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- 1. IB Chemistry Power Points Topic 05 Energeticswww.pedagogics.ca Lesson Exothermic and Endothermic Reactions and Calorimetry
- 2. Much taken fromENTHALPY CHANGES Great thanks toJONATHAN HOPTON & KNOCKHARDY PUBLISHING www.knockhardy.org.uk/sci.htm
- 3. Background and ReviewFirst Law of Thermodynamics (Law of Energy Conservation)Energy can be neither created nor destroyed but it can be converted from oneform to anotherEnergy Changes in Chemical ReactionsAll chemical reactions are accompanied by some form of energy change Exothermic Energy is given out Endothermic Energy is absorbedExamples Exothermic combustion reactions neutralization (acid + base) Endothermic photosynthesis thermal decomposition of calcium carbonate Activity : observing exothermic and endothermic reactions
- 4. Key ConceptThe heat content of a chemical system is calledenthalpy (represented by H). We cannot measure enthalpy directly, only the change in enthalpy ∆H i.e. the amount of heat released or absorbed when a chemical reaction occurs at constant pressure. ∆H = H(products) – H(reactants) ∆Ho (the STANDARD enthalpy of reaction) is the value measured when temperature is 298 K and pressure is 101.3 kPa.
- 5. Enthalpy Diagram -Exothermic Change If ∆H is negative, H(products) < H(reactants) There is an enthalpy decrease and heat is released to the surroundings.enthalpy
- 6. Enthalpy Diagram - Endothermic ChangeIf ∆H is positive, H(products) > H(reactants)There is an enthalpy increase and heat isabsorbed from the surroundings.enthalpy
- 7. Example: Enthalpy change in a chemical reaction Exothermic reactions release heat N2(g) + 3H2(g) 2 NH3(g) ∆H = -92.4 kJ/mol The coefficients in the balanced equation represent the number of moles of reactants and products.
- 8. N2(g) + 3H2(g) 2 NH3(g) ∆H = -92.4 kJ/molState symbols are ESSENTIAL as changes of stateinvolve changes in thermal energy. The enthalpy change is directly proportional to the number of moles of substance involved in the reaction. For the above equation, 92.4 kJ is released - for each mole of N2 reacted - for every 3 moles of H2 reacted - for every 2 moles of NH3 produced.
- 9. The reverse reaction2 NH3(g) N2(g) + 3H2(g) ∆H = +92.4 kJ/mol Note: the enthalpy change can be read directly from the enthalpy profile diagram.
- 10. Thermochemical Standard ConditionsThe ∆H value for a given reaction will dependon reaction conditions.Values for enthalpy changes are standardized :for the standard enthalpy ∆Ho-Temperature is 298 K- Pressure is 1 atmosphere- All solutions involved are 1 M concentration
- 11. Assigned : Review Exercise 1 The ∆H value for a given reaction will depend on reaction conditions. Values for enthalpy changes are standardized : for the standard enthalpy ∆Ho -Temperature is 298 K - Pressure is 1 atmosphere - All solutions involved are 1 M concentration
- 12. Calorimetry – Part 1 Specific Heat CapacityThe specific heat capacity of a substance is a physicalproperty. It is defined as the amount of heat (Joules)required to change the temperature (oC or K) of a unitmass (g or kg) of substance by ONE degree. Specific heat capacity (SHC) is measured in J g-1 K-1 or kJ kg-1 K-1 (chemistry) J kg-1 K-1 (physics)
- 13. Calorimetry – continuedHeat and temperature changeKnowing the SHC is useful in thermal chemistry. Heatadded or lost can be determined by measuringtemperature change of a known substance (water). Q = mc∆T heat = mass x SHC x ∆Temp
- 14. Determining the Specific Heat Capacity of Water Read the background information and lab activity instructions carefullyA kettle (and other electrical heating devices) havepower ratings (given in Watts). This tells you theamount of heat energy supplied by the device. Thewatt is equivalent to 1 J per second. (heat) Q = P x t DON’T FORGET TO CONSIDER UNCERTAINTY
- 15. Determining the Specific Heat Capacity of WaterWhat you need to do (DCP)Plot data and choose a suitable range for analysis(look for constant increase in temperature)Organize the data from the range in a suitable table.You may choose to do further processing at this time.Plot the selected range of raw/processed data. Usethe slope to determine the specific heat capacity ofwater.Present all calculations / data processing clearly DON’T FORGET TO CONSIDER UNCERTAINTY
- 16. Determining the Specific Heat Capacity of WaterWhat you need to do (CE)Write a conclusion paragraph. Don’t forget tocompare data to the accepted value. Mention anyindication of presence (or absence) of random orsystematic error. Is your result valid?Brainstorm a list of procedural/measurementweaknesses or limitations.-do you have evidence (data) that any of thesesignificantly affected the result (as indicated by thedata)? How could you improve the investigation?Be prepared to discuss in class
- 17. Determining Specific Heat Capacity of WaterSample results1300 W kettlesupplies 1300 J ofheat each second.1300 J heat gives Slope tells us∆T of 0.2215 oC temperature increases 0.2215 oC eachA ∆T of 1oC would second.therefore require5870 J of heat.
- 18. Determining Specific Heat Capacity of WaterSample results (continued)Mass of water in kettle was 1405 g. To change the temperatureof this water by 1 degree, 5870 J of heat were required. The amount of heat required to change 1 g of water is therefore 4.18 J. This is the SHC of water (very close to the literature value) SHC of H2O is 4.18 J g-1 K-1
- 19. For exampleWhen 3 g of sodium carbonate are added to 50 cm3 of1.0 M HCl, the temperature rises from 22 °C to 28.5 °C.Calculate the heat required for this temperature change.
- 20. Calorimetry – Part 2ApplicationsA calorimeter is used tomeasure the heatabsorbed or released ina chemical (or other)process by measuringthe temperature changeof an insulated mass ofwater.
- 21. Sample Problem 1When 3 g of sodium carbonate are added to 50 cm3 of1.0 M HCl, the temperature rises from 22.0 C to 28.5 C.Calculate the heat required for this temperature change.
- 22. 3 -3Sample problem 2: 50.0 cm of a 1.00 mol dm HCl solution is mixed with 25.0 3 -3cm of 2.00 mol dm NaOH. A neutralization reaction occurs. The initial otemperature of both solutions was 26.7 C. After stirring and accounting for oheat loss, the highest temperature reached was 33.5 C. Calculate the enthalpychange for this reaction. NaOH HCl both 26.7o . 26.7o 33.5o
- 23. 3 -3Sample problem 2: 50.0 cm of a 1.00 mol dm HCl solution is mixed with 25.0 3 -3cm of 2.00 mol dm NaOH. A neutralization reaction occurs. The initial otemperature of both solutions was 26.7 C. After stirring and accounting for oheat loss, the highest temperature reached was 33.5 C. Calculate the enthalpychange for this reaction.After writing a balanced equation, the molar quantities andlimiting reactant needs to be determined.Note that in this example there is exactly the right amount ofeach reactant. If one reactant is present in excess, the heatevolved will associated with the mole amount of limiting reactant.
- 24. 3 -3Sample problem 2: 50.0 cm of a 1.00 mol dm HCl solution is mixed with 25.0 3 -3cm of 2.00 mol dm NaOH. A neutralization reaction occurs. The initial otemperature of both solutions was 26.7 C. After stirring and accounting for oheat loss, the highest temperature reached was 33.5 C. Calculate the enthalpychange for this reaction.Next step – determine how much heat was released.There are some assumptions in this calculation- Density of reaction mixture (to determine mass)- SHC of reaction mixture (to calculate Q)
- 25. 3 -3Sample problem 2: 50.0 cm of a 1.00 mol dm HCl solution is mixed with 25.0 3 -3cm of 2.00 mol dm NaOH. A neutralization reaction occurs. The initial otemperature of both solutions was 26.7 C. After stirring and accounting for oheat loss, the highest temperature reached was 33.5 C. Calculate the enthalpychange for this reaction.Final step – calculate ΔH for the reaction
- 26. Sample problem 3: to determine the enthalpy of combustion for ethanol (seereaction), a calorimeter setup (below) was used. The burner was lit and allowedto heat the water for 60 s. The change in mass of the burner was 0.518 g and othe temperature increase was measured to be 9.90 C. What is the big assumption made with this type of experiment?
- 27. Sample problem 3: to determine the enthalpy of combustion for ethanol (seereaction), a calorimeter setup (below) was used. The burner was lit and allowedto heat the water for 60 s. The change in mass of the burner was 0.518 g and othe temperature increase was measured to be 9.90 C.First step – calculate heat evolved using calorimetryLast step – determine ΔH for the reaction
- 28. 3 -3Sample problem 4: 100.0 cm of 0.100 mol dm copper II sulphate solution isplaced in a styrofoam cup. 1.30 g of powdered zinc is added and a singlereplacement reaction occurs. The temperature of the solution over time isshown in the graph below. Determine the enthalpy value for this reaction. First step Make sure you understand the graph. Extrapolate to determine the change in temperature.The extrapolation is necessary to compensate for heat loss while the reactionis occurring. Why would powdered zinc be used?
- 29. 3 -3Sample problem 4: 100.0 cm of 0.100 mol dm copper II sulphate solution isplaced in a styrofoam cup. 1.30 g of powdered zinc is added and a singlereplacement reaction occurs. The temperature of the solution over time isshown in the graph below. Determine the enthalpy value for this reaction.Determine the limiting reactantCalculate QCalculate the enthalpy for the reaction. Review Exercise 2

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