem. 108 S09

                                                    
                 290 kg COAL 21.64 kg CO   766 kg CO
                               kg COAL 
                                                 2
                                                                   2
                                                    
         Similarly, for 2 percent sulfur, there are 20 g sulfur per kg of coal. From
         combustion:
                 S + O2  SO2
         Which means 32 g of S (or one mole) react to form 64 g SO2 (or one mole). So
                 1 mol S  1 mol SO2  64 g SO2 
         20 g S                                 40 g SO2
                 32 g S  1 mol S  1 mol SO 
                                            2 


         For 290 kg of coal, SO2 emissions are:
                                         
         290 kg COAL 120 gCOAL   11600 g SO
                       kg
                              SO
                                  
                                      2
                                                          2   or 11.6 kg SO2
                                         


6. Spiro, p.136, problem 21 (a) only.
   Using the relation E = hc/ we can calculate the energy of a photon with  = 1140nm:
   1 nm = 1  107 cm
   So,
   1140 nm = 1.14 104 cm
   E = (4.14  1014 evs)(3  1010 cm/s) / 1.14 104 cm = 1.09 ev
7. Spiro, p.136, problem 22.
   The kernel is composed mainly of starch while the rest of the plant is mostly
   cellulose. Bacteria readily break down starch to the monomer sugar units, but
   cellulose is much more difficult to decompose. Currently, there is no inexpensive
   way to produce ethanol from cellulose. In comparison, production of ethanol from
   starch is relatively inexpensive.
8. Spiro, p.137, problem 25.
   a) For an average annual insolation of 270 W/m2:
   Solar energy/m2yr = (0.270 kW/m2)(24 hrs/day)(365 days/year) = 2.36  103 kWh/m2yr
   At a PV efficiency of conversion of 15%:
         Electricity production = (2.36  103 kWh/m2yr)(0.15) = 354 kWh/m2yr
   Since 1 kWh = 3.6  103 kJ,
         Electricity production = (354 kWh/m2yr)(3.6  103 kJ / 1 kWh) = 1.27  106 kJ/m2yr
   Since the energy embodied in hydrogen is 80% of the collected solar energy,
                                                                                       Dr. Guldan
                                                                                    Chem. 108 S09

       Energy content of H2 = (1.27  106 kJ/m2yr)(0.80) = 1.0  106 kJ/m2yr
   One mole of H2 contains 241 kJ of energy (Table 2.2); therefore:
       Moles H2 produced/ m2yr = (1.0  106 kJ/m2yr)(1 mole / 241 kJ) = 4.1  103 mol/m2yr
   Since 1 mole of H2 = 0.002 kg,
       Weight of H2 = (0.002 kg/mol)(4.1  103 mol/m2yr) = 8.2 kg//m2yr
   b) If PV collectors in the southwestern US can produce 1.0  106 kJ/m2yr stored as H2,
       PV collector land coverage = (40.1 1015 kJ) / (1.0  106 kJ/m2yr) = 40.1 109 m2
       Or 40.1 109 m2(1 km / 1000 m)2 = 40100 km2
       The percent of land area covered:
       [(40100 km2) /(1386370 km2)]  100% = 2.9%
   c) Every mole of H2 produced would required one mole of H2O. In addition, because the
      process of conversion of water to hydrogen is only 80% efficiency,
       water required = (4.1  103 mol/m2yr)/(0.80) = 5.1  103 mol/m2yr
       Given that 1 mole of H2O = 0.018 kg,
       Weight H2O / m2yr = (5.1  103 mol/m2yr)(0.018 kg / mol) = 91.8 kg/m2yr
       Since 1 kg of water = 1 L water,
       (91.8 L/m2yr)(1 106 m2/1 km2) = 91.8  106 L/km2yr
       For an area of coverage of 34500 km2,
       Total water requirements = (91.8  106 L/km2yr)(34500 km2) = 3.2  1012 L/yr
       Percent national water us (1995) = [(3.2  1012 L) / (4.7  1014 L)] 100% = 0.68%
9. Spiro, p.137, problem 26.
   For an ideal heat pump,
       Theoretical efficiency = (heat delivered) / (work) = Th / (Th – Tc)
   Where Th is the final temperature of the water, and Tc is the original temperature.
   Substituting in the values of Th and Tc (in degrees K):
       Theoretical efficiency = (353 K) / (353 K – 293 K) = 5.8
   Thus, the heat pump would have to perform (1/5.8 = 0.172) kJ of work to deliver 1 kJ
   of heat to the water heater.
   Given that the electric heater is 90 percent efficient, (1/0.9 = 1.1) kJ of electric heat is
   required to deliver 1 kJ of heat to the water. The second law efficiency is the ratio of
   the work energy required by the ideal heat pump to the energy required by the electric
   hot water heater, i.e.
   Second law efficiency = (0.172 kJ)/(1.1 kJ) x 100% = 15.6%
   If the electric water heater draws its power from a conventional coal-burning power
   plant, the actual efficiency is about 5 percent, since the efficiency of electric power
                                                                                 Dr. Guldan
                                                                              Chem. 108 S09

   production is around 33%. Thus, about 2 kJ of heat are wasted for every kJ of electric
   power produced.


10. Why is natural gas considered to be an environmentally superior fuel to oil or coal?
    What phenomenon involved in its transmission by pipeline might offset this
    advantage?
   Natural gas is considered to be an environmentally superior fuel since when it is
   burned it produces less CO2 (per 1000 kJ) than oil or coal do, so combustion of
   natural gas contributes less to global warming than combustion of oil or coal would.
   Natural gas also burns much cleaner than coal, since it doesn’t have sulfur, arsenic or
   other impurities. However, methane is also a greenhouse gas, and if it leaks out
   during transportation or extraction, the overall contribution of natural gas to global
   warming could be greater than the contributions of coal or oil.


11. Describe the difference between the two methods of absorbing energy from sunlight.
    What is the difference between active and passive systems?
   The two methods of obtaining energy directly from sunlight are: 1) thermal
   conversion, in which light is absorbed and converted to heat (can be used for cooking,
   heating homes, etc.) and 2) photo-conversion, in which light causes the motion of
   electrons in a material, resulting in electrical current. In passive systems, the heat
   created by absorption of light is used directly. In active systems, the heat created by
   absorption of light is transported. For example water heated by rooftop absorption of
   sunlight may be pumped through radiators to warm the house.


12. Describe how doping a silicon semiconductor can yield a directional current when
    light is absorbed by the silicon.
   Doping the silicon crystal creates regions of positive (Ga+) and negative (As) charge.
   These doped regions attract electrons and holes, respectively, which result from the
   excitation of the silicon by solar radiation (light). The flow of electrons towards Ga+
   and holes towards As constitutes an electrical current.

								
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