Chapter 5Thermochemistry

(The study of the relationships between chemical reactions and energy changes

involving heat)&

Chapter 8 Section 8Strength of Covalent Bonds

Energy

• The ability to do work or transfer heat.

Work

• Energy used to move an object over some distance.

• w = F d,

where w is work, F is the force, and d is the distance over which the force is exerted.

Heat

• Heat is energy transferred between substances due to a difference in temperature (from higher temp to lower temp)

Potential Energy

Energy an object possesses by virtue of its position or chemical composition.

PE = mgh (mass x gravity x height)

Kinetic Energy

Energy an object possesses by virtue of its motion.

12

KE = mv2 (½ x mass x velocity2) (true except at 0 K)

Transferal of Energy Example:

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

c) When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

Units of Energy

• The SI unit of energy is the joule (J).

• An older, non-SI unit is still in widespread use: The calorie (cal).

1 cal = 4.184 J

1 food Calorie = 1kcal

1 J = 1 kg m2

s2

System and Surroundings

• System - the molecules we want to study (here, the hydrogen and oxygen molecules).

• Surroundings - everything else (here, the cylinder and piston).

First Law of Thermodynamicsa.k.a. Law of Conservation of

Energy

• Energy is neither created nor destroyed.

• The total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Internal Energy

The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

Usually we don’t measure E, we just measure the change in E (ΔE)

Internal Energy

By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

E = Efinal − Einitial

(We don’t actually use this equation….we just need to know the change is between

two states!)

• If E is positive (E > 0, Efinal > Einitial)

The system absorbed energy from the surroundings.Energy of the system increasesEnergy of the surroundings decreases

• If E is negative (E < 0, Efinal < Einitial )

The system released energy to the surroundings.

Energy of the system decreasesEnergy of the surroundings increases

Changes in Internal Energy

• When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).

E = q + w.

E, q, w, and Their Signs

State Functions• Depend only on the present state of the

system, not on the path by which the system arrived at that state.

• In the system below, the water could have reached room temperature from either direction.

• E depends only on Einitial and Efinal and is a state function.

State Functions

• However, q and w are not state functions.

• Whether the battery is shorted out or is discharged by running the fan, its E is the same.But q and w are different

in the two cases.

State Functions

• Generally…..

Quantities we look at the change in (E) are state functions

Quantities we don’t look at changes in (q & w) are NOT state functions.

P-V Work

When a chemical reaction occurs, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

PV 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 = −PV

Example #1

Hydrogen & Oxygen gas are in a cylinder with a movable piston. When ignited, the system loses 1150J of heat to the surroundings. The reaction also causes the piston to rise (from the expanding gases) and 480J of work is done as the piston is pushed against the atmosphere. What is the change in internal energy of the system?

Heat is transferred out of the system so q is negative (q=-1150J)

Work is done by the system, so w is negative (w = -480J)

ΔE = q + w = -1150J + -480J = -1630J

1630J of energy has been transferred from the system to the surroundings

Enthalpy at Constant Pressure

• If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.

• Enthalpy is the internal energy plus the product of pressure and volume:

H = E + PV

Enthalpy at Constant Pressure

• When the system changes at constant pressure, the change in enthalpy, H, is

H = (E + PV)• This can be written

H = E + PV

Enthalpy at Constant Pressure• Since E = q + w and w = −PV, we can

substitute these into the enthalpy expression:

H = E + PVH = (q+w)− w

H = qp (p means constant pressure)

• So, at constant pressure the change in enthalpy is the heat gained or lost.

We use H to describe chemical reactions because most reactions happen at constant pressure, and at constant pressure H= q and q is (relatively) easy to find!

Often for chemical reactions, PΔV is very small and ΔE ≈ ΔH

Important!

ΔH and sign:

Endothermic – ΔH is positive (heat transferred to the system from the surroundings)

Phase changes: s l g

Exothermic – ΔH is negative (heat transferred from the system to the surroundings)

Phase changes: g l s

Thermochemical Equations• Must specify physical states of products and reactants

Enthalpy depends on state!

• Must specify enthalpy change

Example:

CaO(s) + CO2(g) CaCO3(s) H298 = -178kJ

H, is called the enthalpy of reaction, or the heat of reaction (HRXN).

(Temp at which the reaction occurs is sometimes included)

How to we figure out H?

Method #1 to find :Calorimetry

When =q, you can calculate q like you did in your first year class when given the proper information!

qp = m Tcp

qp =ccT

qlost = qgained

Specific heat of water (and most aqueous sol’ns):

4.184J/gºC or 4.184J/g K

*On AP Equation Sheet! (q=mc T)

Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, we can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Constant Volume (Bomb) Calorimetry

Reactions can be carried out in a sealed “bomb” calorimeter, such as this one, to measure the heat absorbed by the water.

Bomb Calorimetry• Because the volume

in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.

• For most reactions, the difference is very small.

Temperature changes are converted to heat energy by using heat capacities.

Heat Capacity (C) - The amount of energy required to raise the temperature of a substance by 1 K (J/K)

Molar Heat Capacity (Cm) – The amount of energy required to raise the temperature of 1 mole of a substance by 1 K (J/molK)

Specific Heat Capacity (or simply Specific Heat) (Cs) - The amount of energy required to raise the temperature of 1 g of a substance by 1 K. (J/gK)

Calorimeter Constant (cc) – The heat capacity of a calorimeter(J/K)

*temperatures may also be in °C

Example #2

The specific heat of ammonia is 4.381 J/gK. Calculate the heat required to raise the temperature of 1.50g of ammonia from 213.0K to 218.0K.

q = mΔTcp

q = (1.50g)(218.0K – 213.0K )(4.381J/gK )

q = 33 J

Example #32.15 g of methane is combusted (burned) to form CO2 and

H2O. All of the heat evolved is used to raise the

temperature of 100.0 g of water by 17.8 K. Calculate the heat (q) of combustion per mol of methane.

q = (100.0 g) (17.8 K) (4.184 J/g K) = 7450 J = 7.45 kJ

7.45 kJ----------

2.15 g CH4

16.0 g CH4

X ----------- 1 mole CH4

= 55.4 kJ/mol

Example # 450.00 mL of 0.750 M HCl solution and 50.00 mL

of 0.875 M KOH solution are combined in a calorimeter. The temperature of the calorimeter and both solutions goes from 25.0 oC to 29.6 oC. Assume that the specific heat of each solution is the same as water, 4.184 J/g oC and that the density of each solution is 1.00 g/mL. The heat capacity of the calorimeter is 125 J/oC. Calculate the heat of reaction per mole of water formed.

HCl(aq) + KOH(aq) H2O(l) + KCl(aq)

Hrxn= q= energy change of solutions + energy change of calorimeter

= (m)( T)(cp) + (cc)( T)

(100.00g)(4.6oC)(4.184 J/goC) + (125 J/oC)(4.6oC)

1900 J + 580 J

2500 J or 2.5 kJ

since this reaction is exothermic: Hrxn = - 2500J or - 2.5 kJ

We want the heat of reaction per mole of water!

  Do an excess-limiting problem: .0500L .0500L

.750M .875M ? mol

HCl(aq) + KOH(aq) H2O(l) + KCl(aq)

0.0375 moles of H2O are formed in the reaction  - 2.5 kJ

-------------- = - 67 kJ / mol H2O 0.0375 mol 

Some more about Enthalpy…

1. Enthalpy is an extensive property (it depends on amount).

2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.

3. H for a reaction depends on the state of the products and the state of the reactants.

(Enthalpy: gases > liquids > solids)

ΔH & Stoichiometry

Enthalpy is an extensive property so changing the stoichiometry changes the value of ΔH

CaO(s) + CO2(g) CaCO3(s) H = -178kJ

2CaO(s) + 2CO2(g) 2CaCO3(s) H = -356kJ

½ CaO(s) + ½ CO2(g) ½ CaCO3(s) H = -89.0kJ

Example #5How much heat is released when 4.50g of methane gas

is burned in a constant pressure system? CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = -802 kJ

For every 1 mol of methane burned 802 kJ is released

44

4 4

1 CH 8024.50 CH * * 226

16.0g CH 1 CH

mol kJg kJ

mol

Method #2 to find : Hess’s Law

Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Example #6

Given:

C(s) + O2(g) CO2(g) ΔH = -393.5 kJ

CO(g) + ½ O2(g) CO2(g) ΔH = -283.0 kJ

Find ΔH for the following:

C(s) + ½ O2(g) CO(g)

C(s) + O2(g) CO2(g) ΔH = -393.5 kJ

CO(g) + ½ O2(g) CO2(g) ΔH = -283.0 kJ

C(s) + O2(g) CO2(g) ΔH = -393.5 kJ

+ CO2(g) CO(g) + ½ O2(g) ΔH = +283.0 kJ

C(s) + ½ O2(g) CO(g) ΔH = -110.5 kJ

½

Example #7Given:

C2H2(g) + 5/2 O2(g) 2CO2(g) + H2O(l) ΔH = - 1299.6 kJ

C(s) + O2(g) CO2(g) Δ H = - 393.5 kJ

H2(g) + ½ O2(g) H2O(l) Δ H = - 285.9 kJ

Find:

2 C(s) + H2(g) C2H2(g)

2 CO2(g) + H2O(l) C2H2(g) + 5/2 O2(g)

Δ H = + 1299.6 kJ

2 C(s) + 2 O2(g) 2 CO2(g) Δ H = - 787.0 kJ

H2(g) + ½ O2(g) H2O(l) Δ H = - 285.9 kJ

2C(s) + H2(g) C2H2(g) ΔH = 226.7 kJ

Enthalpy (Heat) of Formation

An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

H2(g) + ½ O2(g) H2O(g) ΔHf = -242.3 kJ/mol

Hf depends on physical state, pressure and temperature

Standard Enthalpy of FormationStandard enthalpies of formation, , are measured under

standard conditions (25°C and 1.00 atm pressure).

The standard enthalpy of formation is the heat of formation of one mole of the compound in its standard state formed from its constituent elements in their standard states

The standard state of formation of any element in its most stable form is by convention taken to be zero

• values are tabulated in Appendix C of your textbook (page 1059)

fH

Method #3 to find The change in enthalpy, H, is the enthalpy of the

products minus the enthalpy of the reactants:

ΔH°RXN = Σ nΔ H°f products –

Σ mΔ H°f reactants

n and m are the stoichiometric coefficients of the chemical equation

**On AP Equation sheet! (without m & n)

Example #8

(a) Calculate the standard enthalpy change for the combustion of 1 mole of benzene, C6H6(l) to CO2(g) and H2O(l) (b) If the ΔH°rxn for the combustion of 1 mole of propane is – 2220 kJ, compare the quantity of heat produced by combustion of 1.00 g propane C3H8(g) to that produced by 1.00 g of benzene

a) C6H6(l) + 15/2 O2(g) 6 CO2(g) + 3 H2O(l)

Δ H°rxn = Σ n Δ H°f (prod.) - Σ m Δ H°f (reactants)

using appendix C

Δ H°rxn = [6(-393.5 kJ) + 3(-285.8 kJ)] –

[(49.04 kJ) + 15/2(0 kJ)]

= - 3267 kJ

(b)

Propane: C3H8(g):

Benzene C6H6(l):

Example #9The standard enthaply change for the reaction

CaCO3(s) CaO(s) + CO2(g) is 178.1 kJ

Given ΔH°f CaO(s) as -635.5 kJ and Δ H°f CO2(g) as -393.5 kJ

calculate Δ H°f CaCO3(s) (without looking it up in the table)

Δ H°rxn = Δ H°f CaO + Δ H°f CO2 - Δ H°f CaCO3

178.1 kJ = -635.5 kJ – 393.5 kJ - Δ H°f CaCO3

solving for Δ H°f CaCO3 gives -1207.1 kJ

which matches the value in the table in Appendix C

Method #4 to find • The energy required to break a bond between

two atoms is called the bond enthalpy.• We use the letter D to represent bond enthalpies.

Ex: D(H-Br) = 366 KJ/mol

Δ Hrxn =

Σ(bond enthalpies of bonds broken) –

Σ(bond enthalpies of bonds formed)

Example #10

• Calculate the enthalpy of the reaction when methane gas and chlorine gas react to form methyl chloride (CH3Cl) and hydrogen chloride.

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

Bonds Broken:

4* D(C-H) + D(Cl-Cl)

Bonds Formed:

3*D(C-H) + D(C-Cl) + D(H-Cl)

Δ Hrxn = 4(413) + 242 – 3(413) – 328 - 431

= 1650 + 242 – 1240 – 328 - 431

= -110 KJ