AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

Energy- the capacity to do work or to produce heat

1st 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.

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 mv2 m = mass in kg v = velocity in m/s units 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 reactants 2C + 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 = PV units: Latm = (atm)(L) 1 Latm = 101.325 J (not tested)

w = 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.

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/molrxn Calculate 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 Capacity C = J or C = J (g)oC (mol)oC -specific heat capacity of H2O is 4.18 J/ g oC

Energy released as heat = (mass of solution ) × (specific heat capacity) × (increase in temp) q = mCT J = ( g) (J/goC ) (T)

Example:. 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 = 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.

q = 104g(4.18J/goC)3.2oC = 1391J absorbed = 1.39 kJ 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 substance intensive property - doesn’t depend on the amount of substance heat of reaction is extensive temperature 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

2C + 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 = 1299.6 kJ/molrxn C + O2  CO2 H = 393.5 kJ/molrxn H2 + ½ O2  H2O H = 285.9 kJ /molrxn 2C + 2O2  2 CO2 H = 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

C + 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  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 4 Horeaction = Hfoproducts - Hforeactants

[0 + 6(271) + 0][2 Hof ClF3 + 2(46)] = 1196 kJ Example: Consider the reaction: 2ClF3(g) + 2NH3(g)  N2(g) + 6HF(g) + Cl2(g)  Ho = 1196 kJ/molrxn. Calculate the Hof for ClF3(g). Compound Hof NH3 -46 kJ/mol HF -271 kJ/mol [0 + 6(271) + 0][2 Hof ClF3 + 2(46)] = 1196 kJ 1626 [2Hof ClF3 92] = 1196 kJ 2Hof ClF3 = 2730 kJ Hof for ClF3 = 1365 kJ

One 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, q Added to the system added to the surroundings Work, w done on the system Work, w done on the surroundings A)  + B) C) D)

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? q w A) + B)  C) − D)