1. Ch. 6: Energy and Thermochemistry Energy: Ability to do work Kinetic Energy: due to motion, ½mv 2 Potential Energy: stored, due to position or composition Thermal Energy: movement of molecules; related to temperature Chemical Energy: positions of nuclei and electrons (bonds) bond breaking energy is required bond making energy is released Heat and Temperature Heat is the flow of energy caused by a temperature difference; thermal energy being transferred. Temperature is a measure of the intensity of heat (thermal energy).

2. Internal Energy -E = internal energy = KE + PE -Energy Changes When reactions occur there is an energy change  E:  E = E final - E initial E final = energy of products E initial = energy of reactants  E = q + w where q = heat w = work sign convention: positive if system gains energy E reactants products

3. Measuring Thermal Energy 1.Units of Energy (see Table 6.1) SI: 1 J = 1 kg m 2 /s 2 (½mv 2 ) 1 cal = 4.184 joule (exactly) 1 kcal = 1 “Cal” 2. Heat Capacity, C --amount of heat required to raise the temp of substance by 1 °C units: energy/temp (J/°C or kJ/°C or cal/°C) quantity of heat = (heat capacity) x  t 3. Specific Heat, C s Heat Capacity of specified mass of substance (1 gram) units: usually J/g °C or cal/g °C e.g. specific heat of water = 1.00 cal/g °C = 4.18 J/g °C quantity of heat = (specific heat) x mass x  t 4. Molar Heat Capacity: –Heat Capacity per mole of a substance –e.g. molar heat capacity of water = 18.0 cal/mole °C = J/g °C x g x °C memorize!

4. Example Problems Problem The temp of 250 g H 2 O (3 sig fig) is raised from 25.0 °C to 30.0 °C. How much heat energy is required?  t = 30.0 - 25.0 = 5.0 ° C Amount of heat = (1.00 cal/g ° C) x (250 g) x (5.0 ° C) = 1,250 cal = 1.25 kcal = 1,250 cal x 4.184 J/cal = 5320 J = 5.3 kJ Problem Identify each energy change as primarily heat or work, and determine whether E sys is positive or negative. a.One billiard ball (the system) hits another one, and stops rolling. b.A book (the system) is dropped on the floor c.A father pushes his daughter on the swing (the daughter & swing are the system) a. work, negative b. work, negative c. work, positive

5. Thermal Energy Transfer Thermal energy flows from matter at higher temperature to matter at lower temp, until thermal equilibrium. q A = –q B Example Problem A 3.35 g iron rod, initially at 22.7 °C, is submerged into an unknown mass of H 2 O at 63.2 °C, in an insulated container. The final temp of the mixture is 59.5 °C. What is the mass of the water? (C s iron = 0.449 J/g∙ °C, C s water = 4.18 J/g ∙ °C) m x C S,Fe x  T Fe = –m x C S,H2O x  T H2O  T Fe = 59.5 – 22.7 = 36.8 °C for Fe  T H2O = 59.5 – 63.2 = –3.7 °C for H 2 O (3.35)(0.449)(36.8) = –m(4.18)(– 3.7) m = (3.35)(0.449)(36.8)/(4.18)(3.7) = 3.6 g

6. Internal Energy and Enthalpy E = internal energy = KE + PE  E = total energy change H = enthalpy = E + PV  H = total heat change (when P is constant) (see book for derivation) Bomb Calorimeter Coffee-Cup Calorimeter constant V; measures  E constant P; measures  H Internal energy and enthalpy are state functions. The energy change (  E) and heat change (  H) of a reaction depend only on the initial and final states of the system -- not on the specific pathway.

7. Enthalpy Changes in Chemical Reactions exothermic reactionheat is a product of the reaction -- gives off heat to the surroundings -- system warms up endothermic reactionheat is essentially a reactant --absorbs heat from the surroundings -- system cools off Enthalpy (H) -- “Heat Content” –The total energy of a chemical system at constant pressure  H = H products - H reactants endothermic reaction  H > 0 (positive) -- heat is absorbed exothermic reaction  H < 0 (negative) -- heat is released

8. Standard Heat of Reaction (  H°)  H = the value of  H for a reaction as written.  H° = the value of  H for a reaction: –Under standard conditions (temp = 25 °C, pressure = 1 atm) –With actual # moles specified by coefficients in balanced eqn e.g. reaction for the combustion of ethylene: C 2 H 4(g) + 3 O 2(g) --> 2 CO 2(g) + 2 H 2 O (l)  H° = -1411 kJ (very exothermic) i.e. 1411 kJ of heat energy are released in the reaction of 1 mole of C 2 H 4 with 3 moles of O 2 Problem If 10.0 g of C 2 H 4 are burned, how much heat is produced? (10.0 g C 2 H 4 ) x (1 mole C 2 H 4 /28.0 g C 2 H 4 ) x (1411 kJ/mole C 2 H 4 ) = 504 kJ

9. Manipulating Thermochemical Equations If reaction is reversed, change sign of  H°. If reaction is multiplied or divided by a factor, apply same factor to  H°.  H° for overall reaction = sum of  H° values for individual reactions. Problem Given the following thermochemical reactions: (eq 1) C 2 H 4(g) + 3 O 2(g) --> 2 CO 2(g) + 2 H 2 O (l)  H° = -1411 kJ (eq 2) C 2 H 5 OH (l) + 3 O 2(g) --> 2 CO 2(g) + 3 H 2 O (l)  H° = -1367 kJ Calculate  H° for the following reaction: C 2 H 4(g) + H 2 O (l) --> C 2 H 5 OH (l)

10. Example Problem, cont. Reverse 2nd reaction to put C 2 H 5 OH on product side then rewrite 1st equation and add them together. (eq 2) 2 CO 2(g) + 3 H 2 O (l) --> C 2 H 5 OH (l) + 3 O 2(g)  H° = + 1367 kJ (note the sign change!!!) (eq 1) C 2 H 4(g) + 3 O 2(g) --> 2 CO 2(g) + 2 H 2 O (l)  H° = -1411 kJ Net: C 2 H 4(g) + H 2 O (l) --> C 2 H 5 OH (l) {note: 3 O 2, 2 CO 2, and 2 H 2 O cancel out}  H° =  H° 1 +  H° 2 = 1367 + (-1411) = -44 kJ

11. Standard Heat of Formation Standard heat (enthalpy) of formation of a substance:  H° f =  H° for the formation of one mole of substance from the elements in their standard states a “formation” reaction H 2(g) + 1/2 O 2(g) --> H 2 O (l)  H° f (liq water) = -286 kJ/mole  H° f is a property of a substance -- see text for examples Practice writing formation reactions -- e.g. Na 2 SO 4 2 Na (s) + 2 O 2(g) + S (s) --> Na 2 SO 4(s)  H° f = -1385 kJ/mole

12. Hess’ Law of Heat Summation Calculate  H° for a reaction from tabulated  H° f values  H° =   H° f (products) -   H° f (reactants) Problem Determine  H° for the following reaction from  H° f values. 2 H 2 O (l) + CaSO 4(s) --> CaSO 4 2H 2 O (s)  H° =  H° f [CaSO 4 2H 2 O (s) ] - {  H° f [CaSO 4(s) ] + 2  H° f [H 2 O (l) ]}* = (-2021.1) - {(-1432.7) + 2(-285.9)} = -16.6 kJ {*units: e.g., (2 moles) x (285.9 kJ/mole) = kJ} Summary two ways to get  H° for a reaction: –By manipulating 2 or more given equations, then adding their  H°’s –From tabulated  H° f values using Hess’ Law

13. Sample Problems Write a balanced chemical equation that represents the formation reaction for (NH 4 ) 3 BO 3. Given the following thermochemical equations, calculate the standard heat of formation (  H° f ) of Mg 3 N 2(s) in kJ/mole. Mg 3 N 2(s) + 3 H 2(g) --> 3 Mg (s) + 2 NH 3(g)  H° = 371 kJ 1/2 N 2(g) + 3/2 H 2(g) --> NH 3(g)  H° = -46 kJ

14. Sample Problems Write a balanced chemical equation that represents the formation reaction for (NH 4 ) 3 BO 3. 3/2 N 2(g) + 6 H 2(g) + 3/2 O 2(g) + B (s) --> (NH 4 ) 3 BO 3(s) Given the following thermochemical equations, calculate the standard heat of formation (  H° f ) of Mg 3 N 2(s) in kJ/mole. Mg 3 N 2(s) + 3 H 2(g) --> 3 Mg (s) + 2 NH 3(g)  H° = 371 kJ 1/2 N 2(g) + 3/2 H 2(g) --> NH 3(g) Answer: 3 Mg (s) + 2 NH 3(g) --> Mg 3 N 2(s) + 3 H 2(g)  H° = -371 kJ N 2(g) + 3 H 2(g) --> 2 NH 3(g)  H° f = 2(-46 kJ) 3 Mg (s) + N 2(g) --> Mg 3 N 2(s) [the formation rxn for Mg 3 N 2 ]  H° = -371 + 2(-46) = -463 kJ   H° f for Mg 3 N 2(s) = -463 kJ/mole

15. Sample Problem The specific heat of copper is 0.387 J/g °C. The molar heat of fusion of water is 6.0 kJ/mole. If a copper rod weighing 225 g is heated to 80 °C and then immersed in 100 g of ice at 0 °C, how many grams of ice will melt?

16. Sample Problem The specific heat of copper is 0.387 J/g °C. The molar heat of fusion of water is 6.0 kJ/mole. If a copper rod weighing 225 g is heated to 80 °C and then immersed in 100 g of ice at 0 °C, how many grams of ice will melt? Answer: Heat lost by Cu = heat gained by ice (0.387 J/g °C)(225 g)(80 °C) = 6966 J = 6.966 kJ (6.966 kJ)(1 mole ice/6.0 kJ)(18.0 g ice/mole ice) = 21 g ice