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AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY. Energy - the capacity to do work or to produce heat.

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## AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

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**1st Law of Thermodynamics:Law of Conservation of**EnergyEnergy 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.**Potential 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**KE = ½ mv2m = mass in kg v = velocity in m/sunits are J,**since J = kg m2 s2**Heat- involves a transfer of energy between two objects due**to a temperature difference.**Work- force acting over a distance-involves a transfer of**energy**Temperature- a property that reflects random motions of the**particles of a particular substance**Exothermic- reaction which releases heat**• energy flows out of the system • potential energy is changed to thermal energy • products have lower potential energy than reactants OR 2C8H18 + 25O2 16CO2 + 18H2O ΔH=5076 kJ/molrxn Heat term is on the right side of the equation. 2C8H18 + 25O2 16CO2 + 18H2O + 5076 kJ For an exothermic reaction, ΔH is negative.**Endothermic- reaction which absorbs heat**• energy flows into the system • thermal energy is changed into potential energy • products have higher PE than reactants2C + 2H2 + 52.3 kJ C2H4 Heat term is on the left side of the equation. OR 2C + 2H2 C2H4 ΔH = 52.3 kJ/molrxn For an endothermic reaction, the ΔH is POSITIVE!**The system is our reaction. The surroundings are everything**else.**Internal energy (E) of a system is the sum of the kinetic**and potential energies of all the particles in a system.**E = q + w**• E is the change in the system’s internal energy • q represents heat • w represents work • usually in J or kJ**Thermodynamic quantities always consist of a number and a**sign (+ or ). The sign represents the systems point of view. (Engineers use the surroundings point of view)**Exothermic q(systems energy is decreasing)**Endothermic +q(systems energy is increasing)**Example: Calculate E if q = 50 kJ and w = +35kJ.**DE =q + w = 50 + 35 = 15 kJ**For a gas that expands or is compressed, work can be**calculated by:w = PVunits: Latm = (atm)(L) 1 Latm = 101.325 J (not tested)**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. w = PDV w = 1.5 atm (1.985L) w = 3.0L . atm**At constant pressure, the terms heat of reaction and change**in enthalpy are used interchangeably.**Example:For the reaction 2Na + 2H2O 2NaOH + H2 , H**= 368 kJ/molrxnCalculate the heat change that occurs when 3.5 g of Na reacts with excess water. 3.5g Na 1 mol Na 368 kJ = 23.0g Na 2 mol Na DH = 28 kJ (or 28 kJ is released)**Calorimetry - the science of measuring heat flow**• based on observing the temperature change when a body absorbs or discharges heat. • instrument is the calorimeter**Calorimetry 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.**Specific Heat CapacityC = J or C = J**(g)oC (mol)oC-specific heat capacity of H2O is 4.18 J/ goC**Energy released as heat = (mass of solution ) ×(specific**heat capacity) ×(increase in temp)q = mCT J = ( g) (J/goC ) (T)**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 = mCDT q = 150g(4.18J/goC)3.4oC = 2132J 2132J = 110g (C)(72oC) C = 0.27J/goC Drawing pictures may help to answer the question.**Example: 4.00g of ammonium nitrate are added to 100.0 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 = mCDT q = 104g(4.18J/goC)3.2oC = 1391J absorbed = 1.39 kJ 4.00g NH4NO3 x 1 mol NH4NO3 = 0.0500 mol 80.06g NH4NO3 1.39 kJ/0.0500 =28 kJ/mol Drawing pictures may help to answer the question.**extensive property- depends on the amount of**substanceintensive property - doesn’t depend on the amount of substanceheat of reaction is extensivetemperature is intensive**Hess’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.**The 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.**N2 + O2 2NO**2NO + O2 2NO2 N2 + 2O2 2NO2**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 = 1299.6 kJ/molrxnC + O2 CO2H = 393.5 kJ/molrxnH2 + ½ O2 H2O H = 285.9 kJ /molrxn 2C + 2O2 2 CO2H = 2(393.5) kJ/molrxn H2 + ½ O2 H2O H = 285.9 kJ/molrxn 2CO2 + H2O C2H2 + 5/2 O2 H = +1299.6 kJ/molrxn 2C + H2 C2H2 H = 226.7kJ/molrxn**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 CO2 DH = -393.5 kJ CO + 1/2O2 CO2 DH = -283.0 kJ C + O2 CO2 DH = -393.5 kJ CO2 CO + 1/2 O2 DH = +283.0 kJ C + 1/2O2 CO DH = -110.5 kJ**Standard 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.**Standard States-**• for gases, pressure is 1 atm • for a substance in solution, the concentration is 1 M • for 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.**Values of H are found in Appendix 4Horeaction =**Hfoproducts - Hforeactants**Example: Consider the reaction:2ClF3(g) + 2NH3(g) N2(g)**+ 6HF(g) + Cl2(g) Ho = 1196 kJ/molrxn.Calculate the Hoffor ClF3(g). [0 + 6(271) + 0][2 HofClF3 + 2(46)] = 1196 kJ 1626 [2Hof ClF392] = 1196 kJ 2HofClF3 = 2730 kJ HofforClF3 = 1365 kJ**One version of the First Law of Thermodynamics is expressed**as ∆E = q + wWhich gives the sign convention for this relationship that is usually used in chemistry?**When 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?

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