chapter 6 thermochemistry l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 6 Thermochemistry PowerPoint Presentation
Download Presentation
Chapter 6 Thermochemistry

Loading in 2 Seconds...

play fullscreen
1 / 12

Chapter 6 Thermochemistry - PowerPoint PPT Presentation


  • 190 Views
  • Uploaded on

Chapter 6 Thermochemistry. Energy Energy = the capacity to do work or produce heat Conservation of Energy = energy can’t be created or destroyed; it can only change forms Potential Energy = due to position or composition Water behind a dam has energy that can be converted to work

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chapter 6 Thermochemistry' - vadin


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chapter 6 thermochemistry
Chapter 6 Thermochemistry
  • Energy
    • Energy = the capacity to do work or produce heat
      • Conservation of Energy = energy can’t be created or destroyed; it can only change forms
      • Potential Energy = due to position or composition
        • Water behind a dam has energy that can be converted to work
        • Attractive and Repulsive forces (gravity) govern this type
        • Gasoline burns to make heat: forces holding atoms together
      • Kinetic Energy = due to motion of the object
        • KE = ½mv2
        • m = mass and v = velocity of the object
      • Conversion of energy types: PE  KE  PE + frictional heating
slide2
Heat and Work
    • Heat = q = transfer of energy between objects at different temperatures
      • Temperature = KE of particles in random motion
      • Heat is not a substance, but our language often treats it that way
    • Work = w = force acting through a distance = F x d
    • Pathway: energy transfer might be the same with different results
      • A rough surface: mostly heat and little work
      • Smooth, steep surface: little heat and mostly work
      • State Function = property that only depends on present state
        • Value doesn’t depend on the path of how it got that way
        • Energy is a state function: DE = q + w
        • Work and Heat are not
  • Chemical Energy
    • Mechanical Energy = energy of the movement of objects (see above)
    • Chemical Energy = energy of the change in chemical bonds
      • CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g) + heat
      • Heat energy is liberated by rearranging the chemical bonds
slide3
Dividing up the Universe
    • System = the specific reactants and products we are investigating
    • Surroundings = everything else in the Universe
  • Heat flow in chemical reactions
    • Exothermic = energy flows out of the system as heat
      • Products have a lower PE than the reactants
      • Heat can be viewed as a product
      • Heat released results in an increase in KE of surrounding particles
slide4
b. Endothermic = energy flows into the system
        • N2(g) + O2(g) + heat -------> 2NO(g)
        • Heat can be viewed as a reactant
        • Heat absorbed results in less KE of the surrounding particles
        • Products have more PE than the reactants
  • D. Doing thermodynamics problems
    • 1. Thermodynamics = study of energy and its interconversions
    • 2. 1st Law of Thermodynamics = The energy of the universe is constant
    • 3. Internal Energy = E = sum of KE and PE;
slide5
Thermodynamic quantities
    • Consist of two parts: a number (magnitude) and a sign (direction)
    • The sign reflects the system’s point of view
      • Heat flowing into a system has +q (endothermic)
      • Heat flowing out of a system = -q (exothermic)
      • Work is done on the system = +w
      • The system does work on the surrounding = -w
    • DE = q + w sums up the changes in energy during a process
  • Example: Find DE of an endothermic reaction if 15.6 kJ is flowing and 1.4 kJ work is done on the system. DE = q + w = 15.6 + 1.4 = 17.0kJ
    • 6. PV Work
      • Gases do work (expand) or have work done on them (compress)
      • Expansion of gas in engine cylinder provides mechanical work
      • i. Expanding gas = -w = (+P)(+DV)
      • ii. Compressed gas = +w = (+P)(-DV)
slide7
Example: Find w for expansion of a gas from 46L to 64L at 15 atm.
        • w = -PDV = -(15atm)(18L) = -270Latm
      • 8. Example: Find DE if 1.3 x 108 J of heat expands a balloon from
      • 4.00 x 106 L to 4.50 x 106 L at 1.0 atm.
        • DE = q + w = q + (-PDV) = (1.3 x 108 J) - (1atm)(0.50 x 106 L)
        • = (1.3 x 108 J) - (5 x 105 Latm)(101.3 J/Latm) = 8 x 107 J
  • Enthalpy and Calorimetry
    • Enthalpy = H = E + PV
      • E, P, and V are all state functions, so H is also a state function
      • Consider a constant P process where only PV work is allowed
        • DE = q + w = q – PDV → q = DE + PDV
        • DH = DE + D(PV)
        • Since P is constant: D(PV) = PDV →DH = DE + PDV
        • DH = q At constant P, DH of a system = energy flow as heat
        • Heat of reaction = change in enthalpy = DH = Hproducts – Hreactants
          • If DH = + heat will be absorbed by the system = Endothermic
          • If DH = - heat will be released by the system = Exothermic
slide8
Example: a. Find DH if 1 mol CH4 burns at const. P and 890 kJ heat is released. b. Also find DH if 5.8 g of CH4 burns.
      • q = DH = -890 kJ
  • Calorimetry = science of measuring heat
    • Calorimeter = device for measuring heat changes in a chemical reaction
    • Each substance changes temperature at a different rate = Heat Capacity
      • Specific Heat capacity = J/g•oC = Cs
      • Molar Heat capacity = J/mol•oC
      • Water has a high heat capacity
        • Very good coolant
        • Much higher than metals
      • 3. Constant-Pressure Calorimeter (q = DH)
      • Experiment done under atm. Pressure
      • 50.0 ml 1.0M HCl + 50.0 ml 1.0M NaOH at 25 oC (H+ + OH---> H2O)
      • T = 31.9 oC after the reaction
slide9
Example: Calculate DH/mol for the following reaction
    • Ba2+ + SO42- -------> BaSO4(s)
    • 1L 1.00M Ba(NO3)2; 1.00L 1.00M Na2SO4; 25 oC start; 28.1 oC end
  • Constant Volume Calorimetry
    • If DV = 0, then w = 0
    • DE = q + w = q
    • Combustion of 0.5269g octane (C8H18) in a bomb calorimeter with a heat capacity = 11.3 kJ/oC with a DT = 2.25 oC. Find DE.

Example:

slide10
Hess’s Law
    • Recall that Enthalpy is a State Function
      • Path doesn’t matter
      • As long as we know reactants and products, steps don’t matter for DH
      • Example:
        • N2(g) + O2(g) -------> 2NO(g) DH1 = 180 kJ
        • 2NO(g) + O2(g) -------> 2NO2(g) DH2 = -112 kJ
        • N2(g) + 2O2(g) -------> 2NO2(g) DHtotal = 68 kJ
      • Hess’s Law
        • You may sum steps in order to find overall DH
        • The DH for the reverse reaction will simply change signs (+/-)
        • If you multiply a reaction, you must multiply DH the same
slide11
Explanation
      • Sign of DH depends on direction of heat flow. Heat flow is reversed if the overall reaction is reversed.
        • Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ)
        • XeF4(s) + 251kJ -------> Xe(g) + 2F2(g) (DH = +251kJ)
      • DH is an extensive property = depends on the amount of substance (an intensive property = depends only on identity of the substance)
        • Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ)
        • 2Xe(g) + 4F2(g) -------> 2XeF4(s) + 502kJ (DH = -502kJ)
  • B. Examples
    • 1. Example: Calculate DH for the conversion of graphite to diamond using the known DH’s for the combustion of graphite and diamond.
      • Cg(s) + O2(g) -------> CO2(g) DH = -394kJ
      • Cd(s) + O2(g) -------> CO2(g) DH = -396 kJ
      • b. Cg(s) + O2(g) -------> CO2(g) DH = -394kJ
      • CO2(g) -------> Cd(s) + O2(g) (reverse) DH = +396 kJ
      • Cg(s) -------> Cd(s) DH = +2 kJ
slide12
2. Example: Find DH for synthesis of B2H6 from B and H2.
    • Known Reactions
      • 2B(s) + 3/2O2(g) -------> B2O3(s) DH = -1273 kJ
      • B2H6(g) + 3O2(g) -------> B2O3(s) + 3H2O(g) DH = -2035 kJ
      • H2(g) + 1/2O2(g) -------> H2O(l) DH = -286 kJ
      • H2O(l) -------> H2O(g) DH = +44 kJ
    • Combining Reactions
      • 2B(s) + 3/2O2(g) -------> B2O3(s) DH = -1273 kJ
      • B2O3(s) + 3H2O(g) -------> B2H6(g) + 3O2(g) DH = +2035 kJ
      • 2B(s) + 3H2O(g) -------> B2H6(g) + 3/2O2(g) DH = +762 kJ
      • 3[H2(g) + 1/2O2(g) -------> H2O(l)] DH = -858 kJ
      • 3[H2O(l) -------> H2O(g)] DH = +132 kJ
      • 2B(s) + 3H2(g) -------> B2H6(g) DH = +36 kJ
    • c. Let the final needed reaction guide how you combine other reactions

+

=

+

+

=