Energetics

1. Topic 5.1 – 5.2 Energetics

2. Topic 5.1Exothermic and Endothermic Reactions ? ?

3. total energy of the universe is a constant • if a system loses energy, it must be gained by the surroundings, and vice versa

4. H Enthalpy • a measure of the heat content (not temp) • you cannot measure the actual enthalpy of a substance • but you can measure an enthalpy CHANGE because of energy it takes in or releases •  = Greek letter ‘delta’ meaning change • H = heat. • so, H means ‘heat change’.

5. Why a standard? • enthalpy values vary according to the conditions • a substance under these conditions is said to be in its standard state • pressure: 100 kPa (1 atmosphere) • temperature: usually 298K (25°C) • if these were not standardized, then energy could be leaving or entering the system • modify the symbol from Enthalpy Change Standard Enthalpy Change (at 298K)

6. Enthalpy (Heat) of Reaction H = Hproducts−Hreactants lower energy is more stable

7. Exothermic reactions • heat energy is given out by the reaction hence the surroundings increase in temperature (feels hot) • occurs when bonds are formed • new products are more stable and extra energy is given off • Hproducts <Hreactants • H is negative • examples • combustion of fuels • respiration • neutralization reactions (acid reacts with something)

8. ENTHALPY REACTION CO-ORDINATE activation energy energy necessary to get the reaction going energy given out, ∆H is negative reactants products

9. Energy taken in to break bonds. H2+ Cl2  2HCl H, H, Cl, Cl (Atoms) Energy given out when bonds are made. H-H, Cl-Cl energy Reactants Overall energy change, H H-Cl, H-Cl Products

10. H2+ Cl2  2HCl Energy in = +678kJ H, H, Cl, Cl (Atoms) Energy out = -862kJ H-H, Cl-Cl energy Reactants Overall energy change, H = -184kJ H-Cl, H-Cl Products

11. Endothermic reactions • heat energy is taken in by the reaction mixture hence the surroundings decrease in temperature (feels cold) • occurs when bonds are broken • the reactants were more stable (bonds are stronger) • therefore, took energy from the surrounding to break bonds • Hreactants <Hproduct H is positive • examples

12. ENTHALPY REACTION CO-ORDINATE activation energy energy necessary to get the reaction going energy taken in, ∆H is positive products reactants

14. 14

15. Summary Table

16. Topic 5.2Calculation of enthalpy change Notice less sources of error here compared to our lab…

17. calorimetry • measurement of heat flow • calorimeter • device used to measure heat flow • heat • energy that is transferred from one object to another due to a difference in temperature • measures total energy in a given substance • temperature • a measure of the average kinetic energy of a substance regardless how much is there

18. 50 ml water 100 C 100 ml water 100C Temperature vs. Heat 100ml of water contains twice the heat of 50 ml.

19. Heat Capacity/Specific Heat • the amount of energy a substance absorbs depends on: • mass of material • temperature • kind of material and its ability to absorb or retain heat. • heat capacity • the amount of heat required to raise the temperature of a substance 1 oC (or 1 Kelvin) • molar heat capacity • the amount of heat required to raise the temperature of one mole 1 oC (or 1 Kelvin) • specific heat • the amount of heat required to raise the temperature of 1 gram of a substance 1 oC (or 1 Kelvin) 19

20. Specific Heat (c) values for Some Common Substances or kJ kg-1K-1 if multiply by 1000 20

21. Heat energy changeq = m c DT • q = change in heat (same as DH if pressure held constant) • m = mass in grams or kilograms • c = specific heat in J g-1 K-1or kJ kg-1 K-1 (or Celsius which has same increments as Kelvin) • DT = temperature change

22. Measuring the temperature change in a calorimetry experiment can be difficult since the system is losing heat to the surroundings even as it is generating heat. By plotting a graph of time vs. temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings.

23. Heat Transfer Problem 1 Calculate the heat that would be required to heat an aluminum cooking pan whose mass is 402.5 grams, from 20.5oC to 201.5oC. The specific heat of aluminum is 0.902 J g-1oC-1. only 4 sig. figs. q = mcDT = (402.5 g) (0.902 J g-1oC-1)(181.0oC) = 65,712.955 J = 65,710 J with correct sig. figs.

24. Heat Transfer Problem 2 What is the final temperature when 50.15 grams of water at 20.5oC is added to 80.65 grams water at 60.5oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1oC-1 Solution:Dq(Cold) = Dq(hot) so… mCDT = mCDT Let T = final temperature (50.15 g) (4.184 J g-1oC-1)(T- 20.5oC) = (80.65 g) (4.184 J g-1oC-1)(60.5oC- T) (50.15 g)(T- 20.5oC) = (80.65 g)(60.5oC- T) 50.15T -1030 = 4880 – 80.65T 130.80T = 5830 T = 44.6oC

25. Heat Transfer Problem 3 On complete combustion, 0.18g of hexane raised the temperature of 100.5g water from 22.5°C to 47.5°C. Calculate its enthalpy of combustion. • Heat absorbed by the water… • q = mcDT • q = 100.5 (4.18) (25.0) = 10,500 J which is same as 10.5 kJ • Moles of hexane burned = mass / molar mass = 0.18 g / 86 g/mol = 0.0021 moles of hexane • Enthalpy change means find heat energy / mole = 10.5 kJ/ 0.0021 mol = 5000 kJ mol -1 or 5.0 x 103kJ mol-1 hexane is C6H14