1. Chapter 10 Energy

2. Definitions Energy: Ability to do work –Kinetic Energy: energy due to an object’s motion (KE= 1/2mv 2 ) –Potential energy: energy due to position or composition Law of Conservation of Energy: Energy can be converted from one form to another but neither created nor destroyed

3. Definitions (cont.) Ball rolls down hill and hits other ball work: force acting over a distance state function: property of a system that changes independently of pathway. –Examples: bank acct.

4. Temperature and Heat Temperature: measure of random motions of a substance heat: flow of energy due to a temperature difference

5. Figure 10.2: Equal masses of hot and cold water.

6. Figure 10.3: H 2 O molecules in hot and cold water.

7. Figure 10.4: H 2 O molecules in same temperature water.

8. Exothermic and Endothermic system: the thing that is the focus of our attention surroundings: everything else in the universe exothermic- energy flows out of the system (it gets hot) endothermic- energy flows into the system from the surroundings (it feels cold)

9. Thermodynamics thermodynamics: study of energy 1 st law of thermo: conservation of energy –“The energy in the universe is constant” Internal energy: sum of KE + PE E can be affected by heat and work –  E = q(heat) + w(work) Thermo quantities have 2 parts –magnitude –sign

10. Practice problem Calculate the energy for the following cases: a. q = +51 KJ; w = -15 KJ b. q = +100 KJ; w = -65 KJ c. q = -65 KJ; w = -20 KJ

11. Measuring Energy changes units of energy: calorie and joule calorie: amt. of energy required to raise the temperature of 1 gram of water by 1 degree celsius. 1000 calories = 1 kilocalorie= 1 Calorie 1 calorie = 4.184 joules

12. Converting energy units Convert 1.69 Joules to calories

13. Specific heat capacity Specific heat capacity: amt. of energy required to change temp. of 1g of a substance 1 o C (also known as specific heat Q = s x m x  T (NO PHASE change!!!) Q = energy or heat s = specific heat  T = change in Temp. (T f – T i )

14. Sample specific heat problem A 1.6g sample has the appearance of gold and requires 5.8J of energy to change its temp. from 23 o C to 41 o C. If the specific heat of gold is 0.13 J/g o C, is this gold?

15. Heating/cooling curve for water

16. Summary of equations NO Phase change: Q=MCAT Phase change Q= moles x  h fus/vap  h fus = solid to liquid or liquid to solid  h vap = liquid to gas or gas to liquid

17. Thermo problem 1. How much heat is required to change... a) 32.6 g of ice at 0.0˚C into water at 0.0˚C? b) 32.6 g of water at 100.0˚C into steam at 100˚C? specific heat of water= 4.18 J/g o C  h v = 540. cal/g  h f = 80. cal/g

18. Thermochemistry Rxns can be exo- or endothermic Chemists like to know how much energy is required or released during a chemical rxn. Chemists use a new term to make finding out the energy of reactions more convenient. enthalpy (H) is heat at constant pressure

19. Calorimeter A calorimeter is a device used to measure the changes in heat of a rxn.

20. Hess’s Law Enthalpy is a property that is a state function and independent of the pathway. It doesn’t matter if a reaction takes place in one steps or multiple steps, the  H will be the same.This principle is known as Hess’s law and can be seen with the following reaction: N 2 + 2O 2  2NO 2  H = 68 KJ

21. Hess’s Law continued The reaction can also be carried out in 2 distinct steps: N 2 + O 2  2NO  H = 180 KJ 2NO + O 2  2NO 2  H = -112 KJ Add both reactions together to get the original rxn and see that  H is still 68 KJ. Some reactions are difficult to calculate energies for, but by using this stepwise approach we can calculate the energy of even difficult reactions.

22. Hess’s Law When using reactions you can: –multiply the whole reaction by a number 2(N 2 + O 2  2NO  H = 180 KJ) 2N 2 + 2O 2  4NO  H = 360 KJ –turn the equation around 2NO  N 2 + O 2  H = -180 KJ

23. Hess’s Law problem Find  H  for the reaction 2H 2 (g) + 2C(s) + O 2 (g)  C 2 H 5 OH(l), using the following thermochemical data. C 2 H 5 OH (l) + 2 O 2 (g)  2 CO 2 (g) + 2 H 2 O (l)  H = -875. kJ C (s) + O 2 (g)  CO 2 (g)  H = -394.51 kJ H 2 (g) + ½ O 2 (g)  H 2 O (l)  H = -285.8 kJ

24. Energy as a driving force. So far, we have 4 driving forces for reactions: 1. Formation of a liquid (water) 2. Formation of a gas 3. Formation of a solid(percipitate) 4. Transfer of electrons Formation of energy is also driving for a reaction

25. Entropy Entropy(S) is the measure of disorder (randomness) 2 nd Law of thermodynamics: The entropy of the universe is always increasing.