AP Chemistry 2014-2015

CH 6 THERMOCHEMISTRY

Energy is the ability to do work or produce heat. The sum of all of the potential and kinetic energy in a system is known as the internal energy of the system.

Potential energy is the energy of an object by virtue of its position (which can mean more than one thing). In chemistry, potential energy is usually the energy

stored in bonds. When bonded atoms are separated, the PE increases because energy must be added to overcome the coulombic attraction between each nucleus and the shared electrons. When atoms bond, this coulombic attraction results in energy being released and a subsequently lower PE.

6.1 THE NATURE OF ENERGY

Kinetic energy is energy of motion (translational, rotational, and vibrational motion of particles in our case), proportional to Kelvin temperature. KE depends on the mass and velocity of an object.

NEED TO KNOW FOR AP EXAM

Rotational motion (motion due to the rotation of bonds) is typically studied using microwave spectroscopy (polar molecules) and Raman scattering (nonpolar molecules).

Vibrational motion (motion due to bond vibration, see bottom figure on right) is typically studied using infrared spectroscopy.

The law of conservation of energy states that any change in the energy of a system must be balanced by the transfer of energy either into or out of the system—this is the First Law of Thermodynamics—the energy of the universe is constant.

Heat (q): two systems with different temperatures that are in thermal contact will exchange thermal energy (which we call heat). This transfer is a process and a measurement of energy.+q heat is absorbed -q heat is released

Temperature (T) is proportional to the average KE of molecules in whatever you’re looking at. It is not a measure of energy.

HEAT AND TEMPERATURE

THERMODYNAMIC EQUILIBRIUM

(not in notes)

Heat is transferred from warmer objects to cooler ones until thermodynamic equilibrium is reached—that is, until both are at the same temperature.

At thermodynamic equilibrium, there is zero net transfer of heat between the two objects.

WHAT ABOUT HEAT CAPACITY

If the masses of these objects are equal, what can we say about their heat capacities?

The ending temperature will NOT always be the average of the two starting temperatures.

ANOTHER EXPLANATION. THANKS NASA.

Enthalpy (H) is the flow of energy at constant pressure when two systems are in contact. Change in enthalpy is written ΔH. Types

Enthalpy of reaction ΔHrxn

Enthalpy of combustion ΔHcomb

Enthalpy of formation ΔH f

Enthalpy of fusion ΔH fus

Enthalpy of vaporization ΔHvap

Endothermic: +ΔH, net absorption of energy by the system, energy is a “reactant”

Exothermic: -ΔH, net loss of energy by the system, energy is a “product”

ENTHALPY

In thermochemistry, the system is the area of the universe you are focusing on, and the surroundings are everything outside of it.

Entropy (S) is a measure of the dispersal of matter and energy (entropy = disorder)Increased disorder/dispersal: +ΔS (natural tendency of the universe, we’ll see more on this)

Increased disorder/dispersal: -ΔS

ENTROPY

Entropy increases when…Solids are converted to liquids or gasesLiquids are converted to gasesThe number of moles is greater on the products side than the reactants side (add the coefficients)

Entropy decreases when the opposite happens. For example: 2 H2(g) + O2(g) → 2 H2O(l)

Entropy decreases in this equation because a) the number of moles decreases and b) gases are being “converted” to liquid form in a roundabout way.

ENTROPY IN EQUATIONS (NOT IN THE NOTES)

Gibbs Free Energy (G) is the criteria for determining thermodynamic favorability and calculating the theoretical amount of energy to do work. ΔG = ΔH – TΔSSpontaneous reaction/process: - ΔG

Example: a ball rolling down a hill, ice melting on a hot surface

Nonspontaneous reaction/process: +ΔG Example: a ball rolling up a hill, water freezing

into ice on a hot surface

GIBBS FREE ENERGY

Work is a force acting over a distance (w = -PΔV = -pressure multiplied by change in volume) in physics where gases are involved; expressed in J or kJ Positive work = decrease in volume = work done on system

= energy added to system Negative work = increase in volume = work done by system

= energy lost by system 1 L atm = 101.3 J = 0.101325 kJ

STP 0°C = 273 K 1 atm = 760 torr = etc….see Ch 5

Standard (lab) conditions 25°C = 298K1 atm etc. 1.0 M for solutions

Add a ° to the symbol—for example ΔH f° is enthalpy of formation at 298 K and 1 atm

New convention is that ΔH° is expressed in kJ/mol rxn, where “molrxn” is “moles of reaction”

Energy is often defined as the ability to do workΔE = q + w = heat + work

Calculate the ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows into the system and where 1.4 kJ of work is done on the system. (answer = 17.0 kJ)

EXERCISE 6.1 INTERNAL ENERGY

ΔE = q + wEndothermic, so q = +15.6 kJWork is done on the system, w = + 1.4 kJ

ΔE = 15.6 kJ + 1.4 kJ = 17.0 kJ

Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. (answer = -270 Latm = -27 kJ)

EXERCISE 6.2 PV WORK

w = -PΔV

P = 15 atm ΔV = 64 L - 46 L = 18 L

w = - 15 atm * 18 L = -270 Latm (work done BY system)

-270 Latm * (101.3 J/1 Latm) = -27360 J = -27 kJ

A balloon is being inflated to its full volume by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate the ΔE for this process. (answer = 8.0 x 107 J)

EXERCISE 6.3 INTERNAL ENERGY, HEAT, AND WORK

ΔE = q + wAddition of heat, so q = +1.3 x 108 JWork is done by the systemw = -PΔV = - 1.0 atm * (4.50 x 106 – 4.00 x 106) L = - 5.00 x 105 Latm = - 5.07 x 107 J

ΔE = 1.3 x 108 J - 5.07 x 107 J = 8.0 x 107 J

We can only measure the change in enthalpy, ΔH, the difference between the potential energies of the products and the reactants

ΔH is a state function (does not depend on path, only on beginning and end)

ΔH = q at constant pressure (ex. atmospheric pressure)

Enthalpy can be calculated from stoichiometry, calorimetry, tables of standard values, Hess’s Law, and bond energy

6.2 ENTHALPY AND CALORIMETRY

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

KOH(s) KOH(aq) + 43 kJ/molSuppose 14.0 g of KOH are added to water.

Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the

14.0 g of KOH? (answer = -10.7 kJ/molrxn, what is the sign?)

EXERCISE 6.4 ENTHALPY

14.0 g KOH 1 mol KOH -43 kJ ____ = -10.7 kJ/molrxn

56.11 g 1 mole KOH

Note: for problems like this, the correct units are kJ/molrxn

Calorimetry is the process of measuring heat based on observations of temperature change when a body absorbs or discharges heat. All of the bullet points below discuss constant-pressure calorimetry. In coffee-cup calorimetry, we use a, well, disposable coffee

cup and reactants that begin at the same temperature, then we look for a change in temperature. This is constant-pressure calorimetry (since the container is open), so q = ΔH. After all the data is collected (mass or volume, initial and final temperatures) we can calculate the energy released or absorbed. In order to do this we need to know things like the specific heat capacities of each substance (unless one is an unknown, of course).

Heat capacity is the energy required to raise the temperature of something by 1°C.

CALORIMETRY

Specific heat capacity (Cp) is the energy required to raise one gram of a substance by 1°C, if the process/experiment is carried out at constant pressure. Units = J/gK = J/g°C

Molar heat capacity is the energy required to raise one mole of a substance by 1°C, if the process/experiment is carried out at constant pressure. Units = J/moleK = J/mole°C

HEAT CAPACITY

q released or gained at constant pressure = mCpΔTThe specific heat of water (liquid state) = 4.184 J/g°C = 1.00 cal/g°C; we need this value because the heat lost by the substance is equal to the heat gained by the water (or vice-versa)

THE M-CAT EQUATION

calorie = amount of heat needed to raise the temperature of 1 of water by 1 1°C

Kilocalorie = 1000 cal = 1 Calorie (for calories, capital/lower-case c mean different things)

Joule = SI unit of energy, 1 cal = 4.184 J

UNITS OF ENERGY

In a coff ee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions are originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter. (answer = -5.6 kJ/mol rxn)

EXERCISE 6.5 COFFEE-CUP CALORIMETRY

The energy gained by the water is equal to the energy lost by the reaction, but opposite in sign. SoΔHrxn = - mwater * Cwater * ΔTwater

= - (200.0 g)(4.184 J/g°C)(31.3-24.6°C) = -5600 J/molrxn = -5.6 kJ/molrxn

When one mole of methane is burned at constant pressure, 890 kJ/mol of energy are released as heat. Calculate ΔH for a process in which a 5.8 g sample of methane is burned at constant pressure. (answer = heat flow = -320 kJ/mole)

EXERCISE 6.6 ENTHALPY

5.8 g CH4 1 mol CH4 -890 kJ ____ = -320 kJ/molrxn

16.04 g 1 mole CH4

Note: for problems like this, the correct units are kJ/molrxn

When 1.00 L of 1.00 M barium nitrate solution at 25.0°C is mixed with 1.00 L of 1.00 M sodium sulfate solution at 25.0°C in a calorimeter, the white solid BaSO 4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter only absorbs a negligible quantity of heat, that the specifi c heat capacity of the solution is 4.184 J/g°C, and that the density of the fi nal solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed. (answer = -26 kJ/mol rxn)

EXERCISE 6.7 CONSTANT-PRESSURE CALORIMETRY

The energy gained by the water is equal to the energy lost by the reaction, but opposite in sign. SoΔHrxn = - mwater * Cwater * ΔTwater

= - (2000.0 g)(4.184 J/g°C)(28.1-25.0°C) = -26000 J/molrxn = -26 kJ/molrxn

Take moles into accountA consequence of Hess’s Law is the

summation of individual reactions to yield a total reaction. The basic idea and an example are shown below.

6.3 HESS’S LAW

First decide how the given equations need to be arranged to “add up” and “cancel out”. Make sure your reactants and products end up on the correct sides.

If an equation has to be reversed, also reverse the sign of its ΔH° (more correctly, its ΔHrxn°)

If an equation has to be multiplied by a factor to obtain the correct coeffi cients, also multiply the ΔHrxn° by this factor (it’s per mole)—same goes for division

EQUATIONS, ENTHALPY OF REACTION, AND HESS’S LAW

Calculate the ΔH for the overall reaction 2 H 3BO3(aq) B2O3(s) + 3 H2O(l) given the following equations. (answer = 14.4 kJ/mol rxn)

EXERCISE 6.8 HESS’S LAW

The first equation needs to be doubled—look at the boric acid, H3BO3.The third equation needs to be halved—look at the diboron trioxide, B2O3.Both the first and third equation “run” the same way as the goal equation—boric acid is a reactant in both, diboron trioxide is a product in both.

The second equation needs to be reversed and halved—reversed so that the metaboric acid (HBO2) from the first equation is used up and so that the tetraboric acid (H2B4O7) needed for the third equation is created.

Here’s how that looks:

Note that the changes made to each equation are reflected in their changes in enthalpy. Equation 1 was doubled, so its enthalpy change was doubled. Equation 2 was reversed and halved, so its enthalpy because positive, and was cut in half. Equation 3 was halved, so its enthalpy was halved.

If we add the three new equations together and cancel terms that appear on both sides, we get

And if we add the enthalpy changes for each reaction, we get

ΔH f° = standard enthalpy of formation, which is the enthalpy change associated with the production of one mole of a compound from its elements in their standard states. Zero for elements in their standard states. Remember…standard states = 25°C, 1 atm, 1 M

We can use standard enthalpies of formation and Hess’s Law to calculate the enthalpy change of a reaction.

6.4 STANDARD ENTHALPIES OF FORMATION

Given the information below, calculate the ΔH rxn° for the following chemical reaction.3 Al(s) + 3NH4ClO4(s)

Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

EXERCISE 6.9 ΔHF° AND ΔHRXN° I

ΣΔHf°products= (1 mol)(-1676 kJ/mol)+(1 mol)(-704 kJ/mol) + (3 mol)(90.0 kJ/mol) + (6 mol)(-242 kJ/mol) = -3562 kJ

ΣΔHf°reactants = (3 mol)(0 kJ/mol) + (3 mol)(-295 kJ/mol) = -885 kJ

ΔHrxn= -3562 kJ– (-885 kJ) = -2677 kJ/mol

Occasionally, a value cannot be found in a 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 of the substance must be calculated by working backwards from its heat of combustion. Calculate the ΔH f° of solid glucose given the following combustion reaction and data table. (answer = -1276 kJ/mole)

EXERCISE 6.10 ΔHF° AND ΔHRXN° II

ΣΔHf°products= (1 mol)(-1676 kJ/mol)+(1 mol)(-704 kJ/mol) + (3 mol)(90.0 kJ/mol) + (6 mol)(-242 kJ/mol) = -3562 kJ

ΣΔHf°reactants = (3 mol)(0 kJ/mol) + (3 mol)(-295 kJ/mol) = -885 kJ

ΔHrxn= (6ΔHf°CO2 + 6ΔHf°H2O) - (ΔHf°C6H12O6 + 6ΔHf°O2)

-2800 = ((6*-393.5) + (6*-285.8)) – (ΔHf°C6H12O6 + (6*0))

ΔHf°C6H12O6 = +2800 + ((6*-393.5) + (6*-285.8)) = -1275.8 -1276 kJ/mol

Bond energies Energy must be added/absorbed to break bonds in order to

overcome the coulombic attractions between each nuclei and shared electrons. Energy is released when bonds are formed because the resulting coulombic attractions between the newly bonded atoms lower the potential energy, causing a release. TL, DR: it takes energy to break bonds, and energy is released when bonds form.

ΔH = sum of the energies required to break old bonds plus the sum of the energies released in the formation of new bonds (negative, since energy is lost from the system).

Notice that the first term is related to the reactants, and the second term is related to the products.

Calculate the change in energy that accompanies the following reaction given the data below. (answer = -544 kJ/mol)

EXERCISE 6.11 BOND ENERGIES

Broken = reactants, formed = products

ΔH = ((1 mol*432 kJ/mol) + (1 mol * 154 kJ/mol)) – (2 mol* 565 kJ/mol) = - 544 kJ/mol

Genuinely matters for AP Exam as well as realityFusion (melting), vaporization, and sublimation require an input of energy to overcome attractive forces between their particles (IMFs)—these processes do not break bonds

Freezing, condensation, and deposition release energy as IMFs form since the particles achieve a lower state mainly due to a decrease in temperature—these processed do not form bonds

MORE THINGS ABOUT ENERGY CHANGES