Thermochemistry & Thermodynamics

AP Chemistry

Brief Overview Chemical reactions involve changes

in energy Thermochemistry

Study of relationships of chemical reactions and energy changes

ThermodynamicsStudy of energy and its transformations

Energy Energy is the ability to do work or

produce heat and is the sum of all potential and kinetic energy in a system

Potential energy is energy possessed due to relative position to other objectsCommonly in chemistry this is energy

stored in bondsKinetic energy is the energy of motion,

usually of particles

KEPEETotal

Work & Heat Heat (q) transfer of energy from object

with higher KE to object with lower KEHence heat flows from hot to coldHeat transfers because of temperature (KE)

difference, but temperature is not a measure of heat directly

Work (w) is a force acting over a distanceUsually used when discussing gasesWhen related to gases, work is a function of

pressure and volume:

Energy, Work, and Heat

Energy is a state function while heat and work are not

State function is a property that is independent of past or future behavior, only on current conditions (temperature, pressure, etc)Example: Many roads from your home to school,

but you have the same starting point and ending point

)()( workwheatqE

Signs of Work & Heat Signs of q

+q means heat is absorbed (endothermic)

-q means heat is released (exothermic)

Signs of w+w if work done on the system (i.e.

compression)-w if work done by the system (i.e.

expansion)

First Law of Thermodynamics Energy cannot be created or

destroyed, only conservedLaw of Conservation of EnergyEnergy of universe is constant

Energy may only transfers between system (the experiment) and surroundings (universe)

Units of Energy SI unit of energy is Joule

1 Joule = 1 kg*m2/s2

Commonly reported in kilojoules Non-SI unit of energy is calories

1 cal = 4.184 JNutrional Calorie (Cal) = 1 kcal =

1000 calories

Enthalpy (∆H) Describes heat gained or lost at constant

pressure

H = qpFollowing the derivation, E = H + P∆V or H = E - P∆V

Enthalpy is a state function Can be calculated from multiple sources:

CalorimetryStoichiometryHeats of formation (tables of standard values)Hess’ LawBond Energies

Enthalpy of Reaction Enthalpy of a reaction can be

utilized stoichiometrically based on coefficients

For endothermic processes ∆H is positive and should be included as a reactant in the rxn

For exothermic processes ∆H is negative and should be included as a product in the rxn

Enthalpy of Reaction Example

Upon adding solid potassium hydroxide pellets to water the following reaction takes place:

Answer the following questions regarding the addition of 14.0 g of KOH to water:

1. Does the beaker get warmer or colder?2. Is the reaction endothermic or exothermic?3. What is the enthalpy change for the dissolution

of the 14.0 g of KOH?

kJKOHKOH aqs 43)()(

Thermochemical Equations Guidelines

Enthalpy is an extensive propertyEnthalpy of the reverse reaction is

equal in magnitude but opposite in signEnthalpy change depends on physical

states of reactants and productsIf the thermochemical equation is

multiplied by a factor of n, then ∆H must also change by the same factor

Calorimetry Process of measuring heat based on

observing the temperature change when a body absorbs or discharges energy as heatAssumption is that no heat is lost to

surroundings (aka closed system) 2 types of Calorimetry:

Constant pressure(coffee-cup setup)Constant volume (bomb calorimeter)

Used in industry for determining food calories

Calorimetry Commonly used in lab to determine

specific heats of metals Can also measure the heat

released/absorbed of many types of reactions:NeutralizationsIonization (Dissolution/Dissolving)Reaction (ppt, combustion, etc.)

Terms to know (and love!) Heat capacity– energy required to raise the

temperature of an object by 1 degree C (J/°C)

Specific heat capacity– same as above but specific to 1 gram of the substance (J/ g °C)

Molar heat capacity – Same as specifc heat capacity, but specific to one mole of a substance (J/mol K or J/mol °C)

Relationships of Heat Heat capacities are extensive properties, while

specific heat is an intensive property Heat capacity can be determined by

multiplying a substances mass by its specific heat:

J/°C = J/g°C x g Specific heat of water is 4.184 J/g°C and is the

same for all dilute aqueous solutions Heat of substance + heat of solution = 0

Therefore, qsubstance = -qsoln

Calculating Specific Heat

q=mCp∆T q= heat transferred (J)

Recall: at constant pressure q=∆H m = mass (g) Cp = specific heat of material at constant

pressure ∆T = Tf – Ti (final – initial)

Calculating Specific Heat Example

How much heat is needed to raise 10.0 grams of aluminum from 22.0 °C to 42.0 °C? (Specific heat of aluminum is 0.90 J/ g K)

Also, what is the molar heat capacity of aluminum?

Constant Pressure Calorimetry Example (Specific Heat of Metal)

A lead (Pb) pellet having a mass of 26.47 g at 89.98C was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50 °C to 23.17 °C. What is the specific heat of the lead pellet?

0.158 J/g˚C

Constant Pressure Calorimetry Example (Molar Heat of Neutralization)

A quantity of 1.00 x 102 mL of 0.500 M HCl was mixed with 1.00 x 102 mL of 0.500 M NaOH in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solutions was the same, 22.50 °C, and the final temperature of the mixed solution was 25.86 °C. Calculate the heat change for the neutralization reaction on a molar basis (that is, the molar heat of neutralizaton)

-56.2 kJ/mol

Constant Volume Calorimetry In order to perform calculations using a constant

volume calorimeter, the bomb calorimeter must be calibrated so that the heat capacity of the calorimeter is known

qcal = Ccal∆T(where Ccal is the heat capacity of the calorimeter)

The Ccal is determined by burning a substance with an accurately known heat of combustion This is a constant that is generally given!

Finally, qcal= -qrxn

Constant Volume Calorimetry Example (Molar Heat of Combustion)

A quantity of 1.435 g of naphthalene (C10H8) was burned in a bomb calorimeter. The temperature of the water rose from 20.28 °C to 25.95 °C. If the heat capacity of the bomb plus water is 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis.

-5.151x103 kJ/mol

Constant Volume Calorimetry Example (Heat Capacity of Bomb Calorimeter)

Camphor (C10H17O) has a heat of combustion of 5903 kJ/mol. When a sample of camphor with mass of 0.1204 g is burned in a bomb calorimeter, the temperature increases by 2.28 °C. Calculate the heat capacity of the calorimeter

Standard Enthalpy of Formation & Reaction Heat (or enthalpy) of formation, ∆Hf, is the heat

required to form the elements Standard enthalpies of formation (∆H°f) can be

used as reference points for determining the standard enthalpy of a reaction (∆H°rxn) Standard state conditions are 1 atm and 25 °C Standard enthalpies of formation of any element in its

most stable form is zero The greater the heat of formation of a molecule,

the less stable the molecule Higher ∆H°f implies more energy to form the bonds

Standard Enthalpy of Formation & Reaction Change of enthalpy that occurs in a chemical

reaction can be given by:

Where n and m represent coefficients Two methods exist for determining the ∆H°rxn for a

reaction Direct Method (using ∆H°f) Indirect Method (using Hess’ Law)

)()( rctsHmprodsHnH ffrxn

Standard Enthalpy of Formation & Reaction When to use Direct Method?

If all of the standard heats of formation are known for each participant in a chemical equation, then plug them into the equation on the last slide

When to use Indirect Method? The heat of a reaction is independent of the steps it

takes to get there (enthalpy is a state function) If only heats of reaction for a series of reactions is

known, then these can be manipulated to determine the heat of the reaction

This is known as Hess’s Law

Direct Method Example (Enthalpy of Reaction)

Calculate the ∆H°rxn for the following:

Given the following values:Substance ∆H°f (kJ/mol)

NH4ClO4 (s) -295

Al2O3 (s) -1676

AlCl3 (s) -704

NO (g) 90.0

H2O (g) -242

Answer: -2680 kJ/mol (exo)

3 Al (s) + 3 NH4ClO4 (s) → Al2O3 (s) + AlCl3 (s) + 3 NO (g) + 6 H2O (g)

Direct Method Example (Enthalpy of Formation)

Sometimes all values are not found in the table of thermodynamic data. For most substances it is impossible to go into a lab and directly synthesize a compound from its free elements. The heat of formation for the substance must be found by working backwards from its heat of combustion. Find the ∆Hf of C6H12O6 (s) from the following information:

C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) + 2800 kJSubstance ∆H°f (kJ/mol)

CO2 (g) -393.5

H2O (l) -285.8

Answer: -1276 kJ.mol for glucose

Hess’s Law (of heat summation) Guidelines

First decide how to rearrange equations so reactants and products are on appropriate sides of the arrows

Manipulate the equations using the thermochemical equation guidelines (mentioned previously)

Check to ensure that everything cancels out to give you the exact equation you want

Note: It is often helpful to begin your work backwards from the answer that you want!

Hess’s Law Given the following equations:H3BO3 (aq) → HBO2 (aq) + H2O (l)

∆Hrxn = -0.02 kJ/mol

H2B4O7 (aq) + H2O (l) → 4 HBO2 (aq)

∆Hrxn = -11.3 kJ/mol

H2B4O7 (aq) → 2 B2O3 (s) + H2O (l)

∆Hrxn = 17.5 kJ/mol

Find the ∆H for this overall reaction:

2 H3BO3 (aq) → B2O3 (s) + 3 H2O (l)

Answer: 14.4 kJ/mol (endothermic)

Thermodynamics! Study of energy changes in chemistry Involves three major players: ∆H, ∆S, ∆G One of the major objectives of thermodynamics is

to predict whether or not a reaction will occur when reactants are brought together (spontaneous vs. nonspontaneous)

First Law of Thermodynamics (stated previously) is that energy cannot be created or destroyed, only conserved

Thermodynamics! Enthalpy can be used to predict spontaneity of a

reaction

In general, lower energy states are preferred, meaning exothermic reactions are favorable (-∆H)

Entropy (∆S) Enthalpy (∆H) is not the only predictor of

spontaneity Entropy (∆S) can be used, and is the measurement

of disorder in a system Second Law of Thermodynamics states that the

entropy (or disorder) of the universe is constantly increasing Examples: Ice cube melting, your room at the end of a

week. Nature tends towards chaos!

universegssurroundinsystem SSS

Entropy (∆S) + ∆S = more disorder in the system

Increase in disorder is favorable, indicates higher chance to be spont.

- ∆S = less disorder in the systemDecrease in disorder is not favorable, indicates lower

chance to be spont. Thermodynamics can predict spontaneity, but

does NOT predict the rate (speed) of the reaction!Talking state functions: we do not look at pathways,

only beginning and end states!

Twice is extra nice! Entropy, like enthalpy, is a state function Entropy can be calculated for reactions exactly like

enthalpy:

Units are J/K*mol (where ∆H is kJ/mol)

)()( rctsSmprodsSnS rxn

Changes in Entropy Entropy increases as one goes from solid to a

liquid

Ssolid Sliquid Sgas

Changes in Entropy Entropy increases as a substance divides into

partsDissolving solids into liquids (exception: carbonates!)

Changes In Entropy Entropy will increase in reactions in

which number of product molecules are greater

Entropy increases with temperature

Third Law of Thermodynamics States that the entropy of a perfect

crystal at 0 K is zeroThis implies no movement, and no

randomness existsNot many perfect crystals out there, so

entropy values rarely ever zero For determination of standard values of

entropy (as in ∆S°), this allows an absolute standard to exist which all other values can be based on

Free Energy To ultimately decide the spontaneity of a reaction,

we use Gibb’s free energy (G) For a constant temperature process, change in

free energy can be given by:

The most important thermodynamic equation, and one of the most beneficial equations in chemistry!

STHG

Free Energy & Spontaneity Free energy is the energy available to do

work - ∆G means forward reaction is

spontaneous + ∆G means forward reaction is non-

spontaneousReverse reaction would be spontaneous!

∆G = 0 means reaction is at equilibrium (more on this later)

Third Time’s the Charm! Standard free enthalpies of reaction can be

calculated using the following equation:

∆G has units of energy (J or kJ) ∆G is a state function (like ∆H and ∆S)

)()( rctsGmprodsGnG ffrxn

Predicting Reaction Spontaneity Both enthalpy and entropy must be known to

predict spontaneous reactionsS T H G ∆H ∆S Result

Negative (-) Positive (+) Always spontaneous

Positive (+) Positive (+) Spontaneous at high temperatures only

Negative (-) Negative (-) Spontaneous at low temperatures

Positive (+) Positive (+) Never spontaneous