Chapter 8 Review (sections 1-3, 5-11)

  1. Energy
    1. Reactions occur if the products are more stable (have less energy) than the reactions.  Note that water and nitrogen gas are two very stable species.  When they are made the reactions are often very exothermic.  Check out this explosion making water - Watch this video - click on activities then formation of water movie. 
    2. Kinetic energy is energy of motion.  Potential energy is stored energy.
    3. Units for energy are kg m2 / s2 which = J (Joule).  Note that a calorie = 4.184 J.  Note that food Cal = 1000 calories.
    4. Snickers is 345 Cal.  How many kJ is this?  345 Cal (1000 cal / Cal)(4.184 J/cal)(kJ/1000J) = 1440 kJ
  2. DE
    1. First Law of Thermodynamics - Energy cannot be created nor destroyed, only transferred. 
    2. Some E forms are kinetic, potential, heat, chemical, light, sound
  3. More DE
    1. Exothermic - when heat leaves the system and enters the surrounding.  The T of surroundings goes up, it feels hot since heat is exiting the system and entering your hand, and DH is negative.
    2. Endothermic - when heat enters the system and leaves the surrounding.  The T of surroundings goes down, it feels cool since heat is entering the system and leaving your hand, and DH is positive.
    3. The combustion of methane is an exothermic reaction, like all combustions. If one mole of methane burns DH is -802 kJ.  If two moles burn, DH is -1604 kJ and so forth. 
    4. State functions depend on the initial and final values only.  Examples:  DE, DH, P, V, T
    5. Path functions depend on the "path" between the initial and final.  Examples:  calories, q (heat), time, work Try problem 8.2
  4. Skip section 4 
  5. Energy and DH
    1. q = heat, DH = enthalpy (change in heat), DH = DE or almost in chemical reactions
  6. Standard State
    1. Standard state for thermodynamics is 1.00 atm, 25oC and 1.00M for solutions
    2. Standard indicated by o symbol called knot.  Such as DHo
  7. DH
    1. Enthalpy for physical changes  Melting is endothermic, freezing is exothermic animation. 
      1. Melting.  DHfus is endothermic.  Must add heat to pull the solid apart.
      2. Freezing.  -DHfus so exothermic.  Must remove heat to form the solid.
      3. Boiling.  DHvap is endothermic.  Must add heat to pull the liquid apart.
      4. Condensation.  -DHvap so exothermic.  Must remove heat to form the liquid.
      5. Sublimation.  DHfus + DHvap is endothermic. Must add heat to make a solid spread apart into a gas.
      6. Deposition.  -DHfus -DHvap so exothermic.  Must remove heat to slow a gas down into a solid.
    2. Enthalpy for chemical reactions
      1. Pretend A g  B.  DHrxn is final minus initial, so, products minus reactants
        1. If A is 50 J and B is 20 J then DH is -30 J and the reaction is exothermic.
        2. If A is 20 J and B is 50 J then DH is +30 J and the reaction is endothermic.
      2. Pretend X  g Y.  DHrxn = -30 J
        1. then 2X  g 2Y.  DHrxn = -60 J  (multiply by 2)
        2. then 3Y  g 3X.  DHrxn = +90 J   (change sign when reverse and multiply by 3)  
      3. Given:  2 H2O(g)  g   2 H2(g)  +  O2(g)      DHrxn = +484 kJ
        1. How much heat would be needed to make 55.5 grams of hydrogen gas?
        2. Answer:  55.5 g H2 ( mol / 2.02 g)( 484 kJ / 2 mol H2 ) = 6650 kJ
    3. Try example 8.3 and problems 8.8 and 8.9
  8. Calorimetry       
    1. A Calorimeter measures the heat gained or lost by the solution when a reaction occurs in solution.  Figure 8.8 (ignore the bomb calorimeter paragraph)
    2. The heat lost/gained by the chemical reaction must equal the heat gained/lost by the surrounding solution in the calorimeter, but is opposite in sign.  So qsoln = - qrxn.  
    3. View a Calorimeter introduction animation here.
    4. Before we do calorimeter problems we must first learn about specific heat (S) which is the heat needed to raise one gram by one degree Celsius and is a constant with units (J / goC) for every different substance.
      1. If a substance conducts heat well it is easy to heat up and S is low.  Cu is 0.385 J / goC
      2. If a substance does not conduct well it is hard to heat up and S is high.  Water is 4.18 J / goC
      3. More values in Table 8.1
      4. Important equation:  q = mSDT where m is mass in grams and DT is Tfinal - Tinitial
      5. Example:  What is the specific heat for Arsenic if it takes 189 J to heat 49.7 grams from 35.5 to 40.9oC?  Plug into q = mSDT.   189 J = (49.7 g)(S)(40.9 - 35.5oC)  and solve for S = 0.70 J/goC
    5. Now a calorimeter problem.  Remember because the temperature change refers to the solution we can only calculate the heat of the solution.  But then the heat of the reaction is equal but opposite to the solution!  Finally we are usually solving for DHrxn so to get that we divide qrxn by the moles of whatever was specified in the question.  DHrxn = qrxn / moles
      1. Calculate DHrxn per mole of acid when 30.0 mL of 1.50M hydrobromic acid reacts with 15.0 mL of 3.00M lithium hydroxide in a calorimeter and the temperature goes from 24.5 to 31.8oC.  Assume the solution is so dilute it has basically the density of water (1.00g/mL) 
      2. Answer:  First find qsoln, then get qrxn, and finally get DHrxn. The mass of water is 45.0 grams because we have 45.0 mL and density is 1 g = 1 mL.   qsoln = (45.0 g)(4.18 J / goC)(31.8 - 24.5oC)(kJ/1000J) = 1.373 kJ.   Now qrxn = - 1.373 kJ.  Now we need moles of acid.  (0.0300 L HBr(aq))(1.50 mol / L) = 0.0450 moles HBr.  Finally DHrxn = - 1.373 kJ / 0.0450 mol = -31 kJ/mol acid.  Exothermic reaction. 
    6. Try examples 8.4 and 8.5 and problems 8.10 to 8.12
  9. Hess's Law
    1. Use reactions with known DHrxn to find the unknown DHrxn of another reaction. It's like a puzzle!  You take the known reactions and rearrange them (reverse them, multiply them by a whole number, etc) so that when they add up everything cancels out but the reactants and products for the other reaction you are trying to find.  If you reverse a reaction, change the sign of DHrxn and if you multiply by two, then multiply DHrxn by two also. 
    2. Pretend we want to find the DHrxn for: the conversion of graphite to diamond
      Given: C(graphite) + O2(g)   g  CO2(g)   DH = -393.5 kJ
      Given:  C(diamond) + O2(g) g CO2(g)        DH = -395.4 kJ
      So our goal reaction is C(graphite)  g  C(diamond).  We put the known reactions together:
      written as is C(graphite) + O2(g) g CO2(g)                          DH = -393.5 kJ
      reversed:        CO2(g)             g C(diamond) + O2(g)      DH = +395.4 kJ
      which adds to:               C(graphite) g C(diamond)                   DH = +1.9 kJ
    3. More interactive examples here They are good!!!  Try them. 
    4. Try example 8.6 and 8.7 and problems 8.13 to 8.15
  10. Heats of formation, DHof 
    1. Where do DHorxn  come from anyway?  Well standard heats of reaction, DHorxn,  come from the standard heats of formation, DHof , of the products and reactants. Heats of formation are the heat needed to form ONE MOLE of the substance from elements in their natural states. Every substance has its own DHof which are listed in tables, see Table 8.2 and Appendix B. So DHof values go with formation reactions. 
    2. Examples of formation reactions.  Remember the product must be one mole thus the strange balancing.
      1. 1/2 N2(g)   +   3/2 H2(g)  g  1 NH3(g)    DHof = - 46.1 kJ/mol
      2. C(s)   +   1/2 O2(g) g   CO(g)                DHof = - 110.5 kJ/mol
    3. What about DHof  for elements themselves?  Well elements don't have to form from themselves, they already exist so DHof for elements are all ZERO.
    4. So DHorxn  is final minus initial like any D.  Therefore DHorxn  = DHof products - DHof reactants. 
    5. Example:  What is the ΔHorxn  in kilojoules for the combustion of 1 mol of ethanol, C2H5OH(l), to form gaseous carbon dioxide and gaseous water?
      1. First write and balance the combustion equation.  C2H5OH(l) + 3 O2(g) g 2 CO2(g) + 3 H2O(g)
      2. Look up DHof values:  C2H5OH(l) = -277.1 kJ/mol,  CO2(g) = -393.5 kJ/mol,  and H2O(g) = -241.8 kJ/mol
      3. ΔHorxn = [(2)CO2(g)    +     (3)H2O(g)]     -    [   C2H5OH(l)   +    (3)O2(g)]
                             = [ (2)-393.5     +      (3)-241.8 kJ/mol]    -    [ -277.1    +    (3)0 kJ/mol]
                             = [-787 -725.4 kJ/mol ] - [ -277.1 kJ/mol ]
                             = -1235.3 kJ/mol
      4. Try examples 8.8 and 8.9 and problems 8.16 and 8.17
  11. Bond Dissociation Energy (D)  
    1. Consider breaking the bond between A and B.   A-B  g A + B   The energy to break the bond is D and is always endothermic since it takes energy to break a covalent or ionic bond.
    2. D = (bonds broken) - (bonds formed)   Note bond energies are in Table 7.1 or online here
    3. Calculate D for this reaction:  N2(g)   +   3 H2(g)  g  2 NH3(g)
      1. D for  N=N =  941 kJ/mol,  H-H = 436 kJ/mol, and  N-H = 393 kJ/mol
      2. bonds broken is one N=N and three H-H = 941 + 3(436) = 2249
      3. bonds formed is six N-H = 6(393) = 2358
      4. D = 2249 - 2358 = - 109 kJ (exothermic)
    4. Try example 8.10 and problems 8.18 and 8.19

 

That's it.  You are almost done with chemistry I. 
Study HARD and LONG for the FINAL