Chem 105 Wed

Transcription

1 10/12/ hem 105 Wed FRIDAY s class will start with a demo of an exothermic reaction. Please assemble Friday 1:00 PM sharp on the West lawn of Reichardt. Be careful of vehicular traffic, and do not trample the plants. 1. Energy and work 2. Enthalpy of reaction, ΔH 3. alorimetry x

2 10/12/ A chemical reaction (the system ) takes place inside the surroundings. Define E = total energy of the system. surroundings system ΔE is the energy change in a chemical reaction ΔE = heat gained or lost + work done by system on surroundings ΔH ΔH is change in enthalpy = change in energy content of atoms, including bond strength, ionic attraction, & solvation forces. PΔV P-V work = work done or by caused by change in volume of gases = P *ΔV

3 10/12/ In exothermic reactions, heat flows out of the system. In endothermic reactions, heat flows into the system. surroundings system

4 10/12/2011 ΔE = heat gained or lost + work done by system on surroundings surroundings PΔV P-V work = work done or by caused by change in volume of gases = P *ΔV system

5 10/12/ A piece of Dry Ice sealed in a Ziplock bag is subliming and raising the book. What is the SIGN of PDV for this process? surroundings > 0 2. < 0 system > 0 < 0 Don t know 3. Don t know 5

6 10/12/ If no gases are produced or consumed, or the system is kept at constant volume such as in a bomb calorimeter, then PΔV = 0 and ΔE = ΔH. This is why we use bomb calorimeters.

7 10/12/ H is the enthalpy (or heat content) of a chemical compound. Enthalpy is the energy stored in: hemical bonds within molecules or ions, Water-ion interactions (if the chemical is dissolved in water), or Other molecule-molecule interactions

8 H 10/12/2011 DH = change in enthalpy in a chemical reaction. 8 ΔH = H final H initial It is the change in energy content of atoms due to a chemical reaction. Endothermic products less stable than reactants (contain more chemical energy) Initial (reactants) Final (products) Exothermic products more stable than reactants (contain less chemical energy)

9 10/12/ The enthalpy of reaction ΔH is proportional to the # of moles of products or reactants. Example: combustion of propane 3 H 8 (g) + 5 O 2 (g) 3 O 2 (g) + 4 H 2 O(l) DH = kj ΔH is enthalpy of reaction Units of J or kj Written to right of equation - for exothermic reaction + for endothermic reaction Applies to molar quantities given in the balanced chemical equation to left

10 10/12/ Why are there no /mole shown in units of ΔH rxn? 3 H 8 (g) + 5 O 2 (g) 3 O 2 (g) + 4 H 2 O (l) DH = kj Because ΔH is a stoiochiometric quantity. It applies to the whole reaction. You can write thermochemical conversion factors for all reactants and products. THESE DO include moles in the units. For example, 2220 kj 1 molpropane 2220 kj 3molO 2 oefficient in the balanced chemical equation as given.

11 10/12/ If we burned 125 g of propane with excess O 2 (per the reaction equation on the previous slide), how much heat (in kj) is produced by this exothermic reaction? Language alert! If you say produced, that already indicates DH < 0. Do not use. 125 g H 8 3? kj 125 g 125 g g 3 8 kj 3H 8 mol 3H g H 3 8 H kj mol H 8 3H 8 3 8?? kj kj 1mol 3H 2220 kj g 3H g mol H 3H 8 3 kj A thermochemical conversion factor

12 10/12/ A really cool website: 6,290 kj per 125 g x 10 5 kj per a 20-lb propane cylinder (verify this conversion on your own..) 1.00 kilojoule = BTU (British Thermal Unit) 6,290 kj per 125 g 4.3 x 10 5 BTU per 20-lb propane cylinder Thus, a 20-lb propane cylinder should last about 10 hours fueling a 40,000- BTU/h space heater. 40# 30# 20# DOT cylinders conform to DOT transportation regulations.

13 10/12/ alorimetry

14 10/12/ In a calorimeter, the reaction may be - physically isolated from water, - or dissolved in the water. rxn Before the reaction, we measure T initial of the water. After the reaction, we measure T final of the water. DT = T final T initial We know m and for water, then calculate q for the calorimeter (q = mc DT). (for the # of moles used..) DH reaction = -q calorimeter

15 Increasing H Increasing H 10/12/ exothermic rxn DH rxn < 0 downhill endothermic rxn DH rxn > 0 uphill products reactants reactants products

16 10/12/ Initial T (T i ) Final T (T f ) Initial T (T i ) Final T (T f ) calorimeter calorimeter calorimeter calorimeter rxn rxn rxn rxn For an exothermic reaction T f > T i q cal > 0 DH < 0 (alorimeter warms up.) For an endothermic reaction T f < T i q cal < 0 DH > 0 (alorimeter cools off.)

17 10/12/ What s the difference between specific heat capacity and heat capacity? Specific heat capacity (text uses, most others use c ) Has units of: J/g or J/g K This is called an intensive property of a substance. It is a general quality that applies to any amount of it. Heat capacity (text uses m, most others use ) Has units of: J/ or J/K This is called an extensive property of a substance. This applies to a particular item, and depends on amount of stuff.

18 10/12/ Which one of the following is an INTENSIVE PROPERTY of a material? It s density 2. mass volume density mass volume

19 10/12/ A research calorimeter contains water and various other parts, so you use separate terms for calorimeter heat capacity and the water heat capacity (Or, you determine a calorimeter constant which combines these terms). Fig. 5.12, p.228

20 10/12/ Assignment: h 5.6d Homework: alorimetry calorimeter H 2 O T i rxn H H calorimeter H H H H H 2 O T f H H ΔT=T f - T i rxn -q rxn = +q water+calorimeter = q H2O + q calorimeter - q rxn =[(1025 g)x (4.184J/g- )xδt] + [(802 J/ )x ΔT] D T D T Heat capacity of water in calorimeter q rxn [( ) ] q rxn o J / J / o Heat capacity of calorimeter (i.e. the box) ΔT is equal for water and calorimeter it is factored out, and solved-for algebraically.

21 10/12/ H O 2 10 O 2 (g) + 4 H 2 O (g) ΔH = 5.15 x 10 3 kj mol Nap kj g Nap kj g Nap 1 mol Nap J DT q J ( 9724 J ) o J / rxn o / o

22 10/12/

23 10/12/ We got q rxn from DH rxn and the mass of reactant naphthalene. m w and cal were given. - q rxn =[(1025 g)x (4.184J/g- )xδt] + [(802 J/ )x ΔT] - q rxn = m w w ΔT + cal ΔT DT was unknown. Bomb calorimeter problems all use this same basic equation, but different quantities may be given or the unknown. Practice OWL problems until you are comfortable with different combinations.

24 10/12/ The End