Ch. 6 Thermochemistry. Course Support AP Chemistry Prep-book suggestions (4) Suggested practice problems are a minimum. Form study groups! Support before.

Ch. 6 Thermochemistry

Course Support• AP Chemistry Prep-book suggestions (4)

• Suggested practice problems are a minimum.

• Form study groups!

• Support before and after school

• Readings done prior to class (class notes are a refresher prior to application).

• YOU CAN DO THIS!!

2014 AP Chem Exam

Exam 2 Analysis

• Averages

• Grab a different colored pen/pencil.• Grab out a sheet of paper for notes.

Ch. 6 Thermochemistry TopicsCh. 6 Thermochemistry-• Internal energy, heat and work-keep track of signs and units…1 kJ is 1000 J• Exothermic and endothermic reactions. What is happening at molecular level and how is

this reflected with H and the signs?• Energy profiles• Hess’s Law can be used to calc heat of reaction• Heat of formation (Hf) can be used to calc heat of reaction (in prdts-reactants eqn)• Heat of formation-all elements in natural state are zero. Why?• Know what (Hf) values mean and what they are used for.• Bond energies can be used to calc heat of reaction• Know all heats of reaction symbols and what they mean• Hand warmer lab report. Know how to do the calcs both with a calorimeter constant (qrxn

= -(q soln + q cal) and where heat lost to calorimeter is 0…..(q soln = - q rxn) Watch signs and units!. A change in temperature is not a Joule!

• Heat capacity-describe it. What does it mean to have high or low heat capacity?• Do not forget PV=nRT• Stoichiometry and thermochem. Know what a Hrxn has to do with reactants/prdts and be

able to manipulate simple stoich with heat.

The Nature of EnergyKey Terms• Energy – the capacity to do work or produce heat• Law of conservation of energy – energy can be converted

from one form to another but can be neither created nor destroyed. (The total energy content of the universe is constant.)• Potential energy – energy due to position or composition • Lifting a pencil in the air• Propane burned (attractive and repulsive forces

between charged particles)• Kinetic energy – energy due to motion • KE = ½ mv2

• Energy conversion

• Heat – involves transfer of energy between two objects due to a temperature difference (different from Temperature)

• Work – force acting over a distance• Force x distance

• State function (property): a property that is independent of pathway. • It does not matter how you get there, the difference in the value

is the same. Home to school. It does not matter if you walk, bike, or drive the actual distance between the two locations is the same.

• Energy is a state function – work and heat are NOT.• Thus, the amount of work and heat ARE dependent on the

pathway

The universe is composed of the system and the surroundings.• System• Part of the universe we are focusing on.

• Surroundings• Everything else in the universe, besides the system.

• Exothermic• Energy(as heat) flows out of the system (feels

warmer).• “heat” in product-side of chemical equation

• Endothermic• Energy (as heat) flows into the system (feels colder).• “heat” in the reactant-side of chemical reaction

Thermodynamics• Thermodynamics- the study of energy and its interconversions

• First Law of Thermodynamics • The energy of the universe is constant.

• Internal energy – sum of the kinetic and potential energy of all the “particles” in the system.

• Change in internal energy can be by flow of work, heat, or both.

ΔE = q + w• Thermodynamic quantities always have a number and a sign.

The SI unit for Energy is the Joule (J)

• The sign is from the system’s perspective. • Energy flows into a system (endothermic) + q system is

increasing in heat/energy.• Energy flows out of a system (exothermic) –q system is

decreasing in heat/energy

• 1000 J = 1 kJ

Units of Work • When the system expands, it is doing positive work on the

surroundings, therefore it is doing negative work on the system.

• When the system contracts, the surroundings have done work on the system, therefore there is positive work done on the system.

• Thus, from the point of view of the system,

w = - PΔV 1 L atm = 101.3 J

• For an ideal gas, work can occur only when its volume changes. Thus, if a gas is heated at constant volume, the pressure increases but no work occurs.

Work Example

• Calculate the work associated with the contraction of a gas from 75 L to 30. L (work is done “on the system”) at a constant external pressure of 6.0 atm in:

*Hint: system compression is positive work, and system expansion is negative work.

• L atm• J

Work Practice

• Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650. torr if all the energy of combustion is converted into work to push back the piston?

* Hint: watch units!

First Law of Thermodynamics Example

• Calculate the change in energy of the system if 38.9 J of work is done BY the system with an associated heat loss of 16.2 J.

1st Law Practice

• Calculate ΔE for each of the following. • q = -47 kJ, w = +88 kJ• q = +82 kJ, w = -47 kJ

• Write it out!! In each of the situations above describe if heat is

lost or gained by the system. Also, describe if work is being done by or on the system.

• ΔE = “-” means the system loses energy.• ΔE = “+” means the system gains energy.

Combining Heat and WorkExample

• A piston is compressed from a volume of 8.3 L to 2.8 L against a constant pressure of 1.9 atm. In the process, there is a heat gain by the system of 350 J. Calculate the change in energy of the system.

Heat and Work Practice• A piston expands against 1.00 atm of pressure, from 11.2

L to 29.1 L. This is done without any transfer of heat.

• Calculate the change in energy of the system.

• Calculate the change in energy for the above change if, in addition, the system absorbs 1,037 J of heat from the surroundings.

• A: -1.81 kJ ; -773J

Ch. 6 Section 1 Practice Problems• Beginning on page 266• # 9, 19, 21, 22, 23, 25, and 27

Enthalpy• A property of a system equal to E + PV, where E is the internal

energy of the system, P is the pressure of the system, and V is the volume of the system. At constant pressure the change in enthalpy equals the energy flow as heat.

H = E + PV or

ΔH = ΔE + PΔV

• Enthalpy is a state function. A change in enthalpy does not depend on the pathway between two states. • ΔH = qp • Change in enthalpy = Heat of reaction (used interchangeably)

Enthalpy Continued…• The change in enthalpy of a system has no easily interpreted

meaning except at constant pressure, where ΔH = heat.

• For a chemical reaction, the enthalpy change is given by the equation:

ΔH = Hproducts – Hreactants

(Think Δ ; Final – Initial)

• At constant pressure, • exothermic means ΔH is negative (heat given off by

system); • endothermic means ΔH is positive (heat absorbed by

system)

Enthalpy ExampleWhen solid sodium hydroxide pellets are added to water, the following reaction occurs:

NaOH(s) NaOH(aq)

For this reason at constant pressure, ΔH = -43kJ/mol. Answer the following questions regarding the addition of 14 g of NaOH to water:• Does the beaker get warmer or colder?• Is the reaction exo- or endothermic?• What is the enthalpy change for the dissolution?

Enthalpy PracticeThe overall reaction in a commercial heat pack can be represented as

4Fe(s) + 3O2(g) 2Fe2O3(s) ΔH= -1652kJ

• How much heat is released when 4.00 mol of iron (Fe) is reacted with excess O2?

• How much heat is released when 1.00 mol of Fe2O3 is produced?

• How much heat is released when 1.00 g iron is reacted with excess O2?

• How much heat is released when 10.0 g Fe and 2.00 g O2 are reacted?

Calorimetry• The science of measuring heat.• Based on observing (experimental) the temperature change when

a body absorbs or discharges energy as heat (q).• At constant pressure, q= ΔH• At constant volume, q = ΔE (work is zero)

• Heat is either gained or lost. The amount of heat exchange in a reaction is dependent upon:

1. The net temperature change during the reaction.2. The amount of substance. The more you have, the more heat can

be exchanged (direct relationship).3. The heat capacity (C) of a substance.

Calorimetry• Some substances can absorb more heat than others for a

given temperature change.

• Three ways of expressing heat capacity: 1. Heat capacity = J/°C

2. Specific heat capacity = heat capacity per gram of substance (J/°C g) or (J/K g)

3. Molar heat capacity = heat capacity per mole of substance (J/°C mol) or (J/K mol)

Conceptual Understanding

See Table 6.1 on page 237 in textbook.• Based on the values for specific heat capacity,

which conducts heat better, water or aluminum? Why is this important in cooking?

• Share out responses.

Calorimetry under constant pressure…• If two reactants at the same temperature are mixed and the

resulting solution gets warmer, this means the reaction taking place is exothermic. An endothermic reaction cools the solution.

• Energy released/absorbed by the reaction • = energy absorbed/released by the solution• = specific heat capacity x mass of solution x change in temperature

• = s x m x ΔT

• Heat of reaction is an extensive property.• What does this mean?

• Temperature is an intensive property.• What does this mean?

Calorimetry• Mentally keep track of the direction of the energy flow and

assign the correct sign at the end of the problem.• Exothermic

• Released• Evolved• Given off• Temperature increase

• Endothermic• Absorbed• Gained• Temperature decrease

• Enthalpies often expressed in terms of moles of reacting substances. kJ/mol or J/mol (determine moles of sample and divide by energy)

Calorimetry Example

• If 596 J of heat are added to 29.6 g of water at 22.9°C in a coffee cup calorimeter, what will be the final temperature of the water?• A: 27.7°C

Calorimetry Practice

• The specific heat of aluminum is 0.89 J/g°C. How much energy is required to raise the temperature of a 15.0 g aluminum can 18°C ?

• A:240J

Constant Volume Calorimetry• “Bomb calorimeter”

• ΔV = 0 thus, ΔE = qv (zero work)

• Energy released (gained) by the reaction • = temperature increase(decrease) x energy required to change

temperature by 1 degree C• =ΔT x heat capacity of bomb calorimeter

• Enthalpies often expressed in terms of moles of reacting substances. kJ/mol or J/mol (determine moles in sample and divide by energy)

Calorimetry Example

• The heat of combustion of glucose, C6H12O6, is 2800 kJ/mol. A sample of glucose weighing 5.00 g was burned with excess oxygen in a bomb calorimeter. The temperature of the bomb rose 2.4°C . What is the heat capacity of the calorimeter?

• A 4.40 g sample of propane (C3H8) was then burned with excess oxygen in the same bomb calorimeter. The temperature of the bomb increased 6.85°C. Calculate ΔEcombustion of propane.

• A: 32.4 kJ/°C ; -2200 kJ/mol

Calorimetry PracticeThe heat capacity of a bomb calorimeter was determined by burning 6.79 g of methane (heat of combustion = -802 kJ/mol) in the bomb. The temperature changed by 10.8°C.

• What is the heat capacity of the bomb?

• A 12.6 g sample of acetylene, C2H6, produced a temperature increase of 16.9°C in the same calorimeter. What is the heat of combustion of acetylene (kJ/mol)?

Independent Practice Continued…

• Page 267 #35, 37, 51, 52, 53, 55, 56

Hand-warmer Lab• Green Lab Book – Investigation 12 page 97

• Central Challenge

• Simulation/Questions• With lab group

• Materials

• Procedure• Works best to use 45.0mL of water to 5.0g of solid• Skip Calorimeter Constant (#1 in Data Collection) 7%

Pre-Lab Assignment• Purpose

• Procedure

• Data Tables

Lab Partners • Group 1:

• Group 2:

• Group 3:

• **Assigned Solids to Investigate**• Will share class data.

Post – Lab Assessment

• Safety!

• Complete Data Collection and Computation Questions

• Argumentation and Documentation

• Complete numbers 1, 2, 3, 4b, and 5 (omit 4a)