2Energy- the capacity to do work or to produce heat
31st Law of Thermodynamics: Law of Conservation of Energy Energy can be converted from one form to another but it can be neither created nor destroyed. The total amount of energy in the universe is constant.
4Potential energy- energy due to position or composition Kinetic energy- energy due to the motion of an object -depends on mass and velocity of an object
6KE = ½ mv2 m = mass in kg v = velocity in m/s units are J, since J = kg m2 s2
7Heat- involves a transfer of energy between two objects due to a temperature difference.
8Work- force acting over a distance -involves a transfer of energy
9Temperature- a property that reflects random motions of the particles of a particular substance
10Exothermic- reaction which releases heat energy flows out of the systempotential energy is changed to thermal energyproducts have lower potential energy than reactantsOR2C8H O2 16CO2 + 18H2O ΔH= 5076 kJ/molrxnHeat term is on the rightside of the equation.2C8H O2 16CO2 + 18H2O kJFor an exothermic reaction, ΔH is negative.
12Endothermic- reaction which absorbs heat energy flows into the systemthermal energy is changed into potential energyproducts have higher PE than reactants 2C + 2H kJ C2H4Heat term is on the leftside of the equation.OR2C + 2H2 C2H ΔH = 52.3 kJ/molrxnFor an endothermicreaction, the ΔH is POSITIVE!
20Example: Calculate E if q = 50 kJ and w = +35kJ. DE = q + w= = 15 kJ
21For a gas that expands or is compressed, work can be calculated by: w = PV units: Latm = (atm)(L) 1 Latm = J (not tested)
22w = PDV w = 1.5 atm (1.985L) w = 3.0L . atm Example: Calculate the work if the volume of a gas is increased from 15 mL to 2.0 L at a constant pressure of 1.5 atm.
23At constant pressure, the terms heat of reaction and change in enthalpy are used interchangeably.
24Example: For the reaction 2Na + 2H2O 2NaOH + H2 , H = 368 kJ/molrxn Calculate the heat change that occurs when 3.5 g of Na reacts with excess water.3.5g Na 1 mol Na kJ =23.0g Na 2 mol NaDH = 28 kJ (or 28 kJ is released)
25Calorimetry - the science of measuring heat flow based on observing the temperature change when a body absorbs or discharges heat.instrument is the calorimeter
26Calorimetry can be used to find the ΔH for a chemical reaction, the heat involved in a physical change, or the specific heat of a substance.
27Specific Heat Capacity C = J or C = J (g)oC (mol)oC -specific heat capacity of H2O is 4.18 J/ g oC
30Energy released as heat = (mass of solution ) × (specific heat capacity) × (increase in temp) q = mCT J = ( g) (J/goC ) (T)
31Example:. A coffee cup calorimeter contains 150 g H2O at 24. 6 oC Example: A coffee cup calorimeter contains 150 g H2O at 24.6 oC. A 110 g block of molybdenum is heated to 100oC and then placed in the water in the calorimeter. The contents of the calorimeter come to a temperature of 28.0oC. What is the heat capacity per g of molybdenum?q = mCDTq = 150g(4.18J/goC)3.4oC = 2132J2132J = 110g (C)(72oC)C = 0.27J/goCDrawing pictures may help to answer the question.
32q = 104g(4.18J/goC)3.2oC = 1391J absorbed = 1.39 kJ Example: 4.00g of ammonium nitrate are added to mL of water in a polystyrene cup. The water in the cup is initially at a temperature of 22.5°C and decreases to a temperature of 19.3°C. Determine the heat of solution of ammonium nitrate in kJ/mol. Assume that the heat absorbed or released by the calorimeter is negligible.q = mCDTq = 104g(4.18J/goC)3.2oC = 1391J absorbed = 1.39 kJ4.00g NH4NO3 x 1 mol NH4NO3 = mol80.06g NH4NO31.39 kJ/ = 28 kJ/molDrawing pictures may help to answer the question.
33extensive property - depends on the amount of substance intensive property - doesn’t depend on the amount of substance heat of reaction is extensive temperature is intensive
34Hess’s Law -the change in enthalpy (H) is the same whether the reaction occurs in one step or in several steps. H is not dependent on the reaction pathway.
35The sum of the H for each step equals the H for the total reaction. 1. If a reaction is reversed, the sign of H is reversed. 2. If the coefficients in a reaction are multiplied by an integer, the value of H is multiplied by the same integer.
372C + 2O2 2 CO2 H = 2(393.5) kJ/molrxn Example: Given the following reactions and their respective enthalpy changes, calculate H for the reaction: 2C + H2 C2H2. C2H2 + 5/2 O2 2CO2 + H2O H = kJ/molrxn C + O2 CO H = 393.5 kJ/molrxn H2 + ½ O2 H2O H = 285.9 kJ /molrxn2C + 2O2 2 CO H = 2(393.5) kJ/molrxnH2 + ½ O2 H2O H = 285.9 kJ/molrxn2CO2 + H2O C2H2 + 5/2 O H = kJ/molrxn2C + H2 C2H2 H = 226.7kJ/molrxn
38C + O2 CO2 DH = -393.5 kJ CO + 1/2O2 CO2 DH = -283.0 kJ Example: The heat of combustion of C to CO2 is 393.5 kJ/mol of CO2, whereas that for combustion of CO to CO2 is 283.0 kJ/mol of CO2. Calculate the heat of combustion of C to CO.C + O2 CO DH = kJCO + 1/2O2 CO DH = kJC + O2 CO DH = kJCO2 CO + 1/2 O2 DH = kJC + 1/2O2 CODH = kJ
39Standard enthalpy of formation (Hof) -change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25oC.
40Standard States- for gases, pressure is 1 atm for a substance in solution, the concentration is 1 Mfor a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid.for an element, the standard state is the form in which the element exists under conditions of 1 atm and 25oC.
41Values of H are found in Appendix 4 Horeaction = Hfoproducts - Hforeactants
43One version of the First Law of Thermodynamics is expressed as One version of the First Law of Thermodynamics is expressed as ∆E = q + w Which gives the sign convention for this relationship that is usually used in chemistry?Heat, qAdded to the systemadded to the surroundingsWork, w done on the systemWork, wdone on the surroundingsA)+B)C)D)
44When one mole of liquid compound X is vaporized, it is observed that the flexible container containing X expands. What is the sign of q and the sign of w?qwA)+B)C)−D)