ch. 10: thermochemistry dr. namphol sinkaset chem 200: general chemistry i

Ch. 10: Thermochemistry

Dr. Namphol Sinkaset

Chem 200: General Chemistry I

I. Chapter OutlineI. Introduction

II. Energy

III. 1st Law of Thermodynamics

IV. Quantifying Heat and Work

V. Reactions at Constant V or P

VI. Thermochemical Equations/Hess’s Law

VII. Formation Reactions

VIII. Using Bond Energies

IX. Born-Haber Cycle

I. Chemistry and Energy

• We have said before that chemistry is all about energy – energy lost, gained, and transferred.

• Things always seek the lowest energy state which is equivalent to seeking highest stability.

• We start to quantify energy in this chapter.

II. What is Energy?

• Energy is the capacity to do work.

II. Kinetic Energy

• Kinetic energy involves motion. A moving car possesses kinetic energy.

• Thermal energy is the energy associated w/ temp of an object. A type of KE because it arises from

motions of atoms/molecules in the substance.

II. Potential Energy

• Potential energy is the energy associated with the position or composition of an object. An object raised above the floor possess PE

due to its position in a gravitational field.• Chemical energy is the energy

associated with relative positions of e-’s and nuclei in atoms and molecules.

II. Summary of Energy

II. Exchanging Energy

• Objects possess energy (sometimes both kinetic and potential).

• Objects exchange energy through one of two ways: work: force acting

through a distance heat: flow of energy

caused by temp. difference

II. Units of Energy

• The SI unit of energy is the joule (J), 1 J = 1 kg·m2/s2.

• An older unit is the calorie (cal), defined as the energy needed to raise the temp. of 1 g H2O by 1 °C. Note that 1 cal = 4.184 J

• Nutritional Calorie is actually the kcal.

II. Energy Conservation

• The law of conservation of energy states that energy cannot be created or destroyed.

• But, we know it can be transferred via work or heat.

• Energy can also change forms.

II. Energy Transformations

II. System and Surroundings

• We define points of view to observe where energy goes.

• system: part of universe being studied

• surroundings: everything else with which the system can exchange energy

II. Changes are Equal but Opposite

• If the system loses energy (-), then the surroundings must gain energy (+).

• If the system gains energy (+), then the surroundings must lose energy (-).

III. 1st Law of Thermodynamics

• You already know this law by another name: Law of Conservation of Energy.

• This can be restated as the total energy of the universe is constant, i.e. there’s never a change in Euniverse.

0 gssurroundinsystemuniverse EEE

III. The Internal Energy

• Say we have a system comprised of reactants in solution.

• Every particle moves constantly and feels different attractions/repulsions. There is KE from movement and PE from

position.• The sum of KE and PE for all particles

in a system is called the internal energy (E).

III. State Functions• Internal energy (E) is a state function.

The value of a state function depends only on the state of the system, not on how it arrived at that state.

• Elevation vs. distance to get to that elevation.

III. State Functions

• The E of a system does not depend on how it got there. Same goes for T, P, V. “Independent of path.” “It is what it is.”

• State functions usually given upper-case variables.

• Changes in state functions are easily calculated by subtracting initial value from final value (final – initial).

III. Changes in Internal Energy

• When a rxn occurs, internal energy of products is usually different than internal energy of reactants.

• The change can be easily calculated by taking the difference between the two.

reactantsproducts

initialfinal

EEE

EEE

III. Energy Diagram

• Internal energy changes resulting from a reaction can be displayed in an energy diagram.

• For C(s) + O2(g) CO2(g), we have:

III. Where Does the Energy Go?

• We see that as C(s) and O2(g) transform into CO2(g), energy is lost.

• If reaction is the system, then energy must flow to the surroundings.

III. How Does the Energy Go?

• The total change in internal energy is equal to the energy transferred via heat and/or work.

• The sign convention for q and w is very important!!

III. Sign Conventions

III. Sample Problem

• Identify each energy exchange as heat or work and determine whether the sign of heat or work (relative to the system) is positive or negative.a) An ice cube (system) melts and cools the

surrounding beverage.

b) A metal cylinder (system) is rolled up a ramp.

c) Steam (system) condenses on skin, causing a burn.

III. Sample Problem

• A cylinder and piston assembly (defined as the system) is warmed by an external flame. The contents of the cylinder expand, doing work on the surroundings. If the system absorbs 559 J of heat and does 488 J of work during the expansion, what is value of ΔE?

IV. Measuring Energy Changes

• To find ΔE, we need to measure or calculate both q and w.

• Heat is related to temperature change.• The 2 types of w that are most relevant

are electrical work and pressure-volume work.

IV. Temperature vs. Heat

• Temperature is the measure of thermal energy within a sample of matter.

• Heat is the transfer of thermal energy.• Thermal energy always flows from hot

to cold until thermal equilibrium is reached.

IV. Measuring Heat

• Intuitively, we know temp. is related to heat, and different substances require different amounts of heat to change their temp. q = C × ΔT

• heat capacity (C): the amount of heat needed to change the temp of a substance by 1 °C What’s wrong with this?

IV. Specific Heat Capacity

• specific heat capacity (Cs): the amount of heat required to change the temp. of 1 g of a substance by 1 °C

• Can also have molar heat capacity.

IV. Sample Problem

• Find the heat transferred (in kJ) when 5.50 L of ethylene glycol in a car radiator cools from 37.0 °C to 25.0 °C if the density of ethylene glycol is 1.11 g/mL, and its specific heat capacity is 2.42 J/g·K.

IV. Sample Problem

• A 55.0-g aluminum block (Cs= 0.903 J/g·°C) initially at 27.5 °C absorbs 725 J of heat. What is the final temperature of the aluminum?

IV. Thermal Energy Transfer

• As we know, heat flow from hotter substance to colder substance until both substances have same temperature.

• Assuming no heat loss, then the heat lost by one must equal the heat gained by the other.

IV. Common Thermal Transfer

• Many problems can be written based on this set up.

• Remember the negative sign!

• Expand each side into specific q expressions.

IV. Sample Problem

• A 32.5-g cube of aluminum (Cs= 0.903 J/g·°C) initially at 45.8 °C is submerged into 105.3 g of water (Cs= 4.18 J/g·°C) at 15.4 °C. What is the final temperature of both substances at thermal equilibrium, assuming no heat loss?

IV. Measuring Work

VPw

VPw

hPAw

PADwA

FP

FDw

sense... make to

since

IV. Sample Problem

• When fuel is burned in a cylinder equipped with a piston, the volume expands from 0.255 L to 1.45 L against a constant pressure of 1.02 atm. In addition, 875 J is emitted as heat. What is ΔE for the burning of the fuel? Note that 1 L·atm = 101.3 J.

V. Measuring ΔE

• With the two equations for q and w, we can measure ΔE for reactions.

• Simplifications can be made if specific conditions are chosen.

V. Reactions at Constant V

V

V

qE

PqE

VPqE

wqE

)0(

V. Calorimetry

• Ignite the reaction.• Measure ΔT.• Obtain qcal.

qcal = Ccal × ΔT

• Relate back to qrxn. qcal = -qrxn

• Relate back to ΔErxn

qrxn = ΔErxn

V. Sample Problem

• When 1.550 g of liquid hexane (C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. Find ΔErxn for the reaction in kJ/mole hexane. Note that the heat capacity of the bomb calorimeter is 5.73 kJ/°C.

V. Reactions at Constant P

• At constant P, we introduce a variable called enthalpy (H) that doesn’t necessitate that we calculate or measure q and w separately.

• Enthalpy is defined as H = E + PV.• Typically we are interested in ΔH…

VPEH

PVEH

V. More Useful ΔH Equation

• If only P-V work can be done under constant P…

P

ppP

pP

qH

wwqH

VPwqH

VPEH

V. More Useful ΔH Equation

pqH

• We see that enthalpy change is the heat gained/lost under constant pressure conditions.

• ΔH measures just heat; ΔE measures all energy.• Much easier to measure ΔH than ΔE, and

typically they are close in value because few reactions do P-V work.

V. Constant P Calorimetry

• The important relationships to remember are:

qsystem= -qsurroundings

qrxn= -qsoln

qrxn = ΔHrxn

V. Heats of Reaction

• H is a state function, so ΔH is simply Hfinal – Hinitial.

• The enthalpy of a reaction is known as the heat of reaction (ΔHrxn).

• Reactions can be either exothermic (-ΔHrxn) or endothermic (+ΔHrxn).

V. Sample Problem

• 50.0 mL of 1.00 M NaOH is added to 25.0 mL of 1.25 M H2SO4 in a coffee cup calorimeter. If the initial temp. is 25.00 °C, and the final temp. is 33.83 °C, what is ΔHrxn in kJ? Use d = 1.00 g/mL for all solutions and Cs = 4.184 J/g·°C for the combined solution.

VI. Thermochemical Equations

• A thermochemical equation is a balanced equation that includes the heat of reaction.

• The value of ΔHrxn is stoichiometrically related to the reactants and products.

2H2O(l) 2H2(g) + O2(g) ΔHrxn = 572 kJ

VI. Sample Problem

• How many grams of carbon dioxide are formed when the following reaction is used to produce 1.5 × 103 kJ of heat?

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(g) ΔHrxn = -5316 kJ

VI. Rules for Manipulating Thermochemical Equations

1) If equation is multiplied by a constant factor, then ΔHrxn is multiplied by the same factor.

2) If equation is reversed, then ΔHrxn changes sign.

3) If equation can be written as a sum of a series of steps, then ΔHrxn for the overall is the sum of the ΔHrxn’s of each step.

VI. Hess’s Law

• The last rule is Hess’s law.

• Hess’s Law: the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps

VI. Sample Problem

• Calculate ΔHrxn for the reaction 2NO2(g) + ½ O2(g) N2O5(g) given the information below.

N2O5(g) 2NO(g) + 3/2 O2(g) ΔH = 223.7 kJNO(g) + 1/2 O2(g) NO2(g) ΔH = -57.1 kJ

VI. Sample Problem

• Calculate ΔHrxn for the reaction Ca(s) + ½ O2(g) + CO2(g) CaCO3(s) using the information below.

Ca(s) + ½ O2(g) CaO(s) ΔH = -635.1 kJCaCO3(s) CaO(s) + CO2(g) ΔH = 178.3 kJ

VII. Standard Heats of Reaction

• Thermodynamic variables vary a little with conditions, so we need to define a set of standard conditions.

• Standard conditions are defined at 1 atm and 25 °C.

• Products and reactants at these conditions are said to be in their standard states. Note that solutions must be 1 M.

• The “not” symbol is used to indicate standard conditions, e.g. ΔH°rxn.

VII. Formation Reactions

• A formation reaction is one that creates exactly 1 mole of a substance from its elements.

• Of course, standard heats of formation occur at 1 atm and 25 °C.

Na(s) + ½ Cl2(g) NaCl(s) ΔH°f = -411.1 kJ/mole

VII. Using ΔH°f’s

• Standard heats of formation are very powerful; they allow calculation of ΔH°rxn for anything using a Hess’s Law type of calculation.

ΔH°rxn = (sum ΔH°f products) – (sum ΔH°f reactants)

VII. Visual Example

VII. Standard Heats of Formation

VII. Sample Problem

• Calculate ΔH°rxn for C2H2(g) + 5/2 O2(g) 2CO2(g) + H2O(g) given that the standard heats of formation for acetylene, carbon dioxide, and steam are 227.0 kJ/mole, -393.5 kJ/mole, and -241.8 kJ/mole, respectively.

VIII. BE’s and Chemical Change

• In a reaction involving covalent compounds, bonds are broken and new bonds are formed. Relative strengths of bonds determine

whether rxn is exo or endo.• We can calculate heats of reaction

based on the energy difference between bonds made and bonds broken.

VIII. Bond Energies

• Breaking bonds requires energy… A-B(g) A(g) + B(g) ΔH˚break = BEA-B

• Forming bonds releases energy... A(g) + B(g) A-B(g) ΔH˚form = -BEA-B

• Different bonds have different levels of attraction and thus different BE’s.

VIII. Estimating Rxn Enthalpies

VIII. Calculating w/ BE’s

• A reaction can be considered a two-step process:

1) Heat absorbed to break bonds

2) Heat released to make bonds

• We can sum these to get the heat of reaction.

VIII. Bond Energy Table

VIII. Sample Problem

• Calculate ΔH˚rxn for the reaction shown below. Use the indicated bond enthalpies.

IX. Ionic Bond Formation

• Recall that we envision the creation of ionic bond as an e- transfer from the metal to the nonmetal.

• However, if we look at the energies involved in these two steps, we see something puzzling.

IX. Ion Formation for NaCl

• We break up the e- transfer process into two steps and add up the energies.

Na(g) Na+(g) + e- IE1 = 496 kJ/mole

Cl(g) + e- Cl-(g) EA = -349 kJ/mole

Na(g) + Cl(g) Na+(g) + Cl-(g) IE1 + EA = 147 kJ/mole

IX. Other Energies

• For most ionics, this is typical; ion formation is an endothermic process.

• So why do ionics form at all?• There must be a huge exothermic

process to offset these endothermic processes.

• The strong +/- attractions that are formed are the source of this exo step.

IX. Lattice Energy• lattice energy: the energy associated with

forming a crystalline lattice of alternating cations and anions from gaseous ions.

IX. Born-Haber Cycle

• Although lattice energy is a critical component of ionic bond formation, it cannot be measured directly.

• We use Hess’s Law in the Born-Haber Cycle.

IX. Born-Haber Cycle for NaCl

IX. Calculating Lattice Energy

• Since enthalpy is independent of path, the following equation applies.

ΔH˚f = ΔH˚atom + ΔH˚BE + ΔH˚IE + ΔH˚EA + ΔH˚lattice

IX. Sample Problem

• Calculate the lattice energy for CaCl2 given the following enthalpy values: heat of sublimation for Ca = 179.3 kJ/mole, Cl2 bond energy = 243 kJ/mole, 1st ionization energy Ca = 589.8 kJ/mole, 2nd ionization energy Ca = 1145.4 kJ/mole, 3rd ionization energy Ca = 4912.4 kJ/mole, electron affinity Cl = -349 kJ/mole, heat of formation CaCl2 = -795.8 kJ/mole.