Ch17 – Thermochemistry Ch17.1 Temp vs Heat

Ch17 – Thermochemistry

Ch17.1 Temp vs HeatAll chemical reactions and changes in state either release or absorb energy.

Energy- ability to do work or supply heat.

Heat- a form of energy that always flows from hotter to cooler objects.

Law of Conservation of Energy- in a chemical or physical progress, energy is neither created or destroyed it only changes form.

Exs: 2CO(g) + O2(g) 2CO2(g) + 221,000 Joules of Energy given off

H2O(g) H2O(l) + 40,700 Joules Metric system unit of energy

Heat Capacity - Each substance absorbs or releases heat differently.

Specific Heat Capacity (Cp) - amount of heat necessary too raise the temp

of 1 gram of a substance 1˚C.

Ex: H2O: Requires 4.186 Joules of heat energy to raise the temp of 1g 1˚c.(originally defined as a calorie)

1 Cal = 4.186J1000 cal = 1 kcal = 1 Calorie(food)

Units:

Specific Heat Capacity Formula

Q = m · Cp · ΔTChange in Temp

Heat Energy MassSpecific Heat

Temp ReviewK = C + 273 “K is always bigger”

Ex1) 20˚C = ____ K “Even through C & K differ by 273,Ex2) 100K = _____˚C they use the same increments.” Ex3) How much heat is required to raise the temp of 50.0g of water

from 20˚C to 100˚C?Cp for H2O = 4.186 J/g·˚C

Ex4) How much heat is required to raise the temp of 50.0g of iron from 20˚C to 100˚C?

Cp for iron = 0.46 J/g·˚C

Ex5) What is the specific heat capacity of a piece of copper with a mass of 95.4g, if it absorbs 849J of heat to change its temp from 298K to 321K?

Ch17 HW#1 1 – 9

Ch17 HW#11) 500oC = ____K 2) 500K = ____ oC

3) 20oC = ____ K 4) 100oC = ____ K

5) Law of Cons of Energy

6) 1 Cal = J

Ch17 HW#11) 500oC = 773 K 2) 500K = 227oC

3) 20oC = 293 K 4) 100oC = 373 K

5) Law of Cons of Energy

6) 1 Cal = J

Ch17 HW#11) 500oC = 773 K 2) 500K = 227oC

3) 20oC = 293 K 4) 100oC = 373 K

5) Law of Cons of Energy – Energy cannot be created or destroys, it can only change forms.

6) 1 Cal = J

Ch17 HW#11) 500oC = 773 K 2) 500K = 227oC

3) 20oC = 293 K 4) 100oC = 373 K

5) Law of Cons of Energy – Energy cannot be created or destroys, it can only change forms.

6) 1 Cal = J

JcalJ

CalcalCal 4186

1186.4

110001

7) How much heat to raise temp of 20.0g of wood from 20oC to 100oC? Cp= 1.8 J/ g.˚C

8) How much heat to raise 20.0g iron from 20oC to 100oC? Cp= 0.46 J/ g.˚C

9) Which hot faster – iron or wooden spoon?


Q = m · Cp · ΔT = (20g)(1.8 J/ g.˚C)(80˚C) = 2800J




Q = m · Cp · ΔT = (20g)(1.8J/ g.˚C)(80˚C) = 2800J


Q = m · Cp · ΔT = (20g)(0.46J/ g.˚C)(80˚C) = 736J


Iron

Ch17.2 CalorimetryEx1) 100.00g of an unknown substance is placed in a test tube, which is placed in a breaker of boiling water at 99.5oC. The sample is removed and placed in a calorimeter with 100.00g of water at 20.0oC.The contents come to equilibrium at 22.4oC. What is the Cp ofthe unknown substance?

Ex2) Lab group # ____ masses their unknown at 18.5g. They place it in a test tube in boiling water measured at 94oC. Meanwhile, theymass an empty calorimeter at 2.2g, half–fill it with tap water and re-massit at 186.6g. (Mwater = 184.4g) They measure the temp of the water with the same thermometer to be 23oC. They not-so-carefully place the unknown in the calorimeter, and the temp rises to 24oC. What Cp didthey get? If the actual is 0.700 J/g.oC, will they get unknown points?

% error:

Ch17 HW#2 10 – 13

Ch17 HW#2 10 – 13 10) A 100.00g sample of iron at 0°C absorbs 1350 joules of heat.

What is the final temp? Cp for iron is 0.46 J/g ·°C

11) 50.04g unknown metal heated to 98.2°C. Placed in calorimeter with51.12g water at 20.0°C. Equilibrium at 23.0°C. Find Cp:


What is the final temp? Cp for iron is 0.46 J/g ·°CQ = m · Cp · ΔT Q = m · Cp · (Tf – Ti)

1350J = (20g)(0.46J/g.˚C)(Tf – 0°C) Tf = 29.3°C

11) 50.04g unknown metal heated to 98.2°C. Placed in calorimeter with51.12g water at 20.0°C. Equilibrium at 23.0°C. Find Cp:


What is the final temp? Cp for iron is 0.46 J/g ·°CQ = m · Cp · ΔT Q = m · Cp · (Tf – Ti)

1350J = (20g)(0.46J/g.˚C)(Tf – 0°C) Tf = 29.3°C

11) 50.04g unknown metal heated to 98.2°C. Placed in calorimeter with51.12g water at 20.0°C. Equilibrium at 23.0°C. Find Cp.

Metal Water Qlost = Qgained

m · Cp · ΔT = m · Cp · ΔT (50.0)(Cp)(73.2) = (51.13)(4.186)(3)

Cp = 0.175

12) 23.95g unknown heated to 95.1°C. Placed in calorimeter with 50.01g water at 21.1°C. Equilibrium at 22.2°C. Cp?



Unknown Water Qlost = Qgained

m · Cp · ΔT = m · Cp · ΔT (23.95)(Cp)(72.9) = (50.00)(4.186)(1.1)

Cp = 0.132




m · Cp · ΔT = m · Cp · ΔT (23.95)(Cp)(72.9) = (50.00)(4.186)(1.1)

Cp = 0.132



m · Cp · ΔT = m · Cp · ΔT (34.55)(Cp)(76.4) = (49.98)(4.186)(3.8)

Cp = 0.301

Lab 17.1 Write Up1. Mass of Empty Test Tube: _________ g2. Mass of test tube and metal: _________ g3. Mass of metal: _________ g4. Mass of empty cup: _________ g5. Mass of cup and distilled H20: _________ g6. Mass of water in cup: _________ g7. Temperature of boiling water: _________˚C8. Initial temperature of metal: _________˚C9. Initial temp of H20 in cup: _________˚C10. Temp of H20 with metal: _________˚C11. Change in temp. of metal: _________˚C

How would you calculate heat lost by the metal?Heat lost by metal = (Cp metal)(Mmetal)(ΔTmetal) =

12. Change in temp. H20: _________˚C How do you find the heat gained by the water?Heat gained by water = (Cp water)(Mmetal)(ΔTmetal) =

What is the relationship between the heat lost by the metal and the heat gained by the water?

That’s right!!! They are equal. Then set them equal and solve for Cp metal.

(Cp water)(Mmetal)(ΔTmetal) = (Cp water)(Mwater)(ΔT water)

Scores out of 12 points: %error < 5% = 20 %error < 10% = 18 %error < 15% = 16 %error < 20% = 14 %error < 25% = 12 %error > 25% = 10

And lose points for lack of precise measuring!

))(())()((

metalmetal

waterpwaterwaterp TM

TCMC

Ch17.2 contEx1) 45.05g unknown heated to 92.5°C. Placed in calorimeter with 50.50g

water at 20.5°C. Equilibrium at 23.2°C. Cp?

Ex2) 122.25g unknown heated to 101.2°C. Placed in calorimeter with 75.00g water at 20.1°C. Equilibrium at 22.2°C. Cp?

Ch17 HW#2B1. 50.00g unknown heated to 95.0°C. Placed in calorimeter with 50.00g

water at 20.0°C. Equilibrium at 21.0°C. Cp?

2. 100.00g unknown heated to 100.0°C. Placed in calorimeter with 50.00g water at 20.0°C. Equilibrium at 25.0°C. Cp?

Lab13.1 – Calorimetry

- unknown due at end of period

- lab due in 3 days

- Ch17 HW#2 due at beginning of period

Ch17.3 – Heat & Changes of State

Time

Temp (°C)

120

100

0-10

For H2O:

Ch12.3 – Heat & Changes of State

Time

Temp (°C)

120

100

0-10

s l1) Raise temp of solid: Q = m ∙Cp ∙ ∆T

l g

2) Melt it: Q = n ∙ Hf

3) Raise temp of liquid: Q = m ∙Cp ∙ ∆T moles!

4) Vaporize it: Q = n ∙ Hv

5) Raise temp of gas: Q = m ∙Cp ∙ ∆TFor H2O:

6) Add steps Step 1: Cp for ice = 2.03 J/g.˚CStep 2 is called Heat of Fusion (melting/freezing): Hf = 6010 J/molStep 3: Cp for water = 4.186 J/g.˚CStep 4 is called Heat of Vaporization (evap/condensing): Hv = 40,700 J/molStep 5: Cp for steam = 1.7 J/g.˚C

Step 1: Raise temp of ice from initial temp to melting point (0°C) Q= m·Cp·ΔT Cp ice = 2.03 J/g·°C

Step 2: Melt solid→ Heat of FusionQ=n·Hf You”ll have to convert grams to moles

Hf for H2O= 6010 J/moleDuring step 2, all heat energy (Q) goes into breaking thebonds of the solid into liquid.Use same formula for solidifying a liquid , only Q is (-) Heat given off rather than absorbed.

Step 3; Raise temp of liquid from 0°C to 100°C Q= m·Cp·ΔT Cp water= 4.186 J/g·˚C

Step 4: Boil liquid→ Heat of Vaporization Q=n·Hv Hv for H2O = 40,700 J/mol

Step 5: Raise temp of gas from 100˚C to final temp.Q=m·Cp·ΔT Cp=1.7 J/g·°C

Step 6: Add up steps.

Ex1) How much heat is required to turn 63.25g of ice at -50.4˚C to a vapor at 124˚

Time

Temp (°C)

124

100

0-50.4

Ex1) How much heat is required to turn 63.25g of ice at -50.4˚C to a vapor at 124˚

Ch17 HW#3

Time

Temp (°C)

124

100

0-50.4

Step 1: Q = m.Cp.ΔT = (63.25g)(2.03J/g.˚C)(50.4˚C) = 6694.4J

Step 2: Q = n.Hf = (3.51mol)(6010J/mol) = 21,100JStep 3: Q = m.Cp

.ΔT = (63.25g)(4.186J/g.˚C)(100˚C) = 26,477JStep 4: Q = n.Hv = (3.51mol)(40,700J/mol) = 143,000JStep 5: Q = m.Cp

.ΔT = (63.25g)(1.7J/g.˚C)(24˚C) = 2623JStep 6: Add: = 199,894J

OHmolg

molg2 51.3

0.18125.63

Ch17 HW#314) 100.00g sample of Al at 0oC absorbs 1350J of heat.

Cp = .90J/g.oC. Final temp?

15) How much heat to bring 5.0g ice from -10oC to 120oC?Step 1: Q = m.Cp

.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = Step 2: Q = n.Hf = (.28mol)(6010J/mol) = Step 3: Q = m.Cp

.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = Step 4: Q = n.Hv = (.28mol)(40,700J/mol) = Step 5: Q = m.Cp

.ΔT = (5.0g)(1.7J/g.˚C)(20˚C) = Step 6: Add: =

16) How much heat given off 10.5g steam at 150oC to water at 75oC?Step 5: Q = m.Cp

.ΔT = (10.5g)(1.7J/g.˚C)(-50˚C) =Step 4: Q = n.Hv = (.56mol)(-40,700J/mol) =Step 3: Q = m.Cp

.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) =Step 6: Add: =


Cp = .90J/g.oC. Final temp?Q = m · Cp · ΔT Q = m · Cp · (Tf – Ti)

1350J = (100g)(0.90J/g.˚C)(Tf – 0°C) Tf = 15°C


.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = Step 2: Q = n.Hf = (.28mol)(6010J/mol) = Step 3: Q = m.Cp

.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = Step 4: Q = n.Hv = (.28mol)(40,700J/mol) = Step 5: Q = m.Cp

.ΔT = (5.0g)(1.7J/g.˚C)(20˚C) = Step 6: Add: =



.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) =Step 6: Add: =


Cp = .90J/g.oC. Final temp?Q = m · Cp · ΔT Q = m · Cp · (Tf – Ti) 1350J = (100g)(0.90J/g.˚C)(Tf – 0°C)

Tf = 15°C15) How much heat to bring 5.0g ice from -10oC to 120oC?Step 1: Q = m.Cp

.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = 105JStep 2: Q = n.Hf = (.28mol)(6010J/mol) = 1669JStep 3: Q = m.Cp

.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = 2093JStep 4: Q = n.Hv = (.28mol)(40,700J/mol) = 11,306JStep 5: Q = m.Cp




.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) =Step 6: Add: =


Cp = .90J/g.oC. Final temp?


.ΔT = (5.0g)(2.03J/g.˚C)(10˚C) = 105JStep 2: Q = n.Hf = (.28mol)(6010J/mol) = 1669JStep 3: Q = m.Cp

.ΔT = (5.0g)(4.186J/g.˚C)(100˚C) = 2093JStep 4: Q = n.Hv = (.28mol)(40,700J/mol) = 11,306JStep 5: Q = m.Cp



.ΔT = (10.5g)(1.7J/g.˚C)(-50˚C) = -893JStep 4: Q = n.Hv = (.56mol)(-40,700J/mol) = -23,742JStep 3: Q = m.Cp

.ΔT = (10.5g)(4.186J/g.˚C)(-25˚C) = -1099JStep 6: Add: = -25,734J

Ch17.4 EnthalpyEnthalpy – heat content (H) Every chemical reaction, every change of state,

and any 2 objects at different temperatures will have a change in enthalpy. (ΔH=Q)

Exothermic reaction – a chemical reaction where heat is given off (ΔH=(-))

Ex1) CaO(s) + H2O(l) Ca(OH)2(aq) ΔH = – 82kJ the (-) is just an indicator that heat is given off (feels hot).

- how much heat is given off when 1 mole CaO reacts? - how much heat is given off when 2 moles CaO reacts? - how much heat is given off when 49.36g CaO reacts?

Endothermic reaction – chemical reaction where heat is absorbed (ΔH=(+))Ex2) Ba(OH)(aq) + NH4Cl(s) BaCl2(aq) + NH3(aq) + H2O(l)

ΔH = +50kJ (feels cold)

Write the equation to include ΔH in the equation, and balance.

- how much heat is absorbed when 10.42g of BaCl2 is produced?

Ch17 HW#4 17 – 20

Lab17.2 – Heat of Fusion

- Lab due tomorrow

- Ch17 HW#4 due at beginning of period

Ch17 HW#4 17 – 20 17) How much heat required to raise temp of 56.02g of water from 20.0˚C

to 85˚C?

18) a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH= -890 kJ

b) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH= -1368 kJ


to 85˚C?

Q = m.Cp.ΔT = (56.02g)(4.186J/g.˚C)(65˚C) = 15,242J




to 85˚C?

Q = m.Cp.ΔT = (56.02g)(4.186J/g.˚C)(65˚C) = 15,242J


CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 890 kJ


C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) +1368 kJ

19) 2NaHCO2(s) Na2CO3(s) + H2O(l) + CO2(g) ΔH= +129 kJ a) Rewrite with ΔH

b) How much heat to produce 125.00g Na2CO3

c) How much heat to produce 44.8L C02

20) NH4NO3 NH4+ + NO3

- ΔH= +25.7 kJ

a)Rewrite

b) How much heat required to dissolve 95.00g NH4N03?

19) 2NaHCO2(s) Na2CO3(s) + H2O(l) + CO2(g) ΔH= +129 kJ a) Rewrite with ΔH 2NaHCO2(s) +129 kJ Na2CO3(s) + H2O(l) + CO2(g)



20) NH4NO3 NH4+ + NO3

- ΔH= +25.7 kJ

a)Rewrite


Jmol

Jg

molg836,151

CONa 1000,129

CONa 2.106CONa 1CONa 00.125

3232

3232




20) NH4NO3 NH4+ + NO3

- ΔH= +25.7 kJ

a)Rewrite


Jmol

Jg

molg836,151

CONa 1000,129


3232

3232

Jmol

JL

molL000,1258

CO 1000,129

CO 4.22CO 1CO 8.44

22

22




20) NH4NO3 NH4+ + NO3

- ΔH= +25.7 kJ

a)Rewrite NH4NO3 +25.7 kJ NH4+ + NO3

- b) How much heat required to dissolve 95.00g NH4N03?

Jmol

Jg

molg836,151

CONa 1000,129


3232

3232

Jmol

JL

molL000,1258

CO 1000,129

CO 4.22CO 1CO 8.44

22

22

Jmol

Jg

molg519,30

NONH 1700,25

NONH 0.80NONH 1NONH 00.95

3434

3434

Ch17 Mid Ch Rev27) How much heat will be absorbed by 320g of water when its temp

is raised by 35˚C?

28) How much heat will be given off by 55g of water as it cools from 87˚C to 25˚C?

29) Calculate the specific heat of glass if 65 grams of it increases its temp by 26˚C when it absorbs 840 J of energy.

30) Calc temp change if 160 g of Hg absorbs 1500J of heat. (Cp is 0.14 J/g·˚C)

31) An unknown 15.00g metal sample is heated in a hot water bath until it reaches 96.2˚C. It is then immediately placed in a calorimeter, filled with75.00g of H20 at 19.6˚C. The calorimeter temp levels at 25˚C. Find Cp of unknown.

32) An unknown 25.00g metal sample is heated to 94.3˚C. Placed in calorimeter with 115.00g of H20 at 21.1˚C. Calorimeter temp levels at 22.5˚C. Find Cp of unknown.

33) a) 2K + Br2 2 KBr + 784 kJ ΔH= b) H20 + 286 kJ H2 + ½O2 ΔH=

Ch17.5 – Thermo EquationsIn all chemical processes, heat is absorbed or given off.

Old bonds break (requires energy)New bonds form (releases energy)(One of these 2 will be a greater amount.) These signs areLook at ΔH: Heat given off ΔH = –only indicators of

Heat absorbed ΔH = + which side to putExothermic reaction – heat given off ΔH. DON’T USE!

CaO + H2O Ca(OH)2 + 82,000J




CaO + H2O Ca(OH)2 + 82,000JExothermic energy diagram

Enthalpy (kJ)

Time




CaO + H2O Ca(OH)2 + 82,000JExothermic energy diagram

Enthalpy (kJ)

TimeΔH = Hproducts – Hreactants = (small #) – (big #)ΔH = (–) (Just an indicator that heat is given off)

Endothermic Reaction – heat absorbed (feels cold) Ba(OH)2 + NH4Cl + 50,000J BaCl2 + NH3 + H2O

Endothermic energy diagram

Enthalpy (kJ)

Time

ΔH = Hproducts – Hreactants = (big #) – (small #)ΔH = (+) (Just an indicator that heat is absorbed)

Change of state absorbs or releases energy too.

Heat of fusion: (s l) absorbs energy H (+)(l s) releases energy H (-)

Heat of vaporization: (l g) absorbs, H (+) (g l) releases, H (-)

Physical state of matter too!

Exs: H2O(l) H2(g) + O2(g) + ΔH = +285.8kJ H2O(g) H2(g) + O2(g) + ΔH = +241.8kJ

Ex1) Ethanol, C2H5OH, reacts with oxygen to produce…


CO2 & H2O & 1235 kJ of heat is given off. 1. Write eqn. 2. Draw an energy diagram. 3. How much heat is produced when 12.5g of ethanol reacts?

1) C2H5OH + 3O2 2 CO2 + 3 H2O + 1235 kJ


CO2 & H2O & 1366.7 kJ of heat is given off. 1. Write eqn. 2. Draw an energy diagram. 3. How much heat is produced when 12.5g of ethanol reacts?

1) C2H5OH + 3O2 2 CO2 + 3 H2O + 1366.7 kJ 12.5g ?

exothermic

12.5g C2H5 OH 1mol C2H5OH 1366.7 kJ 46 g C2H5OH 1 mol C2H5OH

2 C @ 12.0 = 24.06 H @ 1.0 = 6.0 46.0 g/mol1 O @ 16.0 = 16.0

= ____ kJ

HW#21) How much heat will be released when 6.44 g of sulfur reacts with excess O2?

2S + 3O2 → 2SO3 ΔH= -791.4 kJ

HW#21) How much heat will be released when 6.44 g of sulfur reacts with excess O2?

2S + 3O2 → 2SO3 ΔH= -791.4 kJ

6.44g ?

6.44g S 1 mol S 791.4 kJ 32.1g S 2 mol S

Ch 17 HW #5 21-26

2S + 3O2 → 2SO3 + 791.4 kJ/kJ

=79.8 kJ

Ch17 HW#5 21 – 2621) (in class)22) How much heat absorbed when 38.2g Br2 reacts:

H2 + Br2 2HBr ΔH= +72.8kJ

23) Heat released when 4.77g ethanol reacts with excess O2? C2H5OH + 3O2 2CO2 + 3H2O + 1366.7 kJ

24) How much heat absorbed when 13.7g N2 reacts? N2 + O2 + 180kJ 2NO





kJmol

kJg

molg4.17

Br 180.78

Br 8.159Br 1Br 2.38

22

22





kJmol

kJg

molg4.17

Br 180.78

Br 8.159Br 1Br 2.38

22

22

kJmol

kJg

molg7.141

OHHC 17.1366

OHHC 0.46OHHC 1OHHC 77.4

2222

2222





kJmol

kJg

molg4.17

Br 180.78

Br 8.159Br 1Br 2.38

22

22

kJmol

kJg

molg7.141

OHHC 17.1366

OHHC 0.46OHHC 1OHHC 77.4

2222

2222

kJmol

kJg

molg1.88

N 1180

N 0.28N 1N 7.13

22

22

25) How much heat released when 11.8g of Fe react? 3Fe + 2O2 2Fe3O4 + 1120.4kJ

26) How much heat released when 18.6g H2 reacts? 2H2 + O2 2H2O + 571.6kJ



kJmol

kJg

molg 0.79Fe 14.1120

Fe 8.55Fe 1Fe 8.11



kJmol

kJg

molg 0.79Fe 14.1120

Fe 8.55Fe 1Fe 8.11

kJmol

kJg

molg2658

H 16.571

H 0.2H 1H 6.18

22

22

Ch17 Review1) If 1Cal = 4.18kJ, how many KJ of energy in a Snickers

that has 250Cal?

2) How much heat is required to raise the temp of 10g of H2O(l) at 50oC to H2O(g) at 150o?

3) How much heat releases if 20g of Pb(s) is consumed in the followingreaction:

Pb(s) + 2Cl2(g) PbCl2(l) + 329.2kJ

4) Given the following reaction: 2 Al2O3(s) 4 Al(s) + 3 O2(g)

ΔH = +3352kJa. Is the reaction endo or exothermic?b. Write the rxn w ΔH.c. How much Al produced if 10,000kJ of heat is absorbed?d. Draw an energy diagram.

5) A 21.9g unknown metal sample is heated to a temp of 92.3oC.It is placed in a calorimeter with 31.5g of water at 20.1oC.After a few minutes, they reach equilibrium at 22.2oC Cp = ?

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Ch17 – Thermochemistry Ch17.1 Temp vs Heat