Chapter 5:thermochemistry

By Keyana PorterPeriod 2

AP Chemistry

What is thermochemistry? Thermodynamics: the study of energy and its

transformations

Thermochemistry is one aspect of thermodynamics

The relationship between chemical reactions and energy changes

Transformation of energy (heat) during chemical reactions

5.1Kinetic & Potential Energy

Energy is the capacity to do work or the transfer heat Objects possess energy in 2 ways

Kinetic energy: due to motion of object Potential/Stored energy: result of its composition or its position

relative to another object

Kinetic energy (Ek) of an object depends on its mass (m) and speed (v):

Ek= ½ mv2

Kinetic energy increases as the speed of an object and its mass increases

Thermal energy: energy due to the substance’s temperature; associated with the kinetic energy of the molecules

5.1Kinetic & Potential Energy

Potential energy is a result of attraction & repulsion Ex: an electron has potential energy when it

is near a proton due to the attraction (electrostatic forces)

Chemical energy: due to the stored energy in the atoms of the substance

5.1Units of Energy

joule (J): SI unit for energy 1 kJ= 1000 J

calorie (cal): non-SI unit for energy; amount of energy needed to raise the temperature of 1 g of water by 1oC 1 cal = 4.184 J (exactly) Calorie (nutrition unit) = 1000 cal = 1 kcal

A mass of 2 kg moving at a speed of 1 m/s = kinetic energy of 1 J

Ek= ½ mv2 = ½ (2 kg)(1 m/s)2 = 1 kg-m2/s2 = 1 J

5.1System & Surroundings

System (chemicals): portion that is singled out of the study

Surroundings (container and environment including you): everything else besides the system

Closed system: can exchange energy, in the form of heat & work, but not matter with the surroundings

5.1Transferring Energy: Work & Heat Energy is transferred in 2 ways:

Cause the motion of an object against a force Cause a temperature change

Force (F): any kind of push or pull exerted on an object Ex: gravity

Work (w): energy used to cause an object to move against force

Work equals the product of the force and the distance (d) the object is moved:

w = F x d heat: the energy transferred from a hotter object to a

colder one Combustion reactions release chemical energy stored in the form

of heat

5.2Internal Energy The First Law of Thermodynamics: Energy is

conserved; it is neither created nor destroyed Internal energy (E): sum of ALL the kinetic and

potential energy of all the components of the system The change in internal energy = the difference between Efinal

– Einitial

ΔE = Efinal – Einitial

We can determine the value of ΔE even if we don’t know the specific values of Efinal and Einitial

All energy quantities have 3 parts: A number, a unit, and a sign (exothermic versus

endothermic)

5.2Relating ΔE to Heat & Work A chemical or physical

change on a system, the change in its internal energy is given by the heat (q) added to or given off from the system:

ΔE = q + w both the heat added to

and the work done on the system increases its internal energy

Sign Conventions Used and the Relationship Among q, w, and ΔE

Sign Convention for q:q > 0: Heat is transferred from the surroundings to the system (endothermic)q < 0: Heat is transferred from the system to the surroundings (exothermic)

Sign convention for w:w > 0: Work is done by the surroundings on the systemw < 0: Work is done by the system on the surroundings

Sign of ΔE = q + w

q > 0 and w > 0: ΔE > 0q > 0 and w < 0: the sign of ΔE depends on the magnitudes of q and wq < 0 and w > 0: the sign of ΔE depends on the magnitudes of q and wq < 0 and w < 0: ΔE < 0

5.2Endothermic & Exothermic Processes Endothermic: system absorbs heat

Ex: melting of ice Exothermic: system loses heat and the heat flows

into the surroundings Ex: freezing of ice

The internal energy is an example of a state function

Value of any state function depends only on the state or condition of the system (temperature, pressure, location), not how it came to be in that particular state ΔE = q + w but, q and w are not state functions

5.3Enthalpy

Enthalpy (H): state function; the heat absorbed or released under constant pressure

The change in enthalpy equals the heat (qP) gained or lost by the system when the process occurs under constant pressure:

ΔH = Hfinal – Hinitial = qP

only under the condition of constant pressure is the heat that is transferred equal to the change in the enthalpy

The sign on ΔH indicated the direction of heat transfer + value of ΔH means it is endothermic - value of ΔH means it is exothermic

5.4Enthalpies of Reaction Enthalpy of reaction (ΔHrxn): the enthalpy change

that accompanies a reaction The enthalpy change for a chemical reaction is

given by the enthalpy of the products minus the reactants:

ΔH = H(products) – H(reactants) Thermochemical equations: balanced chemical

equations that show the associated enthalpy change

The magnitude of ΔH is directly proportional to the amount or reactant consumed in the process

5.4Enthalpies of Reaction The enthalpy change for the reaction is equal in

magnitude but opposite in sign to ΔH for the reverse reaction

CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ΔH= 890 kJ

CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)

ΔH1=

- 890 kJE

nth

alp

y

ΔH2=

890 kJ

Reversing a reaction

changes the sign but not

the magnitude

of the enthalpy

change: ΔH2 = - ΔH1

5.4Enthalpies of Reaction

The enthalpy change for a reaction depends on the state of the reactants and products Ex: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)

ΔH = -890 kJ If the product was H2O (g) instead of H2O (l),

the ΔH would be - 820 kJ instead of - 890 kJ

5.5Calorimetry/Heat Capacity & Specific Heat Calorimetry: the measure of heat flow Calorimeter: measures heat flow Heat capacity: the amount of heat required to raise its

temperature by 1 K The greater the heat capacity of a body, the greater the heat

required to produce a given rise in temperature Molar heat capacity: the heat capacity of 1 mol of a

substance Specific heat: the heat capacity of 1 g of a substance;

measured by temperature change (ΔT) that a known mass (m) of the substance undergoes when it gains or loses a specific quantity of heat (q):

5.5Calorimetry/Heat Capacity & Specific Heat

specific heat = quantity of heat transferred (grams of substance) x (temperature change)

= q m x ΔT

Practice Exercise (B&L page 160)Calculate the quantity of heat absorbed

by 50 kg of rocks if their temperature increases by 12.0 OC. (Assume the specific heat of the rocks is .82 J/g-K.)

5.5Calorimetry/Heat Capacity & Specific Heat

Solving the problem

q = (specific heat) x (grams of substance) x ΔT = (0.82 J/g-K)(50,000 g)(285 K) = 4.9 x 105 J

5.5Constant-Pressure Calorimetry The heat gained by the solution (qsoln) is equal in

magnitude and opposite in sign from the heat of the reaction

qsoln = (specific heat of solution) x (grams of solution) x ΔT = - qrxn

For dilute aqueous solutions, the specific heat of the solution is approx. the same as water (4.18 J/g-K)

Practice Exercise (B&L page 161)When 50 mL of .100M AgNO3 and 50.0 mL of .100 M HCl are

mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30oC to 23.11oC. The temperature increase is caused by this reaction:

AgNO3 + HCl AgCl + HNO3

Calculate ΔH for this reaction, assuming that the combined solution has a mass of 100 g and a specific heat of 4.18 J/g-oC

5.5Constant-Pressure Calorimetry

Solving the problem

qrxn = -(specific heat of solution) x (grams of solution) x ΔT = - (4.18 J/g-oC)(100 g)(0.8 K)

= - 68,000 J/mol

5.5Bomb Calorimetry (Constant-Volume Calorimetry)

Bomb calorimeter: used to study combustion reactions Heat is released when combustion occurs, absorbed

by the calorimeter contents, raising the temperature of the water (measured before and after the reaction)

To calculate the heat of combustion from the measured temperature increase in the bomb calorimeter, you must know the heat capacity of the calorimeter (Ccal)

qrxn = - Ccal x ΔT

5.6Hess’s Law Hess’s Law: if a reaction is carried out in a series

of steps, ΔH for the reaction will be equal to the sum of the enthalpy changes for the individual steps

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) ΔH = -802 kJ

(ADD)2H2O (g) 2H2O (l) ΔH = -88 kJ

CH4 (g) + 2O2 (g) + 2H2O (g) CO2 (g) + 2H2O (g) + 2H2O (l)

ΔH = -890 kJ

5.6Hess’s Law Practice Exercise 5.8 (B&L page 165)

Calculate ΔH for the reaction:2C (s) + H2 C2H2 (g)

Given the following reactions and their respective enthalpy changes:

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

C (s) + O2 (g) CO2 (g) ΔH = -393.5 kJ

H2 (g) + ½ O2 (g) H2O (l) ΔH = -285.8 kJ

5.6Hess’s Law Solving the problem

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

2C (s) + 2O2 (g) 2CO2 (g) ΔH = -393.5 kJ (2)

-787.0 kJ H2 (g) + ½ O2 (g) H2O (l) ΔH = -285.8 kJ

2C (s) + H2 C2H2 (g) ΔH = 226.8 kJ if the reaction is reversed, the sign of ΔH changes if reaction is multiplied, so is ΔH

Enthalpy Diagram

The quantity of heat generated by combustion of 1 mol CH4 is independent of

whether the reaction takes place in one or

more steps:

ΔH1 = ΔH2 + ΔH3

5.7Enthalpies of Formation Enthalpies of vaporization: ΔH for converting liquids to

gases Enthalpies of fusion: ΔH for melting solids Enthalpies of combustion: ΔH for combusting a substance in

oxygen Enthalpy of formation (ΔHf): enthalpy change where the

substance has been formed from its elements Standard enthalpy (ΔH o): enthalpy change when all

reactants and products are at 1 atm pressure and specific temperature (298 K)

Standard enthalpy of formation (ΔHof): the enthalpy change

for the reaction that forms 1 mol of the compound from its elements, with all substances in their standard states

ΔHof = 0 for any element in its purest form at 295 K and 1

atm pressure

5.7Enthalpies of Formation The standard enthalpy change for any reaction

can be calculated from the summations of the reactants and products in the reaction

ΔH orxn = n ΔH o

f (products) - mΔH of

(reactants)

Practice Problem 5.9 (B&L page 169)Calculate the standard enthalpy change for the

combustion of 1 mol of benzene, C6H6 (l), to CO2 (g) and H2O (l).

5.7Enthalpies of Formation

Solving the problemC6H6 (l) + 15/2 O2 (g) 6CO2 (g) + 3H2O (l)

ΔH orxn= [6Δ H o

f (CO2) + 3Δ H of (H2O)] – [ΔH o

f

(C6H6) + 15/2 ΔH of (O2)]

= [6(-393.5 kJ) + 3(-285.8 kJ)] – [(49.0 kJ) + 15/2 (0 kJ)]= (-2361 – 857.4 – 49.0) kJ= -3267 kJ

Extra Equations Force = mass x 9.8 m/s2

Internal energy… ΔE = Efinal – Einitial

Entropy… Δ S = Sfinal – Sinitial

Enthalpy… Δ H = Hfinal – Hinitial

Gibbs Free Energy… Δ G = Gfinal – Ginitial

Δ S = Δ H / T