Presentation on theme: "AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY. Energy- the capacity to do work or to produce heat."— Presentation transcript:
AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY
Energy- the capacity to do work or to produce heat
1 st 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 mv 2 m = mass in kg v = velocity in m/s units are J, since J = kg m 2 s 2
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 2C 8 H O 2 16CO H 2 O ΔH= 5076 kJ/mol rxn 2C 8 H O 2 16CO H 2 O kJ Heat term is on the right side of the equation. 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 + 2H kJ C 2 H 4 2C + 2H 2 C 2 H 4 ΔH = 52.3 kJ/mol rxn Heat term is on the left side of the equation. For an endothermic reaction, the ΔH is POSITIVE! OR
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. q + w = = 15 kJ
For a gas that expands or is compressed, work can be calculated by: w = P V units: Latm = (atm)(L) 1 Latm = 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 = P V 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 + 2H 2 O 2NaOH + H 2, H = 368 kJ/mol rxn 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 H = 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) o C (mol) o C -specific heat capacity of H 2 O is 4.18 J/ g o C
Energy released as heat = (mass of solution ) × (specific heat capacity) × (increase in temp) q = mC T J = ( g) (J/g o C ) ( T)
Example: A coffee cup calorimeter contains 150 g H 2 O at 24.6 o C. A 110 g block of molybdenum is heated to 100 o C and then placed in the water in the calorimeter. The contents of the calorimeter come to a temperature of 28.0 o C. What is the heat capacity per g of molybdenum? q = mC T q = 150g(4.18J/g o C)3.4 o C = 2132J 2132J = 110g (C)(72 o C) C = 0.27J/g o C Drawing pictures may help to answer the question.
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 = mC T q = 104g(4.18J/g o C)3.2 o C = 1391J absorbed = 1.39 kJ 4.00g NH 4 NO 3 x 1 mol NH 4 NO 3 = mol 80.06g NH 4 NO kJ/ = 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.
N 2 + 2O 2 2NO 2 N 2 + O 2 2NO 2NO + O 2 2NO 2
Example: Given the following reactions and their respective enthalpy changes, calculate H for the reaction : 2C + H 2 C 2 H 2. C 2 H 2 + 5/2 O 2 2CO 2 + H 2 O H = kJ/mol rxn C + O 2 CO 2 H = kJ/mol rxn H 2 + ½ O 2 H 2 O H = kJ /mol rxn 2C + 2O 2 2 CO 2 H = 2( 393.5) kJ/mol rxn H 2 + ½ O 2 H 2 O H = kJ/mol rxn 2CO 2 + H 2 O C 2 H 2 + 5/2 O 2 H = kJ/mol rxn 2C + H 2 C 2 H 2 H = 226.7kJ/mol rxn
Example: The heat of combustion of C to CO 2 is kJ/mol of CO 2, whereas that for combustion of CO to CO 2 is kJ/mol of CO 2. Calculate the heat of combustion of C to CO. C + O 2 CO 2 kJ CO + 1/2O 2 CO 2 kJ C + O 2 CO 2 kJ CO 2 CO + 1/2 O 2 kJ C + 1/2O 2 CO kJ
Standard enthalpy of formation ( H o f ) -change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25 o C.
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 25 o C.
Values of H are found in Appendix 4 H o reaction = H f o products - H f o reactants
Example: Consider the reaction: 2ClF 3 (g) + 2NH 3 (g) N 2 (g) + 6HF(g) + Cl 2 (g) H o = 1196 kJ/mol rxn. Calculate the H o f for ClF 3 (g). [0 + 6( 271) + 0] [2 H o f ClF 3 + 2( 46)] = 1196 kJ 1626 [2 H o f ClF 3 92] = 1196 kJ 2 H o f ClF 3 = 2730 kJ H o f for ClF 3 = 1365 kJ Compound HofHof NH kJ/mol HF-271 kJ/mol
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 Heat, q 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? qw A)++ B)+ C)−+ D)−−