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Chapter 5 gen chem

This document discusses thermochemistry and energy changes in chemical reactions. It defines different types of energy, including kinetic energy, potential energy, and internal energy. The first law of thermodynamics states that energy is conserved in chemical reactions, with the change in internal energy of a system equaling the negative of the change in energy of the surroundings. Enthalpy is related to the internal energy and pressure-volume work term. Calorimetry experiments allow measurement of enthalpy changes using calorimeters and bomb calorimeters. Hess's law allows calculation of enthalpy changes from enthalpies of individual reaction steps. Fossil fuels are discussed as a non-renewable energy source.

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Chapter 5 Thermochemistry The energy of chemical reactions Presented by: Logan
Energy • The ability to: • do work • transfer heat. ➢Work: Energy used to cause an object that has mass to move. ➢Heat: Energy used to cause the temperature of an object to rise.
Work • Energy used to move an object over some distance. • w = F  d, w = work, F = force d = distance over which the force is exerted. Note units: F = ma, mass(distance/s2) W = F(d) = mass(distance2/s2) = mv2
Heat • Energy can also be transferred as heat. • Heat flows from warmer objects to cooler objects.
Units of Energy • The SI unit of energy is the joule (J). • An older, non-SI unit is still in widespread use: The calorie (cal). 1 cal = 4.184 J Energy has units of (mass)(velocity)2 Remember kinetic energy was 1/2mv2 1 J = 1 ⎯⎯ kg m2 s2
Kinetic Energy Energy an object possesses by virtue of its motion. 1 2 KE = ⎯ mv2
Potential Energy Energy an object possesses by virtue of its position or chemical composition. More potential E Less P.E. as bike goes down.
Transferal of Energy a) Add P.E. to a ball by lifting it to the top of the wall b) As the ball falls, P.E ------> K. E. (1/2mv2) Ball hits ground, K.E. =0, but E has to go somewhere. So 1. Ball gets squashed 2. Heat comes out.
System and Surroundings • The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). • The surroundings are everything else (here, the cylinder and piston).
First Law of Thermodynamics • Energy is conserved. • In other words, the total energy of the universe is a constant; ESystem = -Esurroundings
Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. Einternal,total= EKE + EPE + Eelectrons + Enuclei +…… Almost impossible to calculate total internal energy Instead we always look at the change in energy (E).
Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal − Einitial
Changes in Internal Energy • If E > 0, Efinal > Einitial ➢Therefore, the system absorbed energy from the surroundings. ➢This energy change is called endergonic.
Changes in Internal Energy • If E < 0, Efinal < Einitial ➢Therefore, the system released energy to the surroundings. ➢This energy change is called exergonic.
Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). • That is, E = q + w.
E, q, w, and Their Signs +q-q hot plate adds heat to water Surroundings suck heat out of water.
Sign of work Truck pushes block up. Does work on block wtruck- wblock+ block pushes truck down does work on truck wblock- wtruck+
State Functions Total internal energy of a system: K.E. + Eelectrons + Enucleus + P.E.total virtually impossible to measure/calculate
State Functions • However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. ➢In the system below, the water could have reached room temperature from either direction.
State Functions • However, q and w are not state functions. • Whether the battery is shorted out or is discharged by running the fan, its E is the same. ➢But q and w are different in the two cases.
Catch the work, do the same process in a cylinder Process evolves gas, pushes on piston, work done on piston
Catch the work, do the same process in a cylinder w = F*d, F = P*A, d=h w = -P*Ah= -PV Negative because an increase in Volume means that the system is doing work on the surroundings. E = q + w = q - PV qP = E + PV
Example• Gas inside cyclinder with electric heater. • Add 100 j heat with heater. • 1. Piston can go up and down • 2. Piston stuck. • a. What happens to T in each case? • b. What about q and w for each case? • c. What about E in each case? a.1. Piston goes up, some E goes to expand gas, do work. T goes up less a.2 T goes up more, all E goes to q. b.1. both q and w positive b.2. w 0, q larger c. E the same & + in each case
Work Now we can measure the work: w = −PV Zn + 2HCl ---------> H2(g) + ZnCl2
Work I mole of Zn reacts. How much work is done (P = 1 atm, density of H2 = 0.0823 g/L)? 1 mole of H2 is produced. Zn + 2HCl ---------> H2(g) + ZnCl2 1mol 1 mol 2. 014 g/mol 2.014 g d=m/V V=m/d V = 2.014g/0.0823g/L = 24.47 L W = -PV = 1atm(24.47L) = -24.47 L(atm)
Enthalpy焓 • at constant pressure, H, is ( = change in thermodynamics) H = (E + PV) • This can be written (if P constant) H = E + PV H = E + PV
Enthalpy • Since E = q + w and w = −PV (P const.) substitute these into the enthalpy expression: H = E + PV H = (q+w) − w H = q • Note: true at constant pressure • q is a state function at const P & only PV work.
Endothermicity and Exothermicity • A process is endothermic when H is positive. • A process is exothermic when H is negative.
Enthalpies of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants
Enthalpy of reaction example Consider the reaction: 2KClO3 -------> 2KCl + 3O2 H = -89.4 kJ/mol a. What is the enthalpy change for formation of 0.855 moles of O2? 2KClO3 -------> 2KCl + 3O2 H = -89.4 kJ/mol 0.855 mol H = -89.4 kJ/3 mol O2(.855 mol O2) = -25.5 kJ
Heat Capacity and Specific Heat • heat capacity: amount of E required to raise the temperature of something by 1 K • specific heat: amount of E required to raise the temperature of 1 g of a substance by 1 K.
Constant Pressure Calorimetry indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.
Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: q = m  s  T m = mass s = specific heat
Example When a 3.88 g sample of solid ammonium nitrate disolves in 60.0 g of water in a coffee cup calorimeter, the temperature drops from 23.0 °C to 18.4 °C. (a) Calculate H (in kJ/mol ammonium nitrate) for the solution process. Assume that the specific heat is constant and = 1.0 cal/gC. (b) Is this process endothermic or exothermic?
Example When a 3.88 g sample of solid ammonium nitrate disolves in 60.0 g of water in a coffee cup calorimeter, the temperature drops from 23.0 °C to 18.4 °C. (a) Calculate H (in kJ/mol ammonium nitrate) for the solution process. Assume that the specific heat is constant and = 4.184 J/g°C. (b) Is this process endothermic or exothermic? Reaction: NH4NO3(s) ------> NH4 +(aq) + NO3 -(aq) gr 3.88 g MW 80.04 g/mol #Mol 3.88 g/80.04 g/mol = 0.0484 mol Mass of solution = 3.88 g + 60 g = 63.88 g. System: Solid AmNO3 Surroundings: Solution q = s(specific heat)m(mass)T q = s(J/g°C)m(grams)(Tfinal - Tinitial) qsolution = 4.184(J/g°C)(63.88 g)(18.4°C - 23.0°C) = -1229 J qwater=-qammonium nitrate = +1229 J H(per mol NH4NO3) = 1.229 kJ/.0484 mol = 25.39 kJ/mol (b) Endothermic
Bomb Calorimetry Reactions can be carried out separated from the water in a “bomb,” such as this one, And still measure the heat absorbed by the water.
Bomb Calorimetry H = E + PV H = E + PV In a bomb calorimeter, V = 0 For a process that doesn’t evolve gas: P  0 as well. H = E + PV = E
Example • A 50 g sample of gasoline was burned by combustion (with excess oxygen) in a calorimeter with a heat capacity of 10 kJ/°C. The temperature increased 100 °C. Calculate the change in E per g of gasoline. • qsurroundings = CT =10 kJ/°C(100 °C) = 1000 kJ • qsurroundings = -qsystem • qsystem=-1000 • -1000 kJ/50g = -20 kJ/g • Does E = H in this case? • NO! Pressure can’t stay constant in this case.
Hess’s Law  H is known for many reactions. • measuring H can be a pain • Can we estimate H using H values for other reactions?
Hess’s Law Hess’s law: states that: H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. Yes!
Hess’s law, example: • Given: • N2(g) + O2(g) ----> 2NO(g) H = 180.7 kJ • 2NO(g) + O2(g) ----> 2NO2(g) H = -113.1 kJ • 2N2O(g) ----> 2N2(g) + O2(g) H = -163.2 kJ • use Hess’s law to calculate H for the reaction: • N2O(g) + NO2(g)----> 3NO(g) •N2O(g) ----> N2(g) + 1/2O2(g) H = -163.2/2 =-81.6kJ •NO2(g) ----> NO(g) + 1/2O2(g) H = 113.1 kJ/2=56.6kJ •N2(g) + O2(g) ----> 2NO(g) H = 180.7 •N2O(g) + NO2(g) ----> 3NO(g) H = 155.7 kJ
Calculation of H We can use Hess’s law in this way: H =  n Hf(products) -  m Hf(reactants) where n and m are the stoichiometric coefficients.  
Standard Enthalpies of Formation Standard enthalpies of formation, Hf, are measured under standard conditions (25°C and 1.00 atm pressure). 
Calculation of H • C3H8 + 5 O2 -----> 3CO2 + 4H2O H = [3(HfCO2) + 4(HfH2O)] - [(Hf C3H8) + (5Hf O2)] = [3(-393.5 kJ) + 4(-285.8 kJ)] - [(-103.85 kJ) + 5(0) = [-1180.5 kJ + (-1143.2 kJ)] - [(-103.85 kJ)+ 0 kJ = [-2323.7 kJ] - [-103.85 kJ) = -2219.9 kJ
Phase Changes • Conversion from one state of matter to another is a phase change. • Energy is either added or released in a phase change. • .
Heating Curves • A plot of T vs. q • Within a phase: • q= ms • The temperature of the substance does not rise during a phase change. • For the phase changes, the product of mass and the heat of fusion of vaporization is heat.
Fuels The vast majority of the energy consumed in this country comes from fossil fuels.
The Energy cycle: • Us (and almost everything else alive on the earth): • C6H12O6 + 6O2 ----> 6CO2 + 6H2O • Fossil fuel production: • CH4 + 2O2 ----> CO2 + 2H2O (for methane) • Plants: • 6CO2 + 6H2O + light ----> C6H12O6 + 6O2 Net CO2 production could therefore be 0.
The problem with oil • Not “renewable” (will run out) • Pollution (combustion not perfect). • Global warming CO2 absorbs heat. CnH2n+2 + (3n+1/2)O2 -----> nCO2 + (n+1)H2O
Hybrids • Electric motors are way more efficient than gas engines. (94%) • Note, your engine is very hot, • It must be cooled • Flush all that E down drain. No work, only heat. gas engines are 24-30% efficient Problem: batteries suck! Heavy, expensive, limited recharging cycles, limited current etc.
Li ion battery x e- +xLi+ + Li1-xCo(IV)O2 -----> LiCo(III)O2 LixC6 ------> xLi+ + xe- + C6 Lithium is really light. Dissolves in organic solvents which are also light. Li is at the top of the activity series. Means a higher potential (more voltage per battery cell)
Ethanol, where does it come from • Alcoholic fermentation: • C6H12O6 ----> 2CO2 + 2C2H5OH (ethanol) H=-76 kJ/mol • -1270 2(-393) 2(-280) • (anaerobic, bacteria & yeast can do this, we can’t) Exactly the same place it comes from in your beer.
Ethanol • Alcoholic fermentation: • C6H12O6 ----> 2CO2 + 2C2H6O (ethanol) H=-76 kJ/mol • -1270 2(-393) 2(-280) • (anaerobic, yeast can do this, we can’t) only to 10%. • Distillation (requires energy) to purify. bug Alcohol combustion: C2H6O + O2 ---> 2CO2 + 3H2O H = -1367 kJ/mol(1mol/46g)=-29.7kJ/g But why would this be better for global warming?
Ethanol, problems • Lots of land to grow (yield 2-4 tons/acre) • All present agricultural land in U.S. would not be enough for all transportation needs. • requires fertilizer, tractors,etc. for growing (energy) • Distillation requires energy • For every 1.4 kJ need 1.0 kJ, much more than oil • Brazil, however, is approaching 50% ethanol for transportation • Why? Sugar cane, largest starch or sugar yield/acre. • But, you can’t grow sugar cane on the great plains.
Things to consider • Energy yield (how much E out versus E in)? • Break even price (how much/gallon of gas equivalents (present corn ethanol is 2.25/gallon just to make). • Where is the technology NOW? • Is storage required, & if so, how you gonna do it • (solar when the sun doesn’t shine) • Remember, at present Batteries suck!
The Thermite reaction • 2Al + Fe2O3 -------> Al2O3 + 2Fe • What kind of reaction is this? • Why does it happen? • Used for welding railroad tracks • What is the heat of reaction given: • 2Fe + 3/2O2 -----> Fe2O3 H = -825.5 KJ • 2Al + 3/2O2 -----> Al2O3 H = -1675.7 KJ
The Thermite Reaction • 2Al + Fe2O3 -------> Al2O3 + 2Fe • What is the heat of reaction given: • 2Fe + 3/2O2 -----> Fe2O3 H = -825.5 KJ • 2Al + 3/2O2 -----> Al2O3 H = -1675.7 KJ • 2Al + 3/2O2 -----> Al2O3 H = -1675.7 KJ • Fe2O3 -----> 2Fe + 3/2O2 H = 825.5 KJ • 2Al + Fe2O3 -------> Al2O3 + 2Fe H = -850.2 KJ http://www.youtube.com/watch?v=BnHR4cMXiyM A thermite mystery: