AP Chapter 5

ThermochemistryHW: 2 5 6 9 25 37 39 41 45 51 55 61 63 71 77 100 103

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 mu2 where m=mass & u=speed• Potential Energy = Stored energy

• Electrostatic Potential Energy = Energy which arises from the interaction of charged particles

E = kQ1Q2

do k is a constant (8.99 x 109 Jm/C2)o Q is the electrical charge (e-=1.60 x 10-19 C)o d is the distance between the particles o E is positive when same charges repelo E is negative when opposite charges attracto Lower energy = more stable (a very negative E

would be very stable)

Units of Energy

• Joule 1 J = 1 kg*m2/s2

1 calorie = 4.184 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 beexchanged 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 forceow = 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 DE• DE = Efinal – E initial

• + DE = final energy is higher; energy has been absorbed by the system

• -DE = final energy is lower; energy has been released by the system

Energy Diagrams

Water -> Hydrogen + OxygenHydrogen + 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 REACTIONq – 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 Einitial and Efinal.

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).

WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.

w = −PVIf gas expands: V is _________ and w is ________

*Work is done BY the systemIf gas contracts: V is _________ and w is ________

*Work is done ON the system

Enthalpy• Heat flow at constant pressureH = E +PV, when a change occurs:DH = DE + PDV and E = q + w and w = −PVDH = q + w – w = q\ The change in enthalpy when pressure is

constant is equal to the heat change\ DH = 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 ReactionThis quantity, Hrxn, is called the enthalpy of reaction, or the heat of reaction.

Combustion of hydrogen is very exothermicHeat is releasedDH = Hf – Hi is negative

Enthalpies of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Thermochemical Equations• Show both enthalpy and mass (stoichiometry)

changesRules:

1. Stoichiometric coefficients refer to moles2. Reversing the reaction, causes a reverse of sign of enthalpy of reaction3. If a reaction is multiplied, the enthalpy of reaction is multiplied4. 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:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) DH = -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 1oC

• Specific Heat – Cs (sometimes shown as s) = Amount of heat required to raise the temperature of a 1 GRAM sample 1oC

• Cs water = 4.184 J/goC

• q=mCsDT

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/goC), we can measure H for the reaction with this equation:q = m Cs T

Constant Pressure Calorimetry• Unsealed calorimeter (styrofoam cup)

q rxn = DH (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

qsoln = -qrxn

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 21.0oC. The final temperature of the mixed solution was 27.5oC. 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 O2) Known

amount

Known amount

Constant Volume Calorimetry

Example:A 0.5865 g sample of lactic acid (HC3H5O3) is

burned ina calorimeter whose heat capacity is 4.812 kJ/oC. The temperature increases from 23.10oC to 24.95oC. 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) + 2H2(g) -> CH4(g)

From:C(graph) + O2(g) ->CO2(g) DH = -393.5 KJ2H2(g) + O2(g) -> 2H2O(l) =-571.6CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) =-890.4

More Hess’ Law Practice

More Hess’ Law Practice

5.7 – Standard Enthalpy of Formation and Reaction

Standard Enthalpy of Formation (DHof) =

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 25oC-Element in most stable form = 0 kJ

Calculating Enthalpy of Reaction from Enthalpy of Formations:

DHreaction = SnDHproducts – SnDHreactants

Write for this equation:aA +bB ->cC + dD

Enthalpy of Reaction from Enthalpy of Formation Examples:

Combustion of propane gas (C3H8) forming CO2 (g) and H2O (l)

More Examples:

More Examples: