All course materials, including lectures, class notes, quizzes, exams, handouts, presentations, and other materials provided to students or this course are protected intellectual property. As such, the unauthorized purchase or sale o these materials may result in disciplinary sanctions under the Campus Student Code. Thermochemistry deals with the heat involved in chemical and physical changes Two types o energy 2 H2(g) + O2(g) 2 H2O(g) + energy Potential Energy: due to position or composition Examples: gravitational, chemical, electrical, spring, electrochemical Kinetic energy: due to motion o the object KE = ½ m v 2 Potential Energy is a State Function depends only on condition E based on the dierence between initial and inal state = inal initial Value is independent o how the condition was achieved Two skiers: one climbs up mountain, one rides the tram At top o mountain Potential Energy (PE) o each skier is same; PE= m g h Work is a Path Function The amount o work done by the skiers is dierent Skier 2 does more work than Skier 1 We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 1 o 13
Law o Conservation o Energy Potential energy at the molecular level Energy stored in chemical bonds Electrostatic interactions We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 2 o 13
Terminology o Energy Transer System the part o the universe that is the ocus o study Surroundings - everything in the universe that is not part o the system Systems can be open, closed, or isolated Open mass and heat transer with surroundings can occur Closed no mass allowed but heat transer with surroundings can occur Isolated no mass or heat transer with surroundings s Endothermic (q > ) Exothermic (q < ) Phase Changes and Heat Flow Kinetic energy at the molecular level We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 3 o 13
The internal energy, E Sum o KE and PE o all components o the system Change in Internal Energy, E E = q + w q=heat, w=work Pressure- Volume (expansion) work w = P V Where P = pressure, V = change in volume Heat released by burning gasoline in the cylinder o an automobile engine causes the piston to move, converting some o the heat to work Enthalpy, H: Energy Change at Constant Pressure E or reactions that do not involve gases or reactions in which the total moles o gas do not change Enthalpy diagrams or exothermic and endothermic processes In each o the ollowing cases, determine the sign o, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram. H2(g) + ½ O2(g) H2O(l) + 285.8 kj 4.7 kj + H2O(l) H2O(g) We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 4 o 13
Units o Energy The SI unit o energy is the joule (J). 1 J = 1 kg m 2 /s 2 The calorie energy needed to raise the temp o exactly one-gram o water by exactly one-degree Kelvin 1 cal = 4.184 J Heat and temperature change q = c m T, q= heat lost or gained c = speciic heat capacity m = mass in grams The speciic heat capacity (c) o a substance is the quantity o heat required to change the temperature o 1 gram o the substance by 1 K. T = Tinal Tinitial A layer o copper welded to the bottom o a skillet weighs 125 g. How much heat is needed to raise the temperature o the copper layer rom 25 C to 3. C? The speciic heat capacity o Cu is.387 J/g K. A 25.9 g block o lead initially at 25. C receives 2.5 kj o heat. What is the inal temperature o the lead? The speciic heat capacity o Pb is.129 J/g K It takes 11.2 kj o energy to raise the temperature o 145 g o benzene rom 25. C to 7. C. What is the speciic heat o benzene? We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 5 o 13
A heating curve or the conversion o solid water (ice) to gaseous water (steam) Cp, water,(s) = 2.8 J/g K Cp, water,(l) = 4.184 J/g K Cp, water,(g) = 2.1 J/g K us,water = 6.1 kj/mole vap,water = 4.7 kj/mole Calculate the amount o heat (in J) needed to convert 4. g o ice at. ºC to liquid water at 8. ºC? 2.7 1 4 J We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 6 o 13
Calculate the heat involved when 24.3 grams o water is converted rom 85. o C to 117. o C Calorimetry Problems Experimental Determination o Carry out chemical reaction such that heat o reaction is absorbed by a quantity o water The temperature change o the water is measured and is used to calculate the amount o heat involved in the chemical reaction First Law o Thermodynamics A change in the energy o the system must be accompanied by an equal and opposite change in the energy o the surroundings Heat lost + heat gained = We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 7 o 13
A 22.5 g solid is heated in a test-tube to 1. C and added to 5. g o water in a coee-cup calorimeter. The water temperature changes rom 25.1 C to 28.49 C. Find the speciic heat capacity o the solid. What is the inal temperature o the system when a 3.g iron ball at 7. o C is dropped into 25. g water initially at a temperature o 25 o C? Speciic heat capacity o Fe(s) =.449 J/g o C and that o water is 4.184 J/g o C. 25.6 o C We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 8 o 13
5. ml o.25 M NaOH is placed in a coee-cup calorimeter at 25. o C and 25. ml o.5 M HCl is careully added, also at 25. o C. Ater stirring, the inal temperature is 27.21 o C. Calculate rxn in kj/mol o H2O ormed. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 9 o 13
Stoichiometry o Thermochemical Equations A thermochemical equation is a balanced equation that includes rxn The value o can be used in a calculation in the same way as a mole ratio The major source o aluminum in the world is bauxite (mostly aluminum oxide). Its decomposition can be represented by the thermochemical equation given below. How many grams o aluminum can orm when 1. 1 3 kj o heat is transerred? 2 Al2O3(s) 4 Al(s) + 3 O2(g) rxn = 3352 kj Consider the thermochemical equation C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) = 241 kj Is this reaction exothermic or endothermic? Is heat released or absorbed? How much heat is involved when 5. g C3H8 is combusted? Manipulation o Thermochemical Equations Whatever is done to the coeicients must also be done to the value o Given the thermochemical equation 2 N2(g) + O2(g) 2 N2O(g) =+163.14 kj Calculate the standard enthalpy changes or the reaction N2(g) + ½ O2(g) N2O(g) Calculate the standard enthalpy changes or the reaction 8 N2(g) + 4 O2(g) 8 N2O(g) Calculate the standard enthalpy changes or the reaction 2 N2O(g) 2 N2(g) + O2(g) We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 1 o 13
Hess s law states that the enthalpy change o an overall process is the sum o the enthalpy changes o its individual steps overall = 1 + 2 + 3 +...+ n or reaction overall can be calculated i the values or the individual steps are known Using Hess s Law to calculate or an overall process Identiy the target equation, the step whose is unknown Note the amount o each reactant and product Manipulate each equation with known values so that the target amount o each substance is on the correct side o the equation Change the sign o when you reverse an equation Multiply amount and by the same actor Add the manipulated equations and their resulting values to get the target eqn and its All substances except those in the target equation must cancel Given the ollowing inormation, calculate the unknown target Equation A: 2 SO2(g) 2 S(s) + 2 O2(g) A = +593.6 kj Equation B: 2 S(s) + 3 O2(g) 2 SO3(g) B= 791.4 kj Use Hess s law to calculate or the reaction: 2 SO2(g) + O2(g) 2 SO3(g) target =? We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 11 o 13
Two gaseous pollutants that orm in auto exhausts are CO and NO. An environmental chemist is studying ways to convert them to less harmul gases through the ollowing reaction Target: CO(g) + NO(g) CO2(g) + ½N2(g) target =? Given the ollowing inormation, calculate the unknown Equation C: CO(g) + ½ O2(g) CO2(g) C = 283. kj Equation D: N2(g) + O2(g) 2NO(g) D = + 18.6 kj Use Hess s law to calculate or the reaction: 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) target =? Given the ollowing inormation, calculate the unknown target Equation E: N2(g) + O2(g) 2 NO(g) E = +18.6 kj Equation F: N2(g) + 3 H2(g) 2 NH3(g) F = 91.8 kj Equation G: 2 H2(g) + O2(g) 2 H2O(g) G = 483.7 kj We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 12 o 13
Standard Heat (Enthalpy) o Formation, delta H e zero The enthalpy change (heat absorbed or released) when one mole o a compound is ormed rom its elements in their standard states under standard conditions (1 bar or 1 atm and 25 C) Values o Standard State o Element are tabulated and depend on the substance and its physical state Most stable orm and physical state o element under standard conditions By deinition: = or elements in standard state under standard conditions Element Std State Element Std State Element Std State oxygen O2(g) Carbon C(s) carbon C(s,gr) nitrogen N2(g) chlorine Cl2(g) sodium Na(s) helium He(g) bromine Br2(l) sulur S(s) luorine F2(g) iodine I2(s) Zinc Zn(s) For the compounds shown below, write thermochemical equations that represent the given standard enthalpies o ormation. (CO2(g)) = 394 kj/mol (NaCl(s)) = 411 kj/mol,prod = short-hand representation rxn products react,react Calculate the standard enthalpy or the reactions below rom H2S(g) H2(g) + S(s) SO2(g) + 3 H2(g) H2S(g) + 2 H2O(l) rxn = rxn = values H2O(l) 285.5 H2O(g) 241.8 H2S(g) 2.6 SO2(g) 296.8 in kj/mole 2 H2(g) + O2(g) 2 H2O(l) rxn = We are what we repeatedly do. Excellence, then, is not an act, but a habit. Aristotle Page 13 o 13