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THERMOCHEMISTRY PT 2

THERMOCHEMISTRY PT 2. HEAT CAPACITY, SPECIFIC HEAT, ENDOTHERMIC/EXOTHERMIC, ENTHALPY, CALORIMETERY, STANDARD ENTHALPIES, HESS’S LAW, AND HEAT VS. TEMPERATURE, ETC. STANDARD CONDITIONS.

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THERMOCHEMISTRY PT 2

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  1. THERMOCHEMISTRY PT 2 HEAT CAPACITY, SPECIFIC HEAT, ENDOTHERMIC/EXOTHERMIC, ENTHALPY, CALORIMETERY, STANDARD ENTHALPIES, HESS’S LAW, AND HEAT VS. TEMPERATURE, ETC.

  2. STANDARD CONDITIONS • THE AMOUNT OF HEAT ENERGY (ENTHALPY) PRODUCED IN A TYPICAL RXN IS DEPENDENT ON THE CONDITIONS THE LAB IS DONE IN. • THE SAME AMOUNT OF HEAT PRODUCED BY BURNING A PIECE OF PAPER WILL BE DIFFERENT IF IT IS BURNED IN MARCH AS OPPOSED TO AUGUST OR IF IT IS BURNED STANDING ON THE BEACH OF THE PACIFIC OCEAN OR ON THE MOUNTAINTOP OF COLORADO.

  3. BECAUSE OF THIS DEPENDANCE OF THE CONDITIONS OF THE RXN, IT IS IMPORTANT FOR CHEMISTS TO HAVE A REFERENCE POINT FOR VALUES THAT THEY USE IN CALCULATIONS. • IF THE CONDITIONS OF THE REACTION ARE DEFINED AS A PARTICULAR TEMP AND A PARTICULAR PRESSURE THEN THE AMOUNT OF HEAT RELEASED IN THAT PARTICULAR RXN WOULD BE A STANDARD VALUE.

  4. THEREFORE, A SET OF STANDARD CONDITIONS HAVE BEEN DECIDED ON THAT ARE TRUE ALL OVER THE WORLD. • THE STANDARD CONDITIONS HAVE BEEN DEFINED AS: -TEMPERATURE = 298 K OR 25°C - PRESSURE = 1 atmosphere OR 760 mmHg

  5. STANDARD ENTHALPIES • THE AMOUNT OF HEAT ENERGY (ENTHALPY) PRODUCED IN A TYPICAL REACTION UNDER STANDARD CONDITIONS IS DEFINED AS THE STANDARD ENTHALPY. • STANDARD ENTHALPIES OF REACTIONS CAN BE FOUND ON EXTENSIVE TABLES OF DATA COMPILED BY SCIENTISTS ALL OVER THE WORLD.

  6. H° • THE SYMBOL USED TO EXPRESS THE STANDARD ENTHALPY OF A REACTION IS: • MOST OFTEN STANDARD ENTHALPIES ARE GIVEN IN TABLES IN THE FORM OF ENTHALPIES OF FORMATION.

  7. ° Hf • STANDARD ENTHALPIES OF FORMATION ARE MEASURED VALUES FOR THE AMOUNT OF ENERGY NECESSARY TO FORM CHEMICAL COMPOUNDS. • THE SYMBOL USED TO EXPRESS THE STANDARD ENTHALPY OF HEAT OF FORMATION IS:

  8. °= Hf -285.8kJ/mol • WHEN H2 GAS AND O2 GAS ARE MIXED TOGETHER AND IGNITED IT EXPLODES WITH A LARGE BANG PRODUCING WATER. • THE AMOUNT OF ENERGY PRODUCED BY THE RXN IS 285kJ OF ENERGY PER MOLE. • AND CAN BE SHOWN USING THIS CHEMICAL RXN: H2 + 1/202 H2O

  9. THERE ARE LISTS AND LISTS OF COMPOUNDS ON TABLES THAT ARE AVAILABLE FOR SCIENTISTS TO USE TO LOOK UP AND USE THE ENTHALPIES IN CHEMICAL RXNS. • ANY VALUE FOR THE ENTHALPY OF A RXN, OR FOR A CHEMICAL PROCESS ARE CALCULATED USING THE STANDARD ENTHALPIES OF FORMATION FOUND ON A TABLE OF COMPNDS A LOT LIKE THE TABLE:

  10. ° Hf STANDARD ENTHALPIES OF FORMATION

  11. CALORIMETRY • CHANGES IN HEAT ENERGY CAN BE MEASURED USING A DEVICE CALLED A CALORIMETER. • A CALORIMETER USES THE HEAT ABSORBED BY WATER TO MEASURE THE HEAT GIVEN OFF BY A RXN OR AN OBJECT. • THE AMNT OF HEAT ABSORBED BY THE WATER IS EQUAL TO THE AMNT OF HEAT GIVEN OFF BY THE RXN OR OBJECT.

  12. A COFFEE CUP CALORIMETER USED FOR A REACTION IN WATER, OR JUST A TRANSFER OF HEAT. A BOMB CALORIMETER USED WHEN TRYING TO FIND THE AMOUNT OF HEAT PRODUCED BY BURNING SOMETHING.

  13. - QWATER QREACTION - QSUR QSYS = = • WE’RE TALKING ABOUT THE TRANSFER OF HEAT. • THIS CAN BE WRITTEN AS: (REMEMBER Q IS ALSO H) • WRITTEN GENERICALLY: WHERE QSYS IS THE SYSTEM OR WHAT IS TAKING PLACE IN THE MAIN CHAMBER (REACTION ETC.) AND QSUR IS THE SURROUNDINGS WHICH IS GENERALLY WATER.

  14. QSYS - QSUR = • WITH CALORIMETRY WE USE THE SIGN OF WHAT HAPPENS TO THE WATER. WHEN THE WATER LOSES HEAT INTO THE SYSTEM IT OBTAINS A NEGATIVE SIGN. - SIGN MEANS HEAT WAS RELEASED BY WATER + SIGN MEANS HEAT WAS ABSORBED BY THE RXN (ENDOTHERMIC)

  15. HEAT HEAT WATER = SURROUNDINGS - D H water SYSTEM + D H rxn ENDOTHERMIC

  16. - QSYS QSUR = • WHEN THE WATER GAINS HEAT FROM THE SYSTEM IT OBTAINS A POSITIVE SIGN. - SIGN MEANS HEAT WAS RELEASED BY THE RXN + SIGN MEANS HEAT WAS ABSORBED BY WATER (EXOTHERMIC)

  17. HEAT HEAT WATER = SURROUNDINGS +D H water SYSTEM - D H rxn EXOTHERMIC

  18. YOU ALWAYS KNOW THE MASS OF THE WATER, THE SPECIFIC HEAT CAPACITY OF WATER, AND YOU CAN OF COURSE RECORD THE CHANGE IN THE TEMPERATURE OF WATER. • SO YOU CAN CALCULATE THE AMOUNT OF HEAT ABSORBED BY THE WATER, WHICH LEADS TO THE AMOUNT OF HEAT GIVEN OFF BY THE REACTION.

  19. WE WANT TO USE OUR OLD FRIEND THE H EQUATION TO CALCULATE THE CHANGES THAT THE WATER UNDERGOES TO UNDERSTAND THE CHANGE THAT THE OBJECT, REACTION, OR FIRE USES. • TYPICAL SITUATION: A SOLID CHUNK OF Al THAT WEIGHS 72g HEATED TO 100°C IS DROPPED IN A CALORIMETER CONTAINING 120ml OF WATER AT 23°C. THE WATER’S TEMP INCREASES TO 27°C.

  20. WHAT DO WE KNOW? H TFINAL-TINITIAL m C = • MASS OF Al = 72g • TINITIAL OF Al = 100°C • TFINAL OF Al = 27°C • C OF Al = .992J/g°C (FROM TABLE) QSYS • MASS OF H2O= 120g • TINITIAL OF H2O= 23°C • TFINAL OF H2O = 27°C • C OF H2O = 4.18J/g°C(FROM TABLE) QSUR

  21. FOR WATER H H = = 27°C-100°C 27°C-23°C 120g 72g 4.18J/g°C .992J/g°C H H = = 2006J • FOR ALUMINUM -2006J EQUAL BUT OPPOSITE, THE SIGNS MEANS THAT THE ALUMINUM RELEASED HEAT BECAUSE IT DECREASED IN TEMPERATURE CAUSING THE WATER TO INCREASE IN TEMP.

  22. HESS’S LAW • ENTHALPIES OF REACTIONS ARE USEFUL PIECES OF INFORMATION FOR CHEMISTS. • FOR INSTANCE WE COULD ANSWER QUESTIONS LIKE: • WOULD A METHANE, A PROPANE, OR A WOOD BURNING HEATER PRODUCE MORE HEAT TO HEAT A ROOM? • HOWEVER NOT EVERY REACTION’S ENERGY IS MEASURABLE DIRECTLY.

  23. TO GET AROUND THIS CHEMISTS HAVE DISCOVERED THAT IF WE CAN FIND THE ENTHALPIES OF RXNS FOR PIECES OF A RXN WE CAN ADD THEM TOGETHER TO CALCULATE THE ENERGY PRODUCED. • THIS PROCESS OF ADDING TOGETHER SEVERAL RXNS IN ORDER TO CALCULATE A NET ENERGY IS CALLED HESS’S LAW.

  24. FOR EG.THE REACTION FOR THE COMBUSTION OF PROPANE IS: C3H8 + 5O2 3CO2 + 4H2O • WE WANT TO CALCULATE THE ENTHALPY OF THIS REACTION. • WE CAN DO THIS BECAUSE OF HESS’S LAW. • IF WE KNEW HOW MUCH ENERGY IS PRODUCED BY THE FORMATION OF EACH COMPOUND IN THE OVERALL REACTION THEN WE CAN ADD THEM TOGETHER TO FIND OUT HOW MUCH HEAT IS PRODUCED IN THE MAIN REACTION.

  25. 3C + 4H2 C3H8 ONE OF THE COMPOUNDS NEEDED FOR OUR MAIN REACTION. °= Hf -104.7kJ/mol MAIN REACTION: C3H8 + 5O2 3CO2 + 4H2O • ON A TABLE OF STANDARD ENTHALPIES OF FORMATION WE CAN FIND THIS VALUE FOR THE SYNTHESIS OF PROPANE: BUT PROPANE HERE IS A PRODUCT NOT A REACTANT LIKE WE WANT. • WE NEED ONE OF THE RULES FOR USING HESS’S LAW

  26. C3H8 3C + 4H2 + VALUE NOW NOW PROPANE IS ON THE CORRECT SIDE TO FIT INTO OUR MAIN RXN EQUATION. °= Hf 104.7kJ/mol C3H8 + 5O2 3CO2 + 4H2O • RULE #1: IF AN EQUATION IS REVERSED, THE SIGN OF H CHANGES TOO. • THEREFORE IF 104.7KJ OF ENERGY IS RELEASED TO FORM PROPANE (NEGATIVE VALUE), THEN IT TAKES 104.7KJ OF ENERGY TO BREAK IT APART.

  27. H2 + ½ O2 H2O ONE OF THE COMPOUNDS NEEDED FOR OUR MAIN REACTION. °= Hf -241.8kJ/mol C3H8 + 5O2 3CO2 + 4H2O • ON THE SAME TABLE YOU MIGHT FIND ANOTHER OF THE COMPOUNDS THAT MAKE UP THE PUZZLE: IT’S ALREADY ON THE CORRECT SIDE TO FIT INTO OUR EQUATION, BUT WE NEED MORE THEN ONE MOLE OF WATER, WE NEED 4 MOLES. • THAT TAKES ANOTHER RULE

  28. C3H8 + 5O2 3CO2 + 4H2O • RULE #2: IF THE COEFFICIENTS OF AN EQUATION ARE MULTIPLIED BY A FACTOR, THE ENTHALPY CHANGE FOR THE REACTION IS THE MULTIPLIED BY THE SAME FACTOR. • THEREFORE IF 241.5KJ OF ENERGY IS RELEASED TO FORM ONE MOLE OF WATER (NEGATIVE VALUE), THEN IT RELEASES 966KJ OF ENERGY FOR 4 MOLES.

  29. NEED TO MULTIPLY ENERGY BY 4 NEED TO MULTIPLY COEFFIENTS BY 4 C + O2 CO2 °= °= Hf Hf 4(-241.8kJ/mol) -393.6kJ/mol 4H2 + 2 O2 4H2O • NEXT UP IS THE FORMATION OF CARBON DIOXIDE • ON A TABLE: WE NEED MORE THEN ONE MOLE OF WATER, WE NEED 3 MOLES.

  30. NEED TO MULTIPLY ENERGY BY 3 NEED TO MULTIPLY COEFFIENTS BY 3 °= Hf -1180.8kJ/mol 3C + 3O2 3CO2 • WE DON’T HAVE TO DO O2, BECAUSE IT’S NOT A REACTION, ITS AN ELEMENT, SO IT’S AUTOMATIC • SO LET’S REVIEW:

  31. C3H8 3C + 4H2 4H2 + 2 O2 4H2O -967.2kJ/mol 104.7kJ/mol • EQUATION WE’RE TRYING TO BUILD: C3H8 + 5O2 3CO2 + 4H2O • EQUATION #1 • EQUATION #2 • EQUATION #3 3C + 3O2 3CO2 -1180.8kJ/mol • NOW WE ADD THE EQUATIONS TOGETHER

  32. C3H8 3C + 4H2 4H2 + 2 O2 4H2O -967.2kJ/mol 104.7kJ/mol 3C + 3O2 3CO2 -1180.8kJ/mol C3H8+4H2+5O2+3C  3C+4H2+4H2O+3CO2 • NOW WE CAN CROSS OUT ANY COMPOUNDS THAT WE FIND ON BOTH SIDES OF THE EQUATION • NOW REWRITE IT AGAIN AND VOILA! C3H8 + 5O2 3CO2 + 4H2O

  33. °= Hf -2043.3kJ/mol C3H8 + 5O2 3CO2 + 4H2O • ALL WE HAVE LEFT TO DO IS ADD THE RESULTING ENTHALPIES TOGETHER. 104.7kJ/mol + (-967.2kJ/mol) + (-1180.8kJ/mol)= C3H8 + 5O2 3CO2 + 4H2O • THAT’S HESS’S LAW IN ACTION!!!

  34. C2H6  C2H4 + H2 2H2O + 2CO2 C2H4 + 3O2 1323kJ/mol 2H2O  2H2 + O2 484kJ/mol 137kJ/mol • EXAMPLE 2: GIVEN THE FOLLOWING INFORMATION: FIND THE VALUE OF H° FOR THE EQUATION: 2C2H6 + 7O2 4CO2 + 6H2O

  35. C2H6  C2H4 + H2 2H2O + 2CO2 C2H4 + 3O2 1323kJ/mol 2H2O  2H2 + O2 484kJ/mol REVERSE & DOUBLE THIS EQUATION REVERSE THIS EQUATION DOUBLE THIS EQUATION 137kJ/mol ANALYZING THE INFORMATION GIVEN:

  36. 2C2H6  2C2H4 + 2H2 274kJ/mol 2H2 + O2 2H2O-484kJ/mol 2C2H4 + 6O2 4H2O + 4CO2 -2646kJ/mol 2C2H6 + 2H2 + 7O2+2C2H4  2C2H4+6H2O+4CO2+2H2 2C2H6 + 7O2 4CO2 + 6H2O 274kJ/mol+(-484kJ/mol)+(-2646kJ/mol) =-2856kJ/mol

  37. HESS LAW RULES: • RULE #1: IF AN EQUATION IS REVERSED, THE SIGN OF H CHANGES TOO. • RULE #2: IF THE COEFFICIENTS OF AN EQN ARE MULTIPLIED BY A FACTOR, THE ENTHALPY CHANGE FOR THE REACTION IS THE MULTIPLIED BY THE SAME FACTOR.

  38. PHASE CHANGES & HEAT • MELTING AND BOILING ALSO INVOLVES ENERGY TRANSFER, JUST LIKE CHANGING THE TEPMERATURE OF THE LIQUID WATER DOES. • WE CAN FIND TABLES THAT SHOW THE AMOUNTS OF ENERGY REQUIRED TO CHANGE THE PHASE OF A SUBSTANCE.

  39. THERE ARE TWO TYPES OF PHASE CHANGES THAT THESE TABLES INCLUDE -THE AMOUNT OF HEAT NECESSARY TO MELT 1 MOLE OF SUBSTANCE CALLED THE HEAT OF FUSION (Hfus) -THE AMOUNT OF HEAT NECESSARY TO BOIL 1 MOLE OF SUBSTANCE CALLED THE HEAT OF VAPORIZATION (Hvap)

  40. LET’S BACKTRACK FOR A SECOND. WHEN WE TAKE ICE AND PUT IT IN A BEAKER AND HEAT IT UP UNTIL IT BOILS AND GRAPH THE INCREASE IN TEMPERATURE OVER TIME IT WOULD LOOK SOMETHING LIKE THE NEXT SLIDE

  41. vaporization condensation melting freezing Heating Curve GAS 100° TEMPERATURE (C) LIQUID 0° SOLID HEAT ADDED OVER TIME 

  42. THE DIAGONAL LINES MAKE SENSE, THE TEMPERATURE IS INCREASING AS IT IS HEATED. • BUT WHAT ABOUT THE HORIZONTAL LINES? - WHAT’S UP WITH THEM? • WELL LET’S ANALYZE THE DIAGRAM, AND BLOW IT UP AS WE TALK ABOUT IT. • IT LOOKS LIKE WE HAVE 5 SECTIONS WE CAN BREAK THE GRAPH INTO

  43. SECTION #1 • LET’S SAY WE ARE TALKING ABOUT 18 g OF ICE AT –10°C, AND WE PLACE IT ON A HOTPLATE. • HEAT ENERGY FROM THE PLATE GOES INTO THE VIBRATION OF THE MOLECULES, INCREASING ITS KINETIC ENERGY, WHICH IS THE SAME AS RAISING THE TEMP. • THE TEMP. WILL CONTINUE TO INCREASE UNTIL IT REACHES THE POINT AT WHICH IT WILL BEGIN MELTING OR FOR WATER 0°C.

  44. WE CAN CALCULATE THE AMOUNT OF HEAT ENERGY ABSORBED BY THE ICE TO THIS POINT USING THE EQUATION DH=mCDT (ICE HAS A SPECIFIC HEAT OF 2.09J/g°C) 0° -10° (18g)(2.09J/g°C)(0°C-(-10°C)) DH=376.2J

  45. SECTION #2 • AFTER THAT POINT ANY ADDITIONAL HEAT ABSORBED BY THE ICE GOES INTO PARTIALLY BREAKING THE CONNECTIONS BETWEEN THE WATER MOLECULES. • THE HEAT DOES NOT INCREASE THE KINETIC ENERGY OF THE MOLECULES SO THE TEMPERATURE DOES NOT CHANGE, ALL OF THE ENERGY IS NEEDED TO GET OVER THE NEXT HUMP.

  46. AS LONG AS THERE ARE CONNECTIONS TO BREAK OR THERE IS SOLID ICE PRESENT, THE TEMPERATURE CANNOT INCREASE. • THE SOLID AND LIQUID ARE IN EQUILIBRIUM IF THEY ARE BOTH PRESENT AT THE SAME TIME. • THE ENERGY REQUIRED TO CHANGE FROM THE SOLID TO THE LIQUID PHASE IS CALLED THE HEAT OF FUSION AND DEPENDS ON THE SUBSTANCE AND THE QUANTITY (DHfus FOR H2O = 6000J/mol)

  47. ENERGY NEEDED TO MELT ALL OF THE ICE IN THIS EG. CAN BE FOUND USING DH=(mol) DHfus 1 mol H2O =1 mol H2O 18g H2O 18g H2O 0° DH =(1mol)(6000J/mol) DH =6000J DH=376.2J -10°

  48. SECTION #3 • NOW THAT ALL OF THE PARTICLES ARE FREE TO FLOW, THEN THE HEAT ENERGY GAINED GOES INTO THE VIBRATION OF THE MOLECULES. • THE KINETIC ENERGY INCREASES, SO THE TEMPERATURE OF THE LIQUID INCREASES. • THE RATE OF TEMPERATURE INCREASES NOW DEPENDS ON THE HEAT CAPACITY OF LIQUID WATER, WHICH IS 4.180J/g°C.

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