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Transcript
Solid Liquid Gas Heat Heat = AMOUNT of internal energy Temperature = a MEASURE of the average molecular kinetic energy 10 g Pb T = 40 °C 100 g Pb T = 40 °C Both blocks are at the same temperature. Do they both contain the same amount of heat? Which substance requires more heat to increase the temperature by 5 °C? Pb 100 g Specific heat capacity (Cp): amount of heat(q) required to raise 1 g of substance by 1 °C Cp(paraffin) = 2.1 J/g°C Cp(Pb) = 0.126 J/g°C How much heat is required by the 100 g candle to increase the temperature by 5 °C? Cp(paraffin) = 2.1 J/g°C q = Cp(mass)(DT) q = (2.1 J/g°C)(100 g)(5 °C) q = 1050 J If the same amount of heat was used to heat 100 g of water [Cp(liquid water) = 4.184 J/g°C], what would be the DT of the water? q = Cp(mass)(DT) 1050 J = (4.184 J/g°C)(100g)(DT) DT = 2.5 °C For the same amount of heat and mass, DT decreases as the specific heat of the substance increases 100 kg Pb If the temperature of the lead is 327°C before it hits the water, what is the final temperature of the lead after hitting the water? q = mCpDT 200 ft DT = Tf - Ti -qPb = qH2O -(1x105 g)(0.13 J/g°C)(Tf – 327°C) = (1x104 g)(4.18 J/g°C)(Tf – 20°C) 10 kg H2O Ti = 20 °C Baltimore Shot Tower http://www.baltimore.to/ShotTower/ Tf = 93 °C Cp(Pb) = 0.13 J/g°C Cp (H2O) = 4.18 J/g°C Melting one 14-gram Al soda can requires 5.55 kJ of energy. What is its molar heat of fusion? 14 g Al 0.519 mol Al 26.982 g/mol 5.55 kJ 1 can ΔHfus(Al) x 10.7 kJ/mol 1 can 0.519 mol Al 105,000 cans are recycled in the US every minute. How many kJ/s are being used in recycling Al cans? 1.05x105 cans 0.519 mol Al 10.7 kJ x x 9.72x103 kJ/s 60 s 1 can 1 mole Al That’s equivalent to burning 2300 food Calories/s! Experiment: Heat two beakers containing 18 g of water at the same rate, and monitor their temperatures. Question: Will their temperatures increase at the same rate? - 10 0 °C 18 g H2O = 1 mole H2O 100 90 °C Experiment: Heat two beakers containing 18 g of ice and water at the same rate, and monitor their temperatures. Question: Will their temperatures increase at the same rate? Answer: It takes twice as long to increase the temperature of the liquid water by 10 °C than it does to increase the temperature of the ice by the same amount. Heating curve of water Temperature (°C) Gas warming 100 liquid warming 0 liquid + gas present solid + liquid present solid warming Heat (kJ/s) Temperature (°C) Heating curve of water 100 0 boiling/condensation point melting/freezing point Heat (kJ/s) Temperature is constant during phase transitions!! All heat energy goes to changing the state of matter. Temperature (°C) Heating curve of water 100 DHvap (heat of vaporization) 0 DHfus (heat of fusion) Heat (kJ/s) DHfus = the amount of heat needed to covert a solid into its liquid phase DHvap = the amount of heat needed to convert a liquid into its gaseous phase Temperature (°C) Heating curve of water 100 And vice versa 0 A greater DHfus = more time to melt Heat (kJ/s) H2O: DHfus = 6.01 kJ/mol DHvap = 40.7 kJ/mol H2PEw: DHfus = 20.2 kJ/mol DHvap = 10.3 kJ/mol Heating Curve Wrap Up: • The specific heat capacity (Cp)of a substance determines the temperature change observed when heat is added or withdrawn from the substance. • Temperature is INVARIANT during phase transitions. • The amount of heat required to convert one mole of the substance from one phase to another is its molar enthalpy of transition (DHfus, DHvap, DHsub). What is the sign for all three? +DH • The amount of heat given off for one mole of a substance during a phase transition while cooling is its molar enthalpy of transition (DHcond, DHsol, DHdep). What is the sign for all three? -DH • The shape of a heating curve depends upon the heating rate, specific heat capacities of the phases involved, and the enthalpies of transition. 1.0 gram of solid sodium metal is added to 100 g of water. The reaction produces sodium hydroxide and hydrogen gas. Calculate the molar heat of reaction if the water’s temperature increased by 2C. Step 1: Write out the chemical equation and balance it. 2 Na (s) + 2 H2O (l) +1 -1 2 Na OH (aq) + H2 (g) Step 2: Determine if there’s a limiting reagent.* 1H 1.0 g Na x 2 1.26 mol H 2 22.98977 g/mol 2 Na 100.0 g H O 1H 2 x 2 2.77 mol H 2 18.015 g/mol 2 H O 2 Na is the limiting reagent. Only 1.26 mol of H2 will be formed. *Choose a product that has a coefficient of 1 for best results. Step 3: Determine the amount of heat involved in the reaction. q = mCpDT q=? m = 100 g Cp = 4.184 g/JC DT = 2 C q (100 g H O)(4.184 J/gC)(2C) 837 J 2 Step 4: Calculate your molar heat of reaction. If a reaction that produced 1.26 moles of H2 also released 837 J of heat, then the molar enthalpy (heat) change for this reaction would be: ΔHrxn 837 J 664 J/mol 1.26 mol A simpler problem: How much heat is given off when 1.6 g of CH4 are burned in an excess of oxygen if DHcomb = -802 kJ/mol? Step 1: Write the reaction equation. CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) Step 2: Calculate molar amount involved 1.6 g CH 4 0.100 moles CH combusted 4 16.042 g/mol Step 3: Calculate amount of heat given off DHrxn = (-802 kJ/mol)(0.100 mol CH4) = -80.2 kJ Q: Is this an exothermic or endothermic reaction? EX 3: What is the molar heat of combustion of propene (C3H6) if burning 3.2 g releases 156 kJ of heat? Step 1: Write the reaction equation. 2 C3H6 (g) + 9 O2 (g) 6 CO2 (g) + 6 H2O (g) C3H6 (g) + 4.5 O2 (g) 3 CO2 (g) + 3 H2O (g) • This reaction equation involves the combustion of 2 moles of C3H6 and we want to find out what it is for one mole. • Save yourself a headache and simplify future calculations by dividing the reaction equation through by 2. Step 2: Convert grams of propene to moles. 3.2 g C3H6 0.076 mol C3H6 42.081 g/mol Step 3: Divide the heat released by moles of propene. 156 kJ ΔH 2053 kJ/mol comb 0.076 mol C3H6 Ace Ice Melter RAPIDLY MELTS ICE AND SNOW AND PREVENTS RE-FREEZING Ace Ice Melter features a special custom blend of superior ice melting ingredients. Together they melt even the most stubborn ice and snow and work to prevent re-freezing. Melts ice down to 0°F (-18 °C). Active Ingredients KCl NaCl NH2CONH2 (urea) C7O6H14 (methyl-D-glucopyranoside; a surfactant) What have we learned? Sometimes heat is given off during a chemical reaction. This makes it feel hotter. Sometimes heat is absorbed during a chemical reaction. This makes it feel colder. What causes it to be different? • Chemical bonds contain energy. • Add the energy of all of the bonds in the reactants together to find their total energy. • Add the energy of all of the bonds in the products together to find their total energy. • If the two numbers aren’t the same (and they almost never are), then there will be heat energy given off or taken in. 2 H2 + O2 2 H2O 2 H2 + O 2 Energy barrier 2 H2O Energy • If the products contain less energy, energy must have been given off during the reaction. 2 H2O 2 H2 + O2 2 H2 + O2 2 H2O Energy barrier Energy • If the products contain more energy, energy must have been absorbed during the reaction. If heat energy is given off during a reaction, it is called an EXOTHERMIC REACTION. Heat exits = exothermic Exothermic reactions can be recognized by a temperature INCREASE. If heat energy is absorbed during a reaction, it is called an ENDOTHERMIC REACTION. Heat enters = endothermic Endothermic reactions can be recognized by a temperature DECREASE. 2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2 2 AlBr3 + 3 Cl2 But how do we determine the heat content in the first place? 2 AlCl3 + 3 Br2 Energy DHrxn = Heat content of products – heat content reactants DHrxn < 0 Reaction is exothermic Heat of formation, DHf • The DHf of all elements in their standard state equals zero. • The DHf of all compounds is the molar heat of reaction for synthesis of the compound from its elements DHf (AlBr3): 2 Al + 3 Br2 2 AlBr3 DHrxn = 2DHf(AlBr3) DHrxn DHf(AlBr3) = 2 • Since the DHrxn can be used to find DHf, this means that DHf can be used to find DHrxn WITHOUT having to do all of the calorimetric measurements ourselves!! The Law of Conservation of Energy strikes again!! Hess’s Law: DHrxn = S DHf(products) – S DHf(reactants) 6 CO2 (g) + 6 H2O (l) C6H12O6 (s) + 6 O2 (g) DHrxn = [DHf(C6H12O6) + 6 DHf(O2)] – [6 DHf(CO2) + 6 DHf(H2O)] From DHf tables: DHf(C6H12O6) = -1250 kJ/mol DHf(CO2) = -393.5 kJ/mol DHf(H2O) = -285.8 kJ/mol DHrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)] DHrxn = +2825.8 kJ/mol Using Hess’ Law with DHrxn What is the DHcomb for ethane? 2 C6H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g) C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)DHrxn = -1323 kJ/mol C2H4 (g) + H2 (g) → C2H6 (g) DHrxn = -137 kJ/mol H2O (l) H2O (g) H2O (g) H2O (l) DHvap = +40.7 kJ/mol Energy • Water will spontaneously evaporate at room temperature even though this process is endothermic. • What is providing the uphill driving force? a measure of the disorder or randomness of the particles that make up a system • Water will spontaneously evaporate at room temperature because it allows the disorder of the water molecules to increase. • The entropy, S, of gases is >> than liquids or solids. • If Sproducts > Sreactants, DS is > 0 Predict the sign of DS: ClF (g) + F2 (g) CH3OH (l) ClF3 (g) DS < 0 CH3OH (aq) DS > 0 Are all +DS reactions spontaneous? 2 H2O (l) 2 H2 (g) + O2 (g) DS is large and positive… …but DH is large and positive as well. • Gibb’s Free Energy, DG, allows us to predict the spontaneity of a reaction using DH AND DS. If –DG spontaneous reaction 2 H2O (l) 2 H2 (g) + O2 (g) What is DG for this reaction at 25C? DHrxn = SHf(products) –SHf(reactants) DHrxn = [2(0) + 0] - 2(-285.83 kJ/mol) = 571.66 kJ/mol DSrxn = SSf(products) –SSf(reactants) DSrxn = [2(130.58 J/molK) + 205.0 J/molK] - 2(69.91 J/molK) DSrxn = 326.34 J/molK = 0.32634 kJ/molK DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK) DGrxn = +474.41 kJ/mol 2 H2O (l) 2 H2 (g) + O2 (g) What is the minimum temperature needed to make this reaction spontaneous? DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - T(0.32634 kJ/molK) Set DGrxn = 0 to find minimum temperature 0 = 571.66 kJ/mol - T(0.32634 kJ/molK) T = (571.66 kJ/mol)/(0.32634 kJ/molK) = 1751 K T > 1479 C