Calorimetry This equation can be used to determine any of the variables here. You will not have to solve for C, since we will always assume that the energy transfer is being absorbed by or released by a measured quantity of water, whose specific heat is given above. Solving for q Solving for m Solving for DT

Solving for q How many joules are absorbed by 100.0 grams of water in a calorimeter if the temperature of the water increases from 20.0oC to 50.0oC? q = mCDT = (100.0 g)(4.18 J/goC)(30.0oC) = 12500 J

Solving for m A sample of water in a calorimeter cup increases from 25oC to 50.oC by the addition of 500.0 joules of energy. What is the mass of water in the calorimeter cup? q = mCDT, so m = q / CDT = (500.0 J) / (4.18 J/goC)(25oC) = 4.8 g

Solving for DT If a 50.0 gram sample of water in a calorimeter cup absorbs 1000.0 joules of energy, how much will the temperature rise by? q = mCDT, so DT = q / mC = (1000.0 J)/(50.0 g)(4.18 J/goC) = 4.8oC If the water started at 20.0oC, what will the final temperature be? Since the water ABSORBS the energy, its temperature will INCREASE by the DT: 20.0oC + 4.8oC = 24.8oC

Reaction Rate Reactions happen when reacting particles collide with sufficient energy (activation energy) and at the proper angle. Anything that makes more collisions in a given time will make the reaction rate increase. Increasing temperature Increasing concentration (pressure for gases) Increasing surface area (solids) Adding a catalyst makes a reaction go faster by removing steps from the mechanism and lowering the activation energy without getting used up in the process.

Heat of Reaction Reactions either absorb PE (endothermic, +DH) or release PE (exothermic, -DH) Exothermic, PEKE, Temp Endothermic, KEPE, Temp Rewriting the equation with heat included: 4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351 kJ N2(g) + O2(g) +182.6 kJ  2 NO(g)

5.3: Enthalpy, H Since most reactions occur in containers open to the air, w is often negligible. If a reaction produces a gas, the gas must do work to expand against the atmosphere. This mechanical work of expansion is called PV (pressure-volume) work. Enthalpy (H): change in the heat content (qp) of a reaction at constant pressure H = E + PV H = E + PV (at constant P) H = (qp + w) + (-w) H = qp

Sign conventions H > 0 Heat is gained from surroundings + H in endothermic reaction H < 0 Heat is released to surroundings - H in exothermic reaction

5.4: Enthalpy of Reaction (Hrxn) Also called heat of reaction: Enthalpy is an extensive property (depends on amounts of reactants involved). Ex: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l) Hrxn = - 890. kJ Combustion of 1 mol CH4 produces 890. kJ … of 2 mol CH4 → (2)(-890. kJ) = -1780 kJ What is the H of the combustion of 100. g CH4?

Hreaction = - Hreverse reaction CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l) H = - 890. kJ CO2 (g) + 2 H2O (l)  CH4 (g) + 2 O2 (g) H = +890. kJ

5.6: Hess’ Law Ex. What is DHrxn of the combustion of propane? If a rxn is carried out in a series of steps, Hrxn =  (Hsteps) = H1 + H2 + H3 + … Germain Hess (1802-1850) Ex. What is DHrxn of the combustion of propane? C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) 3 C (s) + 4 H2 (g)  C3 H8 (g) H1 = -103.85 kJ C (s) + O2 (g)  CO2 (g) H2 = -393.5 kJ H2 (g) + ½ O2 (g)  H2O (l) H3 = -285.8 kJ C3H8 (g)  3 C (s) + 4 H2 (g) H1 = +103.85 kJ http://en.wikipedia.org/wiki/Germain_Hess Born in Switzerland, moved to St Petersburg at age 3 with his family. His father was an artist. Hess was also interested in geology. 3[ ] 3( ) 4[ ] 4( ) Hrxn = 103.85 + 3(- 393.5) + 4(- 285.8) = - 2219.8 kJ

5.7: Enthalpy of Formation (Hf) Formation: a reaction that describes a substance formed from its elements NH4NO3 (s) Standard enthalpy of formation (Hf): forms 1 mole of compound from its elements in their standard state (at 298 K) C2H5OH (l) Hf = - 277.7 kJ Hf of the most stable form of any element equals zero. H2, N2 , O2 , F2 , Cl2 (g) Br2 (l), Hg (l) C (graphite), P4 (s, white), S8 (s), I2 (s) Ex: 2 N2 (g) + 4 H2 (g) + 3 O2 (g)  2 2 C (graphite) + 3 H2 (g) + ½ O2 (g) 

Hess’ Law (again) Ex. Combustion of propane: C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Given: Compound Hrxn (kJ/mol) C3H8 (g) -103.85 CO2 (g) -393.5 H2O (l) -285.8 H2O (g) -241.82 Hrxn = [3(- 393.5) + 4(- 285.8)] – [1(-103.85) + 5(0)] = - 2219.8 kJ

5.5: Calorimetry q = C DT q = m c DT Measurement of heat flow Heat capacity, C: amount of heat required to raise T of an object by 1 K q = C DT Specific heat (or specific heat capacity, c): heat capacity of 1 g of a substance q = m c DT Ex: How much energy is required to heat 40.0 g of iron (c = 0.45 J/(g K) from 0.0ºC to 100.0ºC? q = m c DT = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC) = 1800 J

Potential Energy Diagrams Steps of a reactions: Reactants have a certain amount of PE stored in their bonds (Heat of Reactants) The reactants are given enough energy to collide and react (Activation Energy) The resulting intermediate has the highest energy that the reaction can make (Heat of Activated Complex) The activated complex breaks down and forms the products, which have a certain amount of PE stored in their bonds (Heat of Products) Hproducts - Hreactants = DH EXAMPLES

Making a PE Diagram X axis: Reaction Coordinate (time, no units) Y axis: PE (kJ) Three lines representing energy (Hreactants, Hactivated complex, Hproducts) Two arrows representing energy changes: From Hreactants to Hactivated complex: Activation Energy From Hreactants to Hproducts : DH ENDOTHERMIC PE DIAGRAM EXOTHERMIC PE DIAGRAM

Endothermic PE Diagram If a catalyst is added?

Endothermic with Catalyst The red line represents the catalyzed reaction.

Exothermic PE Diagram What does it look like with a catalyst?

Exothermic with a Catalyst The red line represents the catalyzed reaction. Lower A.E. and faster reaction time!

19.1: Spontaneous Processes Reversible reaction: can proceed forward and backward along same path (equilibrium is possible) Ex: H2O freezing & melting at 0ºC Irreversible reaction: cannot proceed forward and backward along same path Ex: ice melting at room temperature Spontaneous reaction: an irreversible reaction that occurs without outside intervention Ex: Gases expand to fill a container, ice melts at room temperature (even though endothermic), salts dissolve in water

Entropy Entropy (S): a measure of molecular randomness or disorder S is a state function: DS = Sfinal - Sinitial + DS = more randomness - DS = less randomness For a reversible process that occurs at constant T: Units: J/K

Examples of spontaneous reactions: Particles are more evenly distributed Particles are no longer in an ordered crystal lattice Ions are not locked in crystal lattice Gases expand to fill a container: Ice melts at room temperature: Salts dissolve in water:

19.3: 3rd Law of Thermodynamics The entropy of a crystalline solid at 0 K is 0. How to predict DS: Sgas > Sliquid > Ssolid Smore gas molecules > Sfewer gas molecules Shigh T > Slow T Ex: Predict the sign of DS for the following: CaCO3 (s) → CaO (s) + CO2 (g) N2 (g) + 3 H2 (g) → 2 NH3 (g) N2 (g) + O2 (g) → 2 NO (g) +, solid to gas -, fewer moles produced ?

DG = DH - TDS DG° = DH° - TDS° 19.5: Gibbs free energy, G Represents combination of two forces that drive a reaction: DH (enthalpy) and DS (disorder) Units: kJ/mol DG = DH - TDS DG° = DH° - TDS° (absolute T) Josiah Willard Gibbs (1839-1903) Called “free energy” because DG represents maximum useful work that can be done by the system on its surroundings in a spontaneous reaction. (See p. 708 for more details.)

Determining Spontaneity of a Reaction If DG is: reaction is spontaneous (proceeds in the forward direction Positive Forward reaction is non-spontaneous; the reverse reaction is spontaneous Zero The system is at equilibrium

19.6: Free Energy & Temperature DG depends on enthalpy, entropy, and temperature: DG = DH - TDS DH DS DG and reaction outcome - + Always (-); spontaneous at all T 2 O3 (g) → 3 O2 (g) + - Always +; non-spontaneous at all T 3 O2 (g) → 2 O3 (g) - - Spontaneous at low T; non-spontaneous at high T H2O (l) → H2O (s) + + Spontaneous only at high T ; non-spontaneous at low T H2O (s) → H2O (l)