ThermochemistryThermochemistry

Chapter 6Chapter 6

AP ChemistryAP Chemistry

Seneca Valley SHSSeneca Valley SHS

6.1 The Nature of Energy6.1 The Nature of Energy

Kinetic and Potential EnergyKinetic and Potential Energy

• Energy: the capacity to do work or produce heat.

• Kinetic energy: energy an object has due to its motion

• Potential energy: energy due to position or composition

221 mvEk

6.1 The Nature of Energy6.1 The Nature of Energy

Kinetic and Potential EnergyKinetic and Potential Energy

Potential energy can be converted into kinetic energy.

Example: a ball of clay dropping off a building.

6.1 The Nature of Energy6.1 The Nature of Energy

Temperature: the average of energy of particles in motion.

temp. heat

Heat: the transfer of energy from a hotter object to a colder one.

State Function: property of a system that depends only on its present condition

6.1 The Nature of Energy6.1 The Nature of Energy

Systems and SurroundingsSystems and SurroundingsSystem: part of the universe we are interested in.Surroundings: the rest of the universe.

Exothermic: reaction that involves the release of heat to the surroundingsEndothermic: reaction that absorbs energy from the surroundings

6.1 The Nature of Energy6.1 The Nature of Energy

ThermodynamicsThermodynamics• Def: the study of energy and its interconversions

First Law of Thermodynamics (Law of Conservation of Energy)

Def: Energy cannot be created nor destroyed in ordinary chemical reactions

6.1 First Law of Thermodynamics6.1 First Law of Thermodynamics

Internal EnergyInternal Energy• Internal Energy:

total energy of a system.

• Cannot measure absolute internal energy.

• Change in internal energy, E = Efinal - Einitial

Relating Relating EE to Heat and Work to Heat and Work

•Energy cannot be created or destroyed.•Energy of (system + surroundings) is constant.•when a system undergoes a physical or chemical change, the change in internal energy is given by: the heat added to or absorbed by the system plus the work done on or by the system:

E = q + w

6.1 First Law of Thermodynamics6.1 First Law of Thermodynamics

Relating Relating EE to Heat and Work to Heat and Work

6.1 First Law of Thermodynamics6.1 First Law of Thermodynamics

Thermodynamic quantities are measured from the perspective of the system

Example: in an endothermic process where heat is flowing into the system, q would be positive (+)

Enthalpy: Heat transferred between the system and surroundings carried out under constant pressure.One can only measure the change( in enthalpy (H):

H = Hfinal - Hinitial = qP

6.2 Enthalpy6.2 Enthalpy

For any reaction: Hrxn = H(products) - H (reactants)

Enthalpy is an extensive property (magnitude H is

directly proportional to amount):

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

2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) H = -1604 kJ

When we reverse a reaction, we change the sign of H:

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

Change in enthalpy depends on state:

H2O(g) H2O(l) H = -88 kJ

6.2 Enthalpy6.2 Enthalpy

Heat Capacity and Specific HeatHeat Capacity and Specific Heat

Calorimetry = science of measuring of heat flow.Calorimeter = device that measures heat flow.Heat capacity = the amount of energy required to raise the temperature of an object.Molar heat capacity = heat capacity of 1 mol of a substance.Specific heat capacity = heat capacity of 1 g of a subst.

q = (specific heat) (grams of substance) T.

6.2 Calorimetry6.2 Calorimetry

Constant-Pressure Constant-Pressure CalorimetryCalorimetryAtmospheric pressure is constant!

H = qP

qrxn = -qsoln =

-(specific heat of solution) (grams) of solution) T.

6.2 Calorimetry6.2 Calorimetry

Constant-Pressure CalorimetryConstant-Pressure Calorimetry

Problem 1: When a student mixes 50.0 mL of 12.0 M HCl and 50.0 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0C to 27.5C. Calculate the enthalpy change per mole of water formed, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, the density of the solution is 1.00 g/mL, and its specific heat is 4.18 J/gC.

6.2 Calorimetry6.2 Calorimetry

The reaction is carried out under constant vol.

Usually used to study combustion.

qrxn = -CcalorimeterT.

Bomb Calorimetry (Constant-Volume Calorimetry)Bomb Calorimetry (Constant-Volume Calorimetry)6.2 Calorimetry6.2 Calorimetry

• Hess’s lawHess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is equal to the sum of H’s from each individual step.

• For example:

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

2H2O(g) 2H2O(l) H = -88 kJ

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

6.3 Hess’s Law6.3 Hess’s Law

Notes on Hess’s LawNotes on Hess’s Law1.1. If a reaction is reversed in order to solve a Hess’s Law problem, If a reaction is reversed in order to solve a Hess’s Law problem,

then the sign of the then the sign of the H of the reaction is also reversed.of the reaction is also reversed.

2. The magnitude of H is directly proportional to the quantities of reactants and products. If the coefficients in an equation are multiplied by an integer, the H is also multiplied by that integer

3. Hess’s Law can be used to calculate the H in a reaction; however, the component equations do not necessarily tell us how the overall reaction occurred.

6.3 Hess’s Law6.3 Hess’s Law

In the above enthalpy diagram note that: H1 = H2 + H3

6.3 Hess’s Law6.3 Hess’s Law

Example (ex. 6.8 pg. 258): Diborane (B2H6) is a highly reactive boron hydride.

Calculate H for the synthesis of diborane from its elements, according to the following equation:

2 B(s) + 3 H2 (g) --> B2H6 (g)

Use the following data to help you:

a) 2 B(s) + 3/2 O2(g) --> B2O3 (s) H = -1273 kJ

b) B2H6(g) + 3 O2(g) --> B2O3(s) + 3 H2O(g) H = -2035 kJ

c) H2 (g) + 1/2 O2(g) --> H2O (l) H = -286 kJ

d) H2O(l) --> H2O(g) H = 44 kJ

6.3 Hess’s Law6.3 Hess’s Law

6.4 Enthalpies of Formation6.4 Enthalpies of Formation• If 1 mol of compound is formed from its constituent

elements, then the enthalpy change for the reaction is called the enthalpy of formation, Ho

f .

• Degree symbol in Hof indicates standard conditions…

• Standard conditions 1 atm and 25 oC (298 K). • If there is more than one state for a substance under

standard conditions, the more stable one is used.• Learn the distinction between standard state and

standard conditions---they are not the same!

6.4 Enthalpies of Formation6.4 Enthalpies of Formation• Standard enthalpy of formation of the most stable

form of an element is zero.

6.4 Enthalpies of Formation6.4 Enthalpies of Formation

Using Enthalpies of Formation to Calculate Using Enthalpies of Formation to Calculate Enthalpies of ReactionEnthalpies of ReactionFor a reaction:

of(products) (reactants)f fH n H m H

6.4 Enthalpies of Formation6.4 Enthalpies of Formation

Using Enthalpies of Formation to Calculate Using Enthalpies of Formation to Calculate Enthalpies of ReactionEnthalpies of Reaction

Example 1: (Ex. 6.9)Use the standard enthalpies of formation listed in Table 6.2 to calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This is the first step in the manufacture of nitric acid.

6.4 Enthalpies of Formation6.4 Enthalpies of Formation

Using Enthalpies of Formation to Calculate Using Enthalpies of Formation to Calculate Enthalpies of ReactionEnthalpies of Reaction

Example 2: (Ex. 6.10) Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction, which occurs when a mixture of powdered aluminum and iron (III) oxide is ignited with a magnesium fuse:

2 Al(s) + Fe2O3(s) --> Al2O3 (s) + 2 Fe(s)

End of Chapter 6End of Chapter 6

ThermochemistryThermochemistry

To save time we will not formally cover 6.5-6.6 in class (thus you won’t be tested on it!)