2. Calculating Heat Changes

3. First Law of Thermodynamics: the internal energy of an isolated system is constant Signs (+/-) will tell you if energy is entering or leaving a system + indicates energy enters a system - indicates energy leaves a system

4. A change in internal energy can be identified with the heat supplied at constant volume ENTHALPY (H) (comes from Greek for “heat inside”)

5. The heat supplied is equal to the change in another thermodynamic property called enthalpy (H) i.e.  H = q this relation is only valid at constant pressure As most reactions in chemistry take place at constant pressure we can say that: A change in enthalpy = heat supplied

6. Heat Transfer Specific Heat (C p ) amount of energy required to raise the temp. of 1 kg of material by 1 degree Kelvin units: J/(kg·K) or J/(kg·°C)

7. Heat Transfer Which sample will take longer to heat to 100°C? 50 g Al50 g Cu Al - It has a higher specific heat. Al will also take longer to cool down.

8. Heat Transfer Q = m   T  C p Q:heat (J) m:mass (kg)  T:change in temperature (K or °C) C p :specific heat (J/kg·K)  T = T f - T i – Q = heat loss + Q = heat gain

9. Heat Transfer A 32-g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon? A 32-g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon? GIVEN: m = 32 g T i = 60°C T f = 20°C Q = ? C p = 235 J/kg·K WORK: Q = m·  T·C p m = 32 g = 0.032 kg  T = 20°C - 60°C = – 40°C Q = (0.032kg)(-40°C)(235J/kg·K) Q = – 301 J

10. Thermochemical Equations A chemical equation that includes the heat change Heat of reaction- heat change for the equation exactly as written, represented by ΔH, when pressure is constant.

11. Hess’s law Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps. For example: C + O 2  CO 2 (pg. 165) The book tells us that this can occur as 2 steps C + ½O 2  CO  H  = – 110.5 kJ CO + ½O 2  CO 2  H  = – 283.0 kJ C + CO + O 2  CO + CO 2  H  = – 393.5 kJ I.e. C + O 2  CO 2  H  = – 393.5 kJ Hess’s law allows us to add equations. We add all reactants, products, &  H  values. We can also show how these steps add together via an “enthalpy diagram” …

12. Steps in drawing enthalpy diagrams 1. Balance the equation(s). 2. Sketch a rough draft based on  H  values. 3. Draw the overall chemical reaction as an enthalpy diagram (with the reactants on one line, and the products on the other line). 4. Draw a reaction representing the intermediate step by placing the relevant reactants on a line. 5. Check arrows:Start: two leading away Finish: two pointing to finish Intermediate: one to, one away 6. Look at equations to help complete balancing (all levels must have the same # of all atoms). 7. Add axes and  H  values.

13. C + O 2  CO 2  H  = – 393.5 kJ Reactants Intermediate Products C + O 2 CO 2 CO Enthalpy Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work.  H  = – 110.5 kJ  H  = – 283.0 kJ  H  = – 393.5 kJ + ½ O 2 C + ½ O 2  CO  H  = – 110.5 kJ CO + ½ O 2  CO 2  H  = – 283.0 kJ

14. Bond Strengths Bond strengths measured by bond enthalpy  H B (+ve values) bond breaking requires energy (+ve  H) bond making releases energy (-ve  H) Lattice Enthalpy A measure of the attraction between ions (the enthalpy change when a solid is broken up into a gas of its ions) all lattice enthalpies are positive I.e. energy is required o break up solids

15. Another Example of Hess’s Law Given: C(s) + ½ O 2 (g)  CO(g)  H = -110.5 kJ CO 2 (g)  CO(g) + ½ O 2 (g)  H = 283.0 kJ Calculate  H for:C(s) + O 2 (g)  CO 2 (g)

16. If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation,  H o f. Standard conditions (standard state): Most stable form of the substance at 1 atm and 25 o C (298 K). Standard enthalpy,  H o, is the enthalpy measured when everything is in its standard state. Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states. Enthalpies of Formation

17. If there is more than one state for a substance under standard conditions, the more stable one is used. Standard enthalpy of formation of the most stable form of an element is zero. Enthalpies of Formation

18. Using Enthalpies of Formation to Calculate Enthalpies of Reaction For a reaction Note: n & m are stoichiometric coefficients. Calculate heat of reaction for the combustion of propane gas giving carbon dioxide and water. Enthalpies of Formation

19. Foods 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. Energy in our bodies comes from carbohydrates and fats (mostly). Intestines: carbohydrates converted into glucose: C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O,  H = -2816 kJ Fats break down as follows: 2C 57 H 110 O 6 + 163O 2  114CO 2 + 110H 2 O,  H = -75,520 kJ Fats contain more energy; are not water soluble, so are good for energy storage. Foods and Fuels