309 SP 11-4 HEAT OF FUSION

309 SP 11-4 HEAT OF FUSION• HOW MANY GRAMS OF ICE AT 0°C AND 101.3KJ

COULD BE MELTED BY THE ADDITION OF 2.25 KJ OF HEAT ?

HEAT OF FUSION = 6.01 KJ / 1 MOLE HFUS

2.25 kj1 =

1 MOLE6.01 kj

.37 MOLES 18 g1 MOLE = 6.74 g

1 MOLE OF WATER ( H20) IS 1g + 1g + 16 g = 18g

=

IT TAKE 6.01 KJ ( 6010) SMALL MATCHES TO MELT 18 g

x x

309 SP 11-5 H of VAPORIZATON• HOW MUCH HEAT IN KJ, IS ABSORBED WHEN

24.8g OF WATER AT 100ºC IS CONVERTED TO STEAM AT 100 ºC ?

H2O (L) + 40.7 KJ ----> H20 (g)

IT TAKE 40.7 KJ TO CONVERT 1 MOLE OF LIQUID

WATER AT 100 ºC TO STEAM AT 100 ºC .

24.8 g 1

1 MOLE 18 g

1.38 MOLES= 40.7 kj 1 MOLE

= 56.1 KJ

SP 11-6 HEAT OF SOLUTION• HOW MUCH HEAT ( IN kJ) IS RELEASED WHEN

2.5 MOLES OF Noah IS DISSOLVED IN WATER ?

HEAT OF SOLUTION IS - 445.1 kj / MOLE

H Soln = - 445 .1 kj / MOLE

NaOH (s) ----> Na+1 (aq) + OH - (aq)

WHEN 1 MOLE IS DISSOLVED IN WATER, 445 kj IS RELEASED

2.5 MOLES 1

- 445 kj 1 MOLE

= - 1113 kj

SP 11-6 HEAT OF SOLUTION

HOW MUCH HEAT INKJ IS RELEASED WHEN 2.5 MOLES OF NaOH IS DISSOLVED IN WATER

NaOH (s) -------> Na + + OH - + 445 KJ/ MOLE

2.5 MOLES 1

445 KJ1 MOLE = -113 KJ

HEAT CHANGE IS - 445 KJ BECAUSE THIS IS EXOTHERMIC…. HEAT IS LOST TO THE WATER

START OF 11.1

TERMS FOR TEST 11

• 1. THERMOCHEMICAL EQUATION• 2 JOULE• 3 ENDOTHERMIC PROCESS• 4 CHEMICAL POTENTIAL ENERGY• 5 HEAT OF COMBUSTION• 6 CALORIMETRY• 7 THERMOCHEMISTRY• 8 MOLAR HEAT OF FUSION

SAMPLE PROBLEMS FOR TEST 11• 11-1 FINDING SPECIFIC HEAT

• 11-2 CALCULATING HEAT

• 11-3 THERMOCHEMICAL REACTIONS

• 11-4 HEAT OF FUSION

• 11-5 SKIP

• 11-6 HEAT OF SOLUTION

312 SP11-2 MOLAR HEAT OF SOLUTION

IS THE AMOUNT OF HEAT CREATED BY DISSOLVING 1 MOLE OF A SUBSTANCE IN WATER.

1000g

22º

52º

HEAT OF SOLUTION= g OF WATER x SH x TEMP. CHANGE

1000g

4.18 30º

= 125. 4 kj

CHAPTER 11: THERMOCHEMISTRY

HEAT AND CHEMICAL CHANGE

•11.1 THE FLOW OF ENERGY and HEAT

•11.2 MEASURING AND EXPRESSING HEAT CHANGES •11.3 HEAT IN CHANGES OF STATE

•11.4 CALCULATING HEAT CHANGES

293 TERM for 11.l• 1.THERMOCHEMISTRY

• 2,ENERGY

• 3.CHEMICAL POTENTIAL

ENERGY

• 4.HEAT

• 5.SYSTEM

• 6. SURROUNDING

• 7. UNIVERSE

• 8. LAW OF

CONSERVATION OF

ENERGY

• 9. ENDOTHERMIC

• 10. EXOTHERMIC

11. CALORIE•12. JOULE•13. HEAT CAPACITY•14. SPECIFIC HEAT CAPACITY•15. SPECIFIC HEAT

294 EXOEXOTHERMIC PROCESS

EXOTHERMIC PROCESS•THE WATER (SURROUNDINGS) GET WARMER•HEAT (CALORIES) MOVE INTO THE WATER•THE SYSTEM ( COPPER) GETS COLDER

HEAT IN WATER = 2HEAT IN COPPER = 2TOTAL HEAT = 4


100 -c

10 -c

50-c50 -c

HOTCOPPERPLACED

INCOLD

WATER

HEATaka

CALORIE

294 EXOEXOTHERMIC PROCESS




100 -c

10 -c

50-c50 -c

HOTCOPPERPLACED

INCOLD

WATER

HEATaka

CALORIE

299 SP 11-1 SPECIFIC HEAT

THE TEMPERATURE OF A PIECE OF Cu WITH A MASS OF95.4 g INCREASES FROM 25 -C TO 48 -C WHEN THE METALABSORBS 849 J OF HEAT. WHAT IS THE SPECIFIC HEAT ?

FORMULA

x

SPECIFIC HEAT =HEAT in J or C

MASS TEMP CHANGE

849 J

95.4 g x 23 - C

849j

2194 gxC

.387 j1g x -C

294 EXOTHERMIC PROCESS




100 -c

10 -c

50-c50 -c

HOTCOPPERPLACED

INCOLD

WATER

HEATaka

CALORIE

294 ENDOENDOTHERMIC PROCESS

ENDOTHERMIC PROCESS•THE WATER (SURROUNDINGS) GET COLDER•HEAT (CALORIES) MOVE FROM THE WATER•THE SYSTEM ( COPPER) GET WARMER



10 -c

100 -c

50-c50 -c

OPPOSITE OF EXOTHERMIC PROCESS

HEAT FROMTHE

SURROUNDINGSIS LOST TO THE

SYSTEM

293 TERMS 1-51. THE STUDY OF HEAT CHANGES IN CHEMISTRY.

2. THE CAPACITY TO DO WORK OR SUPPLY HEAT

3. ENERGY STORES IN CHEMICAL SUBSTANCES

THERMOCHEMISTY

ENERGY

CHEMICAL POTENTIAL ENERGY

4. THE ENERGY TRANSFERED FROM ONE OBJECT TO ANOTHER BECAUSE OF TEMPERATURE DIFFERENCES.

5. PART OF THE UNIVERSE WE FOCUS ON

HEAT

SYSTEM

293 TERMS 6-101. IMMEDIATE VICINITY OF SYSTEM

2. SYSTEM + SURROUNDING

3. ENERGY IS NEIHER CREATED OR DESTROYED

SURROUNDINGS

UNIVERSE

LAW OF CONV. OF ENERGY

4. ABSORBS HEAT FROM THE SURROUNDINNGS

5. RELEASES HEAT TO ITS SURROUNDINGS

ENDOTHERMIC

EXOTHERMIC

293 TERMS 11-151. HEAT THAT RAISES 1 g OF WATER 1 -C

2. SI UNIT OF HEAT …… ALSO CALORIE = 4.18 OF THESE

3. NEEDED TO RAISE THE TEMP. OF AN OBECT 1 DEGREE.

CALORIE

JOULE

HEAT CAPACITY

4. RAISES 1 g OF ANY SUBSTANCES 1 DEGREE.

5.

SPECIFIC HEAT CAPACITY

EXOTHERMIC

295 CALORIE vs JOULEUNITS FOR HEAT

1 CALORIELARGE MATCH EQUALS

4,18 JOULES4 SMALL MATCHES

1 CALORIE or 4.18 JOULES IS THE AMOUNT OF HEAT NEEDED TO RAISE THE TEMPERATURE OF 1 g ( 1 mL) OF WATER 1 DEGREE.

1 GRAM OF WATER

56 DEGREESTO

57 DEGREES

295: SPECIFIC HEATWATER ALCOHOL IRON

1 CALORIE O.58 CALORIES 0.11 CALORIES

1 GRAM HEATED 1 DEGREE HOTTER.

SPECIFIC HEAT IS THE AMOUNT OF HEAT ( CAL OR J)NEEDED TO RAISE 1 g OF ANY SUBSTANCE 1 DEGREE

IRON ISVERY EASY

TO HEAT

297UNITS FOR SPECIFIC HEAT

• HEAT CAN BE EXPRESSED AS:– CALORIES– JOULES

SPECIFIC HEAT =HEAT ( C or J)

GRAMS x TEMP. CHANGE

JOULES

GRAMS x -C

296 HEAT CAPACITY• THE HEAT CAPACITY IS THE AMOUNT OF HEAT

NEEDED TO INCREASE THE TEMP. OF AN OBJECT ( OF ANY SIZE ) ONE DEGREE.

34.6 g34.6 g

1 g1 g

HEAT CAPACITY USED FORANY MASS.

SPECIFIC HEAT USED FORA MASS OF 1 GRAM

CALORIES vs kCAL vs JOULES

1 CALORIE = 4.18 J

1KCAL = 1000 CALORIES

299 SP 11-1 SPECIFIC HEAT

THE TEMPERATURE OF A PIECE OF Cu WITH A MASS OF40.0 g INCREASES FROM 25 -C TO 50 -C WHEN THE METALABSORBS 400 J OF HEAT. WHAT IS THE SPECIFIC HEAT ?

FORMULA

x

SPECIFIC HEAT =HEAT in J or C

MASS TEMP CHANGE

400 J

40 g x 25 - C

400

1000 gxC

.40 J1g x -C

299 PP 1 FIND S.H.

WHEN 435 J OF HEAT IS ADDED TO 3.4 g OF OLIVE OIL AT 21 -C, THE TEMPERATURE INCREASES TO 85 -C. WHAT IS THE SPECIFIC HEAT?

SPECIFIC HEAT = HEAT (J)MASS x TEMP

435 J

3.4 g x 64 -C

435 J

218

2.0 J / g x -C= =

299 PP 2 FIND S.H.

A PIECE OF STEEL WEIGHING 1.55 g ABSORBS 141 J OF HEAT WHEN ITS TEMP.

INCREASES 178 -C. FIND THE S.H.IN J. ?

SPECIFIC HEAT = HEAT (J)MASS x TEMP

141 J

1.55 g x 178 -C

141 J

276

.51 J / g x -C= =

START 11.2

300 CALORIMETRYUSING A SODA CAN

• CALORIMETRY IS THE MEASUREMENT OF HEAT CHANGE IN CHEMICAL REACTIONS

• CALORIMETER IS DEVICE USED TO MEASURE HEAT CHANGE ( SODA CAN)

HOW MUCH HEAT, IS NEEDED TO RAISE100g OF WATER 10 DEGREES ? ( CAL)

CALORIES = MASS x SH x TEMP. CHANGE

KILOCALORIES =

JOULES =

g CAL -C=

301 HEAT SIGNS• EXOTHERMIC REACTION

– H IS NEGATIVE– THE WATER GETS HOTTER

• ENDOTHERMIC REACTION– H IS POSITIVE– THE WATER GETS COLDER

312 MOLAR HEAT OF SOLUTION

IS THE AMOUNT OF HEAT CREATED BY DISSOLVING 1 MOLE OF A SUBSTANCE IN WATER.

1000g

22º

52º

HEAT OF SOLUTION= g OF WATER x SH x TEMP. CHANGE

1000g

4.18 30º

= 125. 4 kj

LIKE SP 11-6 HEAT OF SOLUTION

HOW MUCH HEAT ( IN kJ) IS RELEASED WHEN

0.5 MOLES OF NaOH IS DISSOLVED IN WATER ?

LIKE SP 11-6 HEAT OF SOLUTION

HOW MUCH HEAT ( IN kJ) IS RELEASED WHEN

0.5 MOLES OF NaOH IS DISSOLVED IN WATER ?

HEAT OF SOLUTION IS - 445.1 kj / MOLE

H Soln = - 445 .1 kj / MOLE

NaOH (s) ----> Na+1 (aq) + OH - (aq)

WHEN 1 MOLE IS DISSOLVED IN WATER, 445 kj IS RELEASED

0.5 MOLES 1

- 445 kj 1 MOLE

= - 222.5 kj

304 ENDO VS EXO LAB

NaOH

1 MASS OF H2O 21°2 START TEMP ____3 END TEMP ____4 CHANGE ____5 SH IN JOULES 4.18

6 WATER H or C

7 TYPE ENDO or EXO

8 JOULES _____

NaNO3

1 MASS OF H2O ____2 START TEMP ____3 END TEMP ____4 CHANGE ____5 SH IN JOULES 4.18

6 WATER H or C

7 TYPE ENDO or EXO

8 JOULES _____

304 ENDO VS EXO LAB

NaOH NaNO31 10 g2345678

12 21º3 17º45678

10 g

301 HEAT and ENTHALPY• ENTHALPY, IN THE CLASS WILL BE THE SAME

AS HEAT CONTENT.

HEAT = MASS X SH X TEMP. CHANGE

MASS OF WATER

SH OF WATER

CHANGE IN TEMP

302 SP 11-2 HCl + NaOH25 ml OF HCl IS MIXED WITH 25 OF NaOH IN A FOAM CUP.THE STARTING TEMP. IS 25 -C AND THE ENDING TEMP.IS 32 -C. HOW MUCH HEAT IS RELEASED IN JOULES ?

HEAT = MASS x SH x TEMP.

HEAT = 50 g x 4.18 7x 1463 j OR 1.5 kj

25 ml25 ml

50 GRAMS

+

BECAUSE THEDENSITY = 1 ….

302 PP 11 HCl + NaOH50 mL OF HCl IS MIXED WITH 50 OF NaOH IN A FOAM CUP.THE STARTING TEMP. IS 22.5 -C AND THE ENDING TEMP.IS 26 -C. HOW MUCH HEAT IS RELEASED IN KJ ?

HEAT = MASS x SH x TEMP.

HEAT = 100 g x 4.18 3.5x 1463 j OR 1.5 kj

302 SP 11-2 HCl + NaOH • LIKE TEST QUESTION 28

25 mL of 0.025 M HCl IS ADDED TO 25 mL OF 0.25 M NaOHIS A FOAM CUP. AT THE START ALL SOLUTIONS ARE AT25 -C. DURING THE REACTION THE HIGHEST TEMP. IS32 . ASSUMING THE DENSITIES ARE 1.0 g ./ ml, HOW MUCHHEAT IS RELEASED IIN KJ .

HEAT = MASS x SPECIFIC HEAT x TEMP. CHANGE

HEAT = 50 g x 4.18 x 7

HEAT = 1463 J OR 1.5 KJ

303 THERMOCHEMICALEQUATIONS

• THERMOCHEMICAL EQUATION INCLUDES THE HEAT CHANGE.

• HEAT OF REACTION IS THE HEAT CHANGE FOR A REACTION EXACTLY AS IT IS WRITTEN

CaO + H2O ----> Ca(OH)2 + 65.2 kj

1 MOLE OF CaO ADDED TO WATER PRODUCES65.2 kj OF HEAT.

A NEW PART TO A CHEMICAL EQUATIONJOULES OR CALORIES

LAB: COLDER WATER• THE FIRST REACTION WAS ENDOTHERMIC.

• THE WATER GOT COLDER

• HEAT OF REACTION IS POSTIVE

• THE SYSTEM GAINED HEAT /SURROUNDING LOST HEAT

• IN THE EQUATION– HEAT GOES ON THE LEFT

BLUE SALT + 400 kj ----> COLDER WATER

= + 400 kjH

BECAUSE THE SYSTEM( SALT) GAINED HEAT

SYSTEMCALORIE SURROUNDING

LAB: WARMER WATER• THE SECOND REACTION WAS EXOTHERMIC.

• THE WATER GOT WARMER

• HEAT OF REACTION IS NEGATIVE

• THE SYSTEM GAINED HEAT /SURROUNDING LOST HEAT

• IN THE EQUATION– HEAT GOES ON THE RIGHT

BLUE SALT ----> WARMER WATER+ 400 kj

= - 400 kjH

BECAUSE THE SYSTEM( SALT)LOST HEAT

SYSTEMCALORIE SURROUNDING

304 LIKE TEST QUESTION 20

• GIVEN THIS EQUATION, HOW MUCH HEAT IS PRODUCED (kJ) WHEN 320 g OF O2 REACTS ?

C3H8 + 5 O2 --> 3 CO2 + 4 H20 + 2220 kj

320 g 1 MOLE

32 g10 MOLES1

=

CHANGE g TO MOLES

O 2 = 16g + 16g = 32 g

320 g 1 MOLE

32 g1=

CHANGE g TO MOLES

O 2 = 16g + 16g = 32 g

10 M

1

2220 kj

5 MOLES= 4440 kj

FIND HEAT IN kj

304 SP 11-3 H of REACTION

• USING THE EQUATION BELOW, CALCULATE THE KJ OF HEAT NEEDED TO DECOMPOSE 2.24 MOLES OF NaHCO3 (s) ?

2 NaHCO3 + 129 kJ ---> Na2CO3 + H2O + CO 2

129 kJ

2 MOLE

2.24 MOLES1 =

144 kj

304 SP 11-3 H of REACTION

• USING THE EQUATION BELOW, CALCULATE THE KJ OF HEAT NEEDED TO DECOMPOSE 2.24 MOLES OF NaHCO3 (s) ?

2 NaHCO3 + 129 kJ ---> Na2CO3 + H2O + CO2

304 PP 13: HEAT OF REACTION• WHEN CS2 IS FORMED FROM ITS ELEMENTS,

HEAT IS ABSORBED. CALCULATE THE AMOUNT OF HEAT (KJ) ABSORBED WHEN 5.66 g OF C2S IS

FOMMED ?

1 C + 2 S -----> CS2 H= 89.3kj

305 HEAT OF COMBUSTION

• THE HEAT OF COMBUSTION IS THE HEAT OF REACTION FOR THE COMPLETE BURNING F 1 MOLES OF A SUBSTANCE

305 HEAT OF COMBUSTION• THE HEAT OF COMBUSTION IS THE HEAT OF

REACTION FOR THE COMPLETE BURNING OF 1 MOLES OF A SUBSTANCE

CH4 + O2 ----> CO2 + H2O + 890 kJ

CH4 + O2 ----> CO2 + H2O H = - 890 kj

304 SP 11-3 HEAT of REACTIONUSING THE EQUATION FOR THIS REACTION, CALCULATETHE kj OF HEAT REQUIRED TO DECOMPOSE 2.24 MOLESOF NaHCO3 (s).

•2 NaHCO3 + 129 kj ---> NaCO3 + H20+ CO3

129 kj

2 MOLES

2,.24 MOLES

1

= ______ kj

144 kj

START 11.3

309 PP 20 HEAT OF FUSION• HOW MANY GRAMS OF ICE AT 0 -C AND 101.3 kPa

COULD BE MELTED BY THE ADDITION OF 0.4 kj OF HEAT ?

309 PP 20 HEAT OF FUSION• HOW MANY GRAMS OF ICE AT 0 -C AND 101.3KJ


309 PP 20 HEAT OF FUSION• HOW MANY GRAMS OF ICE AT 0 -C AND 101.3KJ


HEAT OF FUSION = 6.01 KJ / 1 MOLE HFUS

.4 kj1 =1 MOLE

6.01 kj..067 MOLES 18 g

1 MOLE = 1.2 g

1 MOLE OF WATER ( H20) IS 1g + 1g + 16 g = 18g

=

310 PHASE DIAGRAMSOLID WARMING

SOLID MELTING

LIQUID WARMING UP

LIQUID BOILING

GAS WARMING UP

HEAT SUPPLIED

TE

MP

ER

AT

UE

HEAT OF FUSION

310 MOLAR HEAT OF FUSIONIT TAKES 6.01 kj OF HEAT TO MELT 1 MOLE (18g) OF WATER AT A CONSTANT TEMP.

18 g1MOLE

Hfus = 6.01 kj / MOLE

ICESTARTS TO MELT

ALL THEICE HASMELTED

HEAT SUPPLIED

TE

MP

ER

AT

UR

E

310 MOLAR HEAT SOLIDIFICATON

1 MOLE (18g) OF LIQUID WATER GIVES OFF6.01 kj OF HEAT WHEN IT SOLIDIFIES

18 g1MOLE

18 g1 MOLE

HSOLID = - 6.01 kj / MOLE

307 HEATS OF FUSION

MOLAR HEAT OF FUSION IS THE AMOUNT OF HEAT NEEDED TO CHANGE 1 MOLE OF A SUBSTANCE FROM A SOLID TO A LIQUID AT A CONSTANT TEMPERATURE

H fus = + 6.01 kj

HEAT OF FUSION

6.01 kj ADDED PER MOLE

310 HEAT OF VAPORIZTIONIT TAKES 40.7 kj OF HEAT TO CHANGE 1 MOLES OF LIQUID WATER TO STEAMAT A CONSTANT TEMPERATURE.

LIQUID STEAM

308 LAB : HEAT OF FUSION• THE AMOUNT OF HEAT NEEDED TO MELT ONE

MOLE OF A SUBSTANCE• HEAT GAINED BY WATER = HEAT LOST BY ICE

1. MASS OF ICE 1____2. MASS OF 1MOLE (H2O) 2 18g3. MOLES OF ICE 3___4. MASS OF WATER 4 100 g5. STARTING TEMP. 5 21°6. ENDING TEMP 6_____7. CHANGE IN TEMP. 7_____8. SH OF WATER 8_____9. HEAT LOST BY WATER 9_____10. HEAT GAINED BY ICE 10 ____

CALCULATIONS.HEAT OF FUSION

10 KJ OF HEAT

3 MOLES OF ICE

DATA

314 HESS’S LAW• IF YOU ADD 2 OR THERMO EQUATIONS:

– THAT GIVE YOU A FINAL EQUATION• YOU CAN ADD THE HEATS OF REATION• TO GIVE YOU A FINAL HEAT OF REATION

1) C + O2 ----> CO2

2) C

Documents

309 SP 11-4 HEAT OF FUSION