1. 5.1 MEASURING ENERGY CHANGES

2. SYSTEM V. SURROUNDINGS SYSTEM – THE ACTUAL CHEMICAL REACTION TAKING PLACE SURROUNDINGS – THE REST OF THE UNIVERSE OUTSIDE OF THE SYSTEM

3. HEAT FLOWS HEAT ALWAYS FLOWS FROM HOT TO COLD HEAT FLOWS (NOT COLD) COLD IS AN ABSENCE OF HEAT

4. CALCULATING HEAT CHANGE, ∆H ∆H = -MC∆T UNITS: JOULES, J WHERE: ∆H IS THE ENTHALPY CHANGE M IS THE MASS OF WATER USED IN GRAMS (SAME AS VOLUME IN CM 3 ) C IS THE SPECIFIC HEAT CAPACITY OF WATER (4.18 JK -1 G -1 ) ∆T IS THE TEMPERATURE CHANGE – YOU SHOULD USE +/- TO SIGNIFY A DECREASE/INCREASE IN TEMPERATURE ∆H CAN EASILY BE CONVERTED TO A MOLAR VALUE BY DIVIDING IT BY THE NUMBER OF MOLES OF REACTANT ASSUMPTIONS: M IS JUST THE TOTAL MASS OF WATER USED THIS IS VALID AS THE MASS OF WATER USED IS MUCH GREATER THAN THE MASS OF ANY OF THE OTHER SUBSTANCES C IS JUST THE SPECIFIC HEAT CAPACITY OF WATER, IGNORING THE REACTANTS THIS IS VALID AS THE SPECIFIC HEAT CAPACITY OF WATER IS MUCH HIGHER THAN MOST OTHER SUBSTANCES, SO THEY ABSORB VERY LITTLE OF THE HEAT

5. How much heat is given off when an 869 g iron bar cools from 94 0 C to 5 0 C? s of Fe = 0.444 J/g 0 C  t = t final – t initial = 5 0 C – 94 0 C = -89 0 C q = ms  t = 869 g x 0.444 J/g 0 C x –89 0 C= -34,000 J 6.5

6. CALORIMETRY WHEN MEASURING HEAT LOST OR GAINED BY THE REACTION, WE MEASURE IT USING WATER AS THE BASIS FOR THE CALCULATIONS. Q w = m x s x  t m of water Q of water s of water  T of water

7. EXOTHERMIC RXN OF C U SO 4 AND Z N METAL ENTHALPY CHANGES IN A REACTION CAN BE CALCULATED BY CARRYING OUT THE REACTION IN AN INSULATED SYSTEM.

8. HEAT IS LOST FROM THE SYSTEM AS SOON AT THE TEMPERATURE RISES ABOUT THE TEMPERATURE OF THE SURROUNDINGS THE MAX RECORDED TEMP WILL BE LOWER THAN THE TRUE TEMP, WE CAN MAKE SOME ALLOWANCES AND EXTRAPOLATE THE COOLING DATA BACK TO WHEN THE RXN STARTED

9. ∆H REACTION =-∆H WATER IN ORDER TO UNDERSTAND THIS WE HAVE TO ASSUME FOUR THINGS 1) NO HEAT LOSS FROM THE SYSTEM 2) ALL THE HEAT GOES FROM THE REACTION TO THE WATER 3) THE SOLUTION IS DILUTE V(CUSO 4 )=V(H 2 O) 4) WATER HAS A DENSITY OF 1.00 G/CM 3

10. ∆H SYSTEM =0 (ASSUMPTION 1) ∆H SYSTEM =∆H REACTION + ∆H WATER (ASSUMPTION 2) THEREFORE ∆H REACTION =-∆H WATER

11. WORKED EXAMPLE THE NEUTRALIZATION REACTION BETWEEN SOLUTIONS OF N A OH AND H 2 SO 4 WAS STUDIED BY MEASURING THE TEMPERATURE CHANGES WHEN DIFFERENT VOLUMES OF THE TWO SOLUTIONS WERE MIXED TOGETHER. THE TOTAL VOLUME WAS ALWAYS 120 CM 3 THE CONCENTRATIONS OF BOTH LIQUIDS WAS ALWAYS 1 MOL DM -3

12. ∆H = - MC ∆T A) DETERMINE THE VOLUMES WHICH PRODUCE THE LARGEST INCREASE IN TEMPERATURE. B) CALCULATE THE HEAT PRODUCED BY THE REACTION WHEN THE MAX TEMPERATURE WAS PRODUCED. C) CALCULATE HEAT PRODUCED BY 1 MOL OF N A OH THE LITERAURE VALUE FOR THE ENTHALPY OF NEUTRALIZATION IS -57.7KJ MOL -1 CALCULATE THE % ERROR AND DEDUCE A REASON FOR THE DISCREPANCY

13. SOLUTIONS A) LOOKING AT THE GRAPH: V(N A OH) 80.0 CM 3 V(H 2 SO 4 ) 40.0 CM 3 B) ASSUMING THAT 120.0 CM3 OF THE SOLUTION CONTAINS 120.0 G OF WATER AND ALL OF THE HEAT PASSES INTO THE WATER. ∆H REACTION = - M (H2O) X C (H2O) X ∆T(H2O) = -120.0 X 4.18 J G -1 K -1 X (33.5-25.0)K = -4264 J

14. SOLUTIONS C) ∆H REACTION = -4264 / N (N A OH) J MOL -1 = -4264/ (1.00 X 80.00)/1000 J MOL -1 = -4264/80.00 KJ MOL -1 = -53.3 KJ MOL -1 D) % ERROR = -57.5 – (-53.3)/-57.5 X100 7% ERROR

15. 5.2 HESS’S LAW

16. HESS’ LAW THE ENTHALPY CHANGE OF A REACTION IS INDEPENDENT OF THE PATHWAY OF THAT REACTION I.E. ALL THAT MATTERS IS THE START AND FINISH POINTS NOTE: ADD WHEN GOING ‘WITH’ AN ARROW, SUBTRACT WHEN GOING AGAINST AN ARROW. WHY MUST THIS BE THE CASE? A + BC + D E + F ∆H 1 ∆H 2 ∆H 3 ∆H 1 = ∆H 2 + ∆H 3 A + B C + D E + F ∆H 1 ∆H 2 ∆H 4 ∆H 1 = ∆H 2 - ∆H 3 + ∆H 4 G + H ∆H 3

17. STATE FUNCTIONS A PROPERTY THAT DEPENDS ON THE PRESENT STATE OF THE SYSTEM, NOT THE PATH TAKEN TO ARRIVE AT THAT STATE. THE ENERGY CONTAINED WITHIN 1.0 MOLE OF WATER AT 25 0 C IS THE SAME WHETHER OR NOT THE ICE WAS MELTED OR CH 4 WAS COMBUSTED.

18. HESS’S LAW IF A REACTION IS CARRIED OUT IN A SERIES OF STEPS, THE OVERALL CHANGE IN ENTHALPY WILL BE EQUAL TO THE SUM OF THE ENTHALPY CHANGES FOR THE INDIVIDUAL STEPS OVERALL ENTHALPY CHANGE WILL BE THE SAME IF A REACTION OCCURS IN ONE STEP OR IN SEVERAL STEPS.

19. HESS’S LAW CALCULATE THE  H FOR THE REACTION 3C(S) + 4H 2(G) C 3 H 8(G) USE THE FOLLOWING INFORMATION: RxnChem equationEnthalpy Change 1 2H 2(g) + O 2(g) 2H 2 O (l)  H= -571.7 kJ 2C 3 H 8(g) + 5O 2(g) 3CO 2(g) + 4H 2 O (l)  H= -2220.1 kJ 3C(s) + O 2 (g) CO 2 (g)  H= -393.5 kJ

20. HESS’S LAW CONT… 1 2 H 2(g) + O 2(g) 2 H 2 O (l)  H= -571.7 kJ 2 C 3 H 8(g) + 5O 2(g) 3CO 2(g) + 4H 2 O (l)  H= -2220.1 kJ 3 C(s) + O 2 (g) CO 2 (g)  H= -393.5 kJ 3C(s) + 4H 2(g) C 3 H 8(g) x2 4 H 2(g) + 2O 2(g) 4 H 2 O (l)  H= -1143.4 kJ X -1  H= +2220.1 kJ x3 3C(s) + 3O 2 (g) 3CO 2 (g)  H= -1180.5 kJ  H= -103.8 kJ

21. Calculate the standard enthalpy of formation of CS 2 (l) given that: C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ rxn S (rhombic) + O 2 (g) SO 2 (g)  H 0 = -296.1 kJ rxn CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g)  H 0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS 2 C (graphite) + 2S (rhombic) CS 2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C (graphite) + O 2 (g) CO 2 (g)  H 0 = -393.5 kJ 2S (rhombic) + 2O 2 (g) 2SO 2 (g)  H 0 = -296.1x2 kJ rxn CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g)  H 0 = +1072 kJ rxn + C (graphite) + 2S (rhombic) CS 2 (l)  H 0 = [1072] – [-393.5 + (2x-296.1)] = 2057 kJ rxn