- 243 Views
- Uploaded on
- Presentation posted in: General

Chapter 6 - Thermochemistry

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 6 - Thermochemistry

Prelab for Exp #2 due 10/7 or 10/9

Online HW Ch 6 Due before: 10/15

Exam #1 On 10/15; Quiz #1 on 10/14 (Monday)

The exam will cover chapters 5 & 6; Fe & Unk Metal Labs

6 11 17 18 19 25 26 38 43 47

51 59 61 67 69 75 77 79 83 85

87 91 103 123 146

Plus: Calculate the ΔHr for: 1N2F2 (g) + 1F2 (g) ---) 1N2F4 (g)

using the bond energy table 9.5 page 367. (answer = -1.30x102 kJ)

- Thermodynamics deals with energy relationships.
- Thermochemistrydeals with energy relationships in chemical reactions.
- Thermochemistry allows us to:
1)predict the spontaneity of a reaction (Chapter 18)

2)calculate the heat lost or gained in a reaction (Ch. 6)

3) calculate the useful work involved in a reaction (Ch. 18)

4)calculate equilibrium constants (Chapters 18 & 19)

5)calculate voltages of redox reactions (Chapter 19)

6)calculate bond energies (Chapter 9 & 6)

Energy = The ability to move matter

- Are several forms of energy associated with chemical reactions: heat, electrical, light & mechanical energies.

- These types of energy can be interconverted and are usually measured as if they were a) heat energy in calories or b) mechanical energy in joules.

- Are two forms of energy: Kinetic & Potential.

Potential Energy (Ep) – Stored energy due to position.

- Ep can be energy stored in the chemical bonds, in the intermolecular forces, and in the subatomic particles.

Kinetic Energy – Energy associated with motion of an object

Ek = ½ Mass x (Velocity)2 = ½ m x v2(know)

If m is in kg & v in meters/sec then E in J; 1J = 1 kg m2 / sec2

B. Examples

- Calculate Ek of 4 kg bowling ball going 2 meters per sec.

Ek = ½ MV2 = ½ (4 kg)(2 m/s)2 = 8 kg m2 / s2 = 8 J (joule)

- a joule is a small unit.

- a watt (a measure of power or rate of using energy) = J/sec

- a 60 watt light bulb uses 60 J of energy per second.

- A typical American home uses ~ 4 billion joules per month in electricity. Note: since types of energies can be interconverted, we can report electrical energy in terms of either mechanical energy (joules) or heat energy (calories).

- Older unit of energy is calorie (cal) = amount of energy required to raise temperature of 1 g of water by 1 oC.

- 1.000 cal = 4.184 J

- How many J are in a large piece of apple pie which releases 500 Cal upon digestion? (Note: 1 dietary calorie = 1 Cal = 1 kcal; typical candy bar has 400 Cal = 400,000 cal !)

500. Cal 1 kcal 1000 cal 4.184 J = 2.09x106 J

1 Cal 1 kcal 1 cal

- If we want to determine the total energy of a substance, then we must consider: 1) its position [Ep]; 2) its overall motion [Ek]; 3) the internal motion of its protons, neutrons and electrons [ek]; and 4) the potential energy of these subatomic particles [ep].
- ek + ep = U (Internal Energy)
- ET = Ek + Ep + U
- Internal Energy = sum of kinetic & potential energies of the subatomic particlesmaking up the substance.

- ET = Ek+ Ep + U
- EkEp & U can vary; however, ET remains constant = Law of Conservation of Energy
- Law of Conservation of Energy = energy may be converted from one form to another, but the total quantity of energy remains constant.
- We will not concern ourselves with U.

- System = the part of the universe you are studying (rxn).
- Surroundings = the rest of the adjacent universe which interacts with the system (calorimeter).
- Heat = energy that flows because of a difference in temperature.
- Heat spontaneously flows from hot to cold substances because of higher molecular motion in hotter substance.
- The amount of heat in chemical reactions is related to the amount, type & temperature change of the substances; temperature is related to molecular motion of the particles of the substance.

- Generally we use Δq or ΔH for heat; Δ = change. H is also called enthalpy & is the heat involved under constant P.
- Δq or ΔH = + when heat absorbed by a chemical reaction.
- Δq or ΔH = - when heat given up by a chemical reaction.
Changes that absorb heat (surroundings cooler) = endothermic(Δq = +)

Changes that give upheat (surroundings hotter) = exothermic(Δq = - )

- When one burns wood, what is the sign of Δq for this rxn?
- When one dissolves NH4NO3 in water, the water drops in temperature. What is sign of Δq for this dissolving?

- KNOW these sign conventions well; many points in both chemistry 122 & 123 depend on knowing these signs.

- ΔHr = heat of reaction = change in enthalpy for a reaction. Be careful, since ΔHrmay be expressed as either 1) kJ per mol of a given chemical or 2) total kJ for the balanced reaction (rxn).
Example: 2H2(g) + 1O2(g) 2H2O(l)

∆Hr = 1) -286 kJ for 1 mol of H2O or

∆Hr = 2) -572 kJ for the balanced rxn

- ∆Hr = Hfinal - Hinitial = ∑Hproducts - ∑Hreactants
Note: This equation works because ΔH is a state function.

State Function – property whose value depends on the initial & final states; not how that change was reached.

Example of a State Function = Altitude change from B to A is 3600 ft regardless of path taken.

- We frequently show ΔHr (heat of reaction) after the reaction:
2 H2 + 1 O2 ---) 2 H2O ΔHr = - 572 kJ for rxn(exothermic)

- 572 kJ for rxn; - 572 kJ/1 mol O2; - 286 kJ/1 mol H2; - 286 kJ/1 mole H2O

- Note: For reverse reaction ΔH has same magnitude, but changes sign = + 572 kJ (+ = heat absorbed).
2 H2O ---) 2 H2 + 1 O2ΔHr = + 572 kJ(endothermic)

- Calculate the ΔH in the top rxn when 5.1g of H2 reacts.
5.1 g H21.0 mol H2 – 572 kJ = - 729 kJ

2.0 g H22.0 mol H2

1) Reactions in Calorimeters (exp): ΔH = C x ΔT

2) Reactions in Calorimeters (exp): ΔH = g x s x ΔT

3) Hess’s Law:Add & Subtract Reactions

4) Bond Energies: (table 9.5, p 367. a) Draw Lewis Structures; b) Consider mols of bonds broken & formed); c) use sign (+) from Table.

∆Hr = ∑ ∆Hbonds broken -∑ ∆Hbonds formed

5)Heats of Formation: ΔHof (table 6.2, p 250; a) use sign (±) from table; b) consider mols of compound)

ΔHr = ∑ ΔHof products- ∑ ΔHofreactants

- Heat Capacity = C = Heat needed to raise T of system 1 oC.
ΔH = C x ΔT where ΔT = Tf – TiC in J/oC or kJ/oC(know)

Be careful: C may be for a) empty or b) full calorimeter

Example: Where C=11 J/oCfor full calorimeter. Calculate ΔHr in J for a rxn in this calorimeter in which the T goes from 25.1 to 30.3 oC.

ΔH = C x ΔT = 11 J/oC x 5.2 oC = 57 J

Example: Calculate C for a full calorimeter if the calorimeter is calibrated and found to absorb 69 J when the T goes from 30.0 to 37.2 oC.

C = ΔH / ΔT = 69J / 7.2 oC = 9.6 J/oC

- Specific Heat Capacity = s = Heat needed to raise T of 1 g of substance by 1 oC; s is characteristic of substance.
ΔH = g x s x ΔT s in units of: J / goC or kJ / goC (know)

Example: How much heat (kJ) is must be added to raise 225 g of H2O from 25.0 to 100.0 oC? Given: s = 4.18x10-3 kJ/(goC) for water

ΔH = g x s xΔT

ΔH = 225 gx4.18x10-3 kJ/(goC) x 75.0 oC = 70.5 kJ

Example: 5.00 g of Ag absorbs 25.0 J of heat when heated by 21.3 oC. Calculate s for Ag. ΔH = g x s x ΔTs = ΔH / [g x ΔT]

s = 25.0 J / [5.00 g x 21.3 oC] = 0.235 J/goC

1) C may be for empty calorimeter or fullcalorimeter – caution

2) Small s = good heat conductor; Large s = good insulator

3) s = 1.00 cal/goC or 4.18 J/goCor 4.18x10-3 kJ/goCfor water

4) May use one or both equations (1&2)in a problem

5) These two are Experimental methods (rest of methods are theoretical)

6) When using ΔH = g x s x ΔTg= mass of solution; get g from density (1.00 g/mL unless told otherwise) & from volume.

Example: 15.5 g of metal (Me) is heated to 98.9 oC and dropped into 25.0 g of water at 22.5 oC in a calorimeter. The water temperature goes to 25.7 oC. Calculate the specific heat capacity (s) of the unknown Me.

Note: ΔT water = (25.7- 22.5)oC = 3.2 oC for water

ΔTmetal = (25.7- 98.9)oC = -73.2 oC for metal

a) qwater = g x s x ∆T = 25.0g x 4.18 J/(goC) x3.2oC =3.3x102 J

b) qwater = - qMe& -qMe = - 3.3x102J = gMexsMexΔTMe

sMe = - 3.3x102J/ [gxΔT] = - 3.3x102J / [15.5g x -73.2oC] = 0.29 J / (goC)

Note: a) When perform a reaction in a calorimeter, the q lost by Rxn = q gained by calorimeter -- will have opposite signs.

b) s must always be a positive value.

- 0.56 g of carbon is placed in a bomb calorimeter at 25.00 oCalong with excess O2. After ignition the T is 25.89oC. The heat capacity for the filled calorimeter, C = 20.7 kJ/oC.Calculate theΔHreaction(ΔHr) in kJ/mol of C. Note: ΔT = (Tf – Ti) = 0.89 oC.
- Note: Use only heat capacity equation:ΔH = C x ΔT
ΔHcal= CxΔT = 20.7 kJ/oC x 0.89oC = +18 kJ (for cal)

ΔHr = - ΔHcal = -18 kJ for burning 0.56 g of C

-18 kJ 12.0 g C = - 3.9x102kJ/mol of C

0.56 g C 1.00 mol C

1C + 1O2 -----) 1CO2ΔHr = - 3.9x102 kJ/mol of C

- Mix 50.0 mL 0.250 M HCl & 50.0 mL 0.250 M NaOH together. The T goes from 19.50 oC to 21.21 oC. Calculate ΔHr in kJ/mol. Given the dsoln = 1.00 g/mL & ssoln = 4.18x10-2 kJ/(goC).
- notes: a) 100.0 g soln; b) use ΔH = s x g x ΔT c) ΔT = 1.71 oC (exo; q = - )
0.250 mol/L x 0.0500L = 0.0125mol of either HCl & NaOH

1HCl + 1NaOH -----) 1NaCl + 1H2O ΔH = s x g x ∆T

ΔHsoln = 4.18x10-3 kJ/(goC) x100.0g x 1.71oC = 0.715 kJ

ΔHr = - 0.715 kJ & is for 0.0125 mol (convert to kJ/mol)

ΔHr = - 0.715 kJ / 0.0125 mol= - 57.2 kJ / mol

Some calorimeter problems use both ΔH = s x g x ΔT (given s for contents) and ΔH = C x ΔT (given C for empty calorimeter)

- AnEmptycalorimeterhas a heat capacity of 0.0111 kJ/oC. A chemical reaction takes place in 32 mL of water inside the calorimeter, and the temperature goes from 25oC to 46oC. Calculate the ΔHr in kJ. ssoln = 4.18x10-3 kJ/goC ; d = 1.00 g/mL
- ΔHgainedbysystem = ΔHcalorimeter + ΔHsolution
ΔHE cal = C x ΔT = 0.0111 kJ/oC x 21 oC = 0.23 kJ

ΔHsoln = s x g x ΔT = 4.18x10-3 kJ/goC x 32 g x 21 oC = 2.8 kJ

ΔH = ΔHcalorimeter + ΔHwater = 0.23 kJ + 2.8 kJ = 3.03 kJ

Since the reaction is exothermic: ΔHr = - 3.0 kJ

Calculate ΔHr in kJ\mol Mg for: 1Mg + 2HCl -----) 1MgCl2 + 1H2

102 g of soln (s= 3.93 J/goC) contains 1 mol of HCl. You add 0.01036 mol of Mg to the soln in a calorimeter (C= 15 J/ oC for the empty calorimeter )& T raises by 12.47 oC. Note: Mg is limiting reagent = 0.01036 m

ΔH = Hsoln + Hcalorimeter= (s x g x ΔT) + (C x ΔT)

ΔH = (3.93 J/goCx 102g x 12.47oC) + (15J/oC x 12.47oC) = 5000+190 = 5190 J

Calorimeter system absorbs 5190 J or 5.19 kJ (-5.19 kJ for reaction)

ΔHrxn = - 5.19 kJ / 0.01036 mol Mg = - 501 kJ/mol Mg

- Hess’s Law:The heat of reaction is the same, whether the reaction is carried out directly in one step or indirectly through a number of steps – add & subtract heats. (Use given rxn as a template.)
- Example 1: Calculate ΔHr for: 2S + 3O2 -----) 2SO3
Given:a) S + O2 -----) SO2ΔH = - 297 kJ

b) 2SO3 -----) 2SO2 + O2ΔH = + 198 kJ

Solve: Multiply # a) by 2; reverse # b); ADD results together

2 x [S + O2 -----) SO2ΔH = -297] = - 594 kJ

2SO2 + O2 -----) 2SO3ΔH = - 198 kJ

2SO2 + 2S + 3O2 ----) 2SO2 + 2SO3ΔHr = - 594 - 198 = -792 kJ

2S + 3O2 -----) 2SO3ΔHr = - 792 kJ

Calculate ΔHr for: 2 Mg + 1 O2 -----) 2MgO From following 3 reactions.

1) Mg + 2HCl -----) MgCl2 + H2H1 = - 501 kJx 2

2)MgO + 2 HCl -----) MgCl2 + H2OH2 = - 151 kJx - 2

3)2 H2 + O2 -----) 2 H2OH3 = - 572 kJOK

2 Mg + 4 HCl -----) 2 MgCl2 + 2 H2H = - 1002 kJ

2 MgCl2 + 2 H2O -----) 2 MgO + 4 HClH = + 302 kJ

2 H2 + O2 -----) 2 H2OH3 = - 572 kJ

2Mg + 4HCl + 2MgCl2 + 2H2O + 2H2 + O2 --) 2MgCl2 + 2H2 + 2MgO + 4HCl + 2H2O

2Mg + O2 --) 2MgO ΔHr = -1002 +302 -572 = -1272 kJ

- Bond Energy = Energy needed to break a bond in gas phase. ∆Hbreak bond= + value (adding energy) (Table 9.5, pg 367 + values)
∆Hform bond= - value (releasing energy) - Know signs

- Typical bond energies ~ 150 to1000 kJ/mol
- Can use bond energies to calculate heats of reaction (∆Hr).
- ∆Hr = ∑ ∆Hbonds broken ∑ ∆Hbonds formed(Know)
1) Need to know # bonds & type of bonds involved; use Lewis Structures and stoichiometry: examples: a) 2BF3 = six B-F bonds

b) CO is CΞO (not C=O & not C-O)

2) The above equation works USING SIGNS GIVEN IN TABLE since

∆Hbonds broken = add energy (+) ∆Hbonds formed = releases energy (-).

- ∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed]
- Example 1: calculate the ∆Hrxn for following:
1 H2 + 1 Cl2 ------) 2 HCl

- Given these bond energies from Table 9.5; pg 367

[ H-H = 432 ; Cl-Cl = 240; H-Cl = 428 kJ/mole ]

1 H2 + 1 Cl2 ------) 2 HCl

1x432 1x240 2x428

∆Hr = [432 + 240] - [856] = - 184 kJ

1 H2 + 1 Cl2 ------) 2 HCl∆Hr = - 184 kJ

- Example 2: Bond energy problem (may not be single bonds).
Calculate the ΔHr for: 1C2H2 + 2Br2 -----) 1CHBr2CHBr2

1) Draw the Lewis Structures; 2) determine bond energies; 3) calculate ΔHr.

Br Br

1) H-C≡C-H + 2 Br-Br -----) H-C - C-H

Br Br

2) C-H= 411 C≡C = 835 Br-Br= 190 C-Br= 285 C-C= 346

3) ∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed]

Bonds BrokenBonds Formed

2C-H 1C≡C 2Br-Br 4C-Br 2C-H 1C-C

∆Hr = [2x411 + 1x835 + 2x190] - [4x285 + 2x411 + 1x346] = -271 kcal

- We have five ways of calculating heats:
1) Reactions in Calorimeters 1: q = C x ΔT

2) Reactions in Calorimeters 2: q = s x g x ΔT

3) Hess’s Law:Adding & Subtracting Rxns

4) Bond Energies:

∆Hr = ∑ [∆Hbonds broken] - ∑ [∆Hbonds formed]

5)Heats of Formation: ΔHof

ΔHr = ∑ ΔHof products– ∑ ΔHofreactants

- In order to use the last way we need some values of ΔHof
- Chemists devised a relative scale of enthalpies since path to answer not important (ΔH is state function).

- ΔHofis a relative number defined using a standard state.
- The Standard State = 25 oC and 1.0 atm; represented by o

- ΔHof = enthalpy change for formation of 1 mol of substance in its standard state from its elements in their standard states. KNOW This Definition (several pts on Ex 1 & Final involved)
- Examples of Heat of formation reactions (use definition):

1) for H2O(l): 1H2(g) + 1/2O2(g) 1H2O(l)ΔHr = ΔHofH2O(l)

2) for N2O4(g) : 1N2(g) + 2O2(g) 1N2O4(g)ΔHr = ΔHofN2O4(g)

3) for AgCl(s) : 1Ag(s) + 1/2Cl2(g) 1AgCl(s)ΔHr = ΔHofAgCl(s)

- ΔHof of a pure element in it’s standard state is 0.00 kJ/mol
- Which are not elements in their standard states?:

H2(g) Br2(l) Ag(s) KF(s) Hg(l)Fe+3(aq) O3(g)

ΔHof = 0.00 kJ/mol for each except: KF(s) Fe+3(aq) O3(g)

- Table 6.2, pg250 & appendix C have tables of ΔHof values.
- ΔHr = ∑ ΔHof products– ∑ ΔHofreactants
(use signs given in Table & Table values are kJ/1Mole)

- Calculate ΔHr for: 2 NO2(g) ----) 1 N2O4(g)
Table 6.2:ΔHof = 33.2 for NO2& 9.16 for N2O4(kJ/1 mol)

2 NO2(g) ----) 1 N2O4(g)

2 x 33.21 x 9.16

ΔHr = ∑ ΔHof products– ∑ ΔHofreactants

ΔHr = [1 x 9.16] - [2 x 33.2] = - 57.2 kJ

- Calculate ΔHr for: 2Cdiamond + 1O2(g) -----) 2CO(g)
Pg 250: ΔHof = 2 kJ/m For Cdiamond & - 111 kJ/m for CO(g)

2Cdiamond + 1O2(g) -----) 2CO(g)

2 x 2 1 x 0 2 x -111

4 0 -222

ΔHr = ∑ ΔHof products – ∑ ΔHofreactants

ΔHr = [ - 222] - [ 4 + 0] = - 226 kJ

Note: ΔHof = 0.00 for O2 (g) & will not be given on exams

- Calculate ΔHr for following at 25oC & 1 atm pressure.
GetHof from Table 6.2, pg 250 (element in Std. St. = 0); Values in kJ/mol.

3 CH4(g) + 2 H2O(l) + 1 CO2(g) -----) 4 CO(g) + 8 H2(g)

Hof : 3(-74.9) 2(-286) 1(-394) 4(-111) 8(0)

-225 -572 -394 -444 0

ΔHr = ∑ ΔHof products– ∑ ΔHofreactants

ΔHr = [-444 + 0] - [-225 + -572 + -394] = + 747 kJ

1) Reactions in Calorimeters: q = C x ∆T

2) Reactions in Calorimeters: q = s x g x ∆T

3) Hess’s Law:Add & Subtract Rxns

4) Bond Energies:[take # mols & # bonds (Lewis Structure) into account; use Table 9.5, pg 367]

∆Hr = ∑ ∆Hbonds broken- ∑ ∆Hbonds formed

5)Heats of Formation, ΔHof: (from tables: pg 250 or pg A-8; take # mols ito account & ΔHof = 0 for element in it’s standard state)

∆Hr = ∑ ∆Hof products– ∑ ∆Hofreactants

Note: Rules 1 and 4 are double checks on your structure.

1. Calculate total # of valence electrons - take into account the charge if it is a polyatomic ion.

2. Place atom that forms most bonds at center (Closest to Group IVA/14 & Lowest if in same group). If is a charge, then take into account on central atom.

3. Arrange other atoms around central atom & allow sharing so that each atom has stable electron configuration. Show bonding pairs as lines & nonbonding valence e- as dots.

4. Double check rule: a) see that each atom has a stable electron configuration and b) check that you still have the same number of valence electrons that you started with in step #1.

Useful generalities: For neutral, Stable Compounds: H, F, Cl, Br have 1 bond; C has 4 bonds; N has 3 bonds; O has 2 bonds. B stable with 6 electrons/3 bonds; H stable with 2 electrons.

Try with: H2Se AlF4- H2CO N2F2 C2H2 C2H4 C2H6