AP Chapter 5 Thermochemistry HW:

5.1 – Nature of Energy Energy = Capacity to do work or transfer heat Work = Energy used to cause an object with mass to move against a force Heat = Energy needed to cause a temperature change Kinetic Energy = Energy of motion E = ½ m  2 where m=mass &  =speed Potential Energy = Stored energy

Electrostatic Potential Energy = Energy which arises from the interaction of charged particles E = k Q 1 Q 2 d o k is a constant (8.99 x 10 9 Jm/C 2 ) o Q is the electrical charge (e-=1.60 x C) o d is the distance between the particles o E is positive when same charges repel o E is negative when opposite charges attract o Lower energy = more stable (a very negative E would be very stable)

Units of Energy Joule  1 J = 1 kg*m 2 /s 2  1 calorie = J  1 Calorie (food) = 1000 cal

System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).

Open System = Matter and energy can be exchanged with the surroundings (styrofoam cup calorimeter) Closed System = Energy can be exchanged with the surroundings Isolated System = Neither energy nor matter can be exchanged. (Like a thermos, but can use a vacuum jacket)

Transferring Energy: Work and Heat Energy can be transferred as either work or heat Work = Energy used to cause an object to move against a force o w = F * d Heat = Energy transferred from a warmer object to a colder object

5.2 – First Law of Thermodynamics Energy is conserved Internal Energy (E) = Sum of all the kinetic and potential energies of the system Measure  E  E = E final – E initial +  E = final energy is higher; energy has been absorbed by the system -  E = final energy is lower; energy has been released by the system

Energy Diagrams Water -> Hydrogen + Oxygen Hydrogen + Oxygen ->Water (hydrogen balloon)

Relating Energy to Work When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w) That is,  E = q + w Energy entering the system is always +

 E, q, w, and Their Signs q + is an ENDOTHERMIC REACTION q – is an EXOTHERMIC REACTION

State Functions The internal energy of a system is independent of the path by which the system achieved that state. – In the system below, the water could have reached room temperature from either direction.

State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. State functions are often CAPITAL Letter Symbols And so,  E depends only on E initial and E final.

5.3 - Enthalpy We more commonly focus on the heat changes in chemical systems, but work can also be done The work done in most chemical systems is associated with a change in the volume of the system This is easiest to see if the system is closed with a moveable piston, but the “atmosphere” is affected the same way in an open system

Pressure – Volume Work When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −P  V If gas expands:  V is _________ and w is ________ *Work is done BY the system If gas contracts:  V is _________ and w is ________ *Work is done ON the system

Enthalpy Heat flow at constant pressure H = E +PV, when a change occurs:  H =  E + P  V and  E = q + w and w = −P  V  H = q + w – w = q  The change in enthalpy when pressure is constant is equal to the heat change  H = q

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system to the surroundings, the process is exothermic.

5.4 - Enthalpies of Reaction This quantity,  H rxn, is called the enthalpy of reaction, or the heat of reaction. Combustion of hydrogen is very exothermic Heat is released  H = H f – H i is negative

Enthalpies of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

Thermochemical Equations Show both enthalpy and mass (stoichiometry) changes Rules: 1. Stoichiometric coefficients refer to moles 2. Reversing the reaction, causes a reverse of sign of enthalpy of reaction 3. If a reaction is multiplied, the enthalpy of reaction is multiplied 4. States of matter must be shown

Thermochemical Example How much heat is released when 4.50 g methane gas is burned in a constant pressure system if: CH 4 (g) + 2O 2 (g) -> CO 2 (g) + 2H 2 O(g)  H = -890 kJ Answer = -250 kJ

5.5 - Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure  H through calorimetry, the measurement of heat flow.

5.5 - Calorimetry Heat Capacity – C = Amount of heat required to raise that sample’s temperature by 1 o C Specific Heat – C s (sometimes shown as s) = Amount of heat required to raise the temperature of a 1 GRAM sample 1 o C C s water = J/g o C q=mC s  T

Calorimetry Example How much heat is needed to warm 250 g of water from 22oC to 98oC? (answer = 79 kJ) What is the molar heat capacity of water? (answer = 75.4 J/mol K)

Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g o C), we can measure  H for the reaction with this equation: q = m  C s   T

Constant Pressure Calorimetry Unsealed calorimeter (styrofoam cup) q rxn =  H (because at constant Pressure) In an exothermic reaction, the heat GAINED by the solution/calorimeter has to be the same quantity as is lost by the reaction In an endothermic reaction, the heat LOST by the solution/calorimeter has to be the same as is gained by the reaction  q soln = -q rxn

Constant Pressure Calorimetry Example 50 mL of 1.00 M HCl is mixed with 50.0 mL of 1.00 M NaOH. The calorimeter loses only a negligible amount of heat. The initial temperature of each solution was 21.0 o C. The final temperature of the mixed solution was 27.5 o C. Calculate the heat change of this reaction and the heat of neutralization. Assume the densities and specific heats of the solutions are equal to that of water.

Solving the problem…

Lab Problems We need to find a CALORIMETER CONSTANT because our calorimeters are NOT efficient Finding the Calorimeter Constant:

Lab Problems Specific heat of a metal/glass:

Constant Volume / Bomb Calorimetry Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water. -Often used to measure heat of combustion (burning in O 2 ) Known amount

Constant Volume Calorimetry Example: A g sample of lactic acid (HC 3 H 5 O 3 ) is burned ina calorimeter whose heat capacity is kJ/ o C. The temperature increases from o C to o C. Calculate the heat of combustion for this reaction and the molar heat of combustion of lactic acid.

Solving the problem…

5.6 - Hess’ Law Many enthalpies of reaction are already recorded Not necessary to do the calorimetry ourselves all of the time When reactants are converted to products, enthalpy is the same whether the process occurs in one step or in many steps

Hess’ Law Example: C(graph) + 2H 2 (g) -> CH 4 (g) From: C(graph) + O 2 (g) ->CO 2 (g)  H = KJ 2H 2 (g) + O 2 (g) -> 2H 2 O(l) = CH 4 (g) + 2O 2 (g) -> CO 2 (g) + 2H 2 O(l)=-890.4

More Hess’ Law Practice

5.7 – Standard Enthalpy of Formation and Reaction Standard Enthalpy of Formation (  H  f ) = Energy change when a compound is formed from its constituent elements in their most stable form Table Page 189 and Appendix C (Page 1112) -At 25  C -Element in most stable form = 0 kJ

Calculating Enthalpy of Reaction from Enthalpy of Formations:  Hreaction =  n  products –  n  reactants Write for this equation: aA +bB ->cC + dD

Enthalpy of Reaction from Enthalpy of Formation Examples: Combustion of propane gas (C 3 H 8 ) forming CO 2 (g) and H 2 O (l)

More Examples: