Chapter 5 Principles of Chemical Reactivity. Basic Principles Thermodynamics: The science of heat...

Chapter 5

Principles of Chemical Reactivity

Basic PrinciplesThermodynamics: The science of heat

and workEnergy: the capacity to do work

-chemical, mechanical, thermal, electrical, radiant, sound, nuclear-affects matter by raising its temperature,

eventually causing a state change-All physical changes and chemical

changes involve energy

Potential Energy: energy that results from an object’s

position -gravitational, chemical, electrostatic

Kinetic Energy: energy of motion

Basic Principles

Law of Energy Conservation:Energy can neither be created nor

destroyed-a.k.a. The first law of thermodynamics

-The total energy of the universe is constant

Temperature vs. Heat:– Temperature is the measure of an object’s

heat energy– Heat ≠ temperature

The Measurement of Heat

Thermal Energydepends on temperature and the amount (mass or volume) of the object

-More thermal energy a substances has the greater the motion its atoms/molecules have-Total thermal energy of an object is the sum of the individual energies of all atoms/molecules/ions that make up that object

SI unit: Joule (J)1 calorie = 4.184 JEnglish unit = BTU

Converting Calories to Joules

Convert 60.1 cal to joules

System: object or collection of objects being studied– In lab, the system is the chemicals inside the

beaker

Surroundings: everything outside of the system that can exchange energy with the system– The surroundings are outside the beaker

Universe: system plus surroundings

Exothermic: heat transferred from the system to the surroundings

Endothermic: heat transferred from the surroundings to the system

Basic Principles

Specific Heat Capacity (C)

amount of heat required to raise the temperature of 1gram of a substance by 1 kelvin

SI Units: Specific heat capacity = J /g.K

Specific heat of water = 4.184J

g.K

Heat Transfer

Heat transfer equationused to calculate amounts of heat (q) in

a substance

T C mq J

g Jg.K

K

q1 + q2 + q3 … = 0 or qsystem + qsurroundings = 0

Heat Transfer

Calculate the amount of heat to raise the temperature of 200 g of water from 10.0 oC to 55.0 oC

Heat Transfer

Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C

Heat Transfer

Specific heat of gold is 0.13

Therefore the metal cannot be pure gold.

A 1.6 g sample of metal that appears to be gold requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold?

Jg.K

Changes of State

occurs when enough energy is put into a substance to over come molecular interactions

Solid-liquid: molecules in a solid when heated move about vigorously enough to break solid-solid molecular interactions to become a liquid

Liquid-gas: molecules in a liquid when heated move about more vigorously enough to break liquid-liquid molecular interactions to become a gas

Note: This happens in reverse by removing heat energy

Energy and Changes of StateHeat of fusion:

heat needed to convert a substance from a solid to a liquid (at its melting/freezing point)

333 J/g for water

Heat of vaporization: heat needed to convert a substance from a liquid to a gas (at its boiling/condensation point)

2256 J/g for water

Example: Calculate the amount of heat involved to convert 100.0 g of ice at -50.0°C to steam at 200.0°C.

The First Law of Thermodynamics

This law can be stated as, “The combined amount of energy in the universe is constant”

Also called-The Law of Conservation of Energy:– Energy is neither created nor destroyed in

chemical reactions and physical changes.

The First Law of Thermodynamics

There are two basic ideas for thermodynamic systems:

1. Chemical systems tend toward a state of minimum potential energy

Some examples of this include:-H2O flows downhill-Objects fall when dropped

Epotential = mg(h)

State FunctionThe value of a state function is independent

of pathway-An analogy to a state function is the energy required to climb a mountain taking two different paths:

E1 = energy at the bottom of the mountainE1 = mgh1

E2 = energy at the top of the mountainE2 = mgh2

E = E2-E1 = mgh2 – mgh1 = mg(h)

Examples of state functions: Temperature, Pressure, Volume, Energy, Entropy, and

enthalpy

Examples of non-state functions: Number of moles, heat, work

The First Law of Thermodynamics

2. Chemical systems tend toward a state of maximum disorder

Common examples of this are:- A mirror shatters when dropped and does

not reform- It is easy to scramble an egg and difficult to

unscramble it- Food dye when dropped into water disperses

The First Law of Thermodynamics

Thermodynamic questions:1. Will these substances react when they

are mixed under specified conditions?2. If they do react, what energy changes

and transfers are associated with their reaction?

3. If a reaction occurs, to what extent does it occur?

Changes in Internal Energy (E or U)

all of the energy contained within a substance– all forms of energy such as kinetic,

potential, gravitational, electromagnetic, etc.

First Law of Thermodynamics states that the change in internal energy, E, is determined by the heat flow (q) and the work (w)

E = q + wbook: U = q + w

Changes in Internal Energy (E)

E = Eproducts – Ereactants

E = q + w at constant pressure:

w = -P x V

q > 0 if heat is absorbed by the systemq < 0 if heat is absorbed by the surroundingsw > 0 if surroundings do work on the systemw < 0 if system does work on the surroundings

Changes in Internal Energy (E)

If 1200 joules of heat are added to a system, and the system does 800 joules of work on the surroundings, what is the:

1. energy change for the system, Esys?

2. energy change of the surroundings, Esurr?

Changes in Internal Energy (E)

E is negative when energy is released by a system

-Energy can be written as a product of the process

kJ 10 3.516- E

kJ 10 3.516 OH 6 CO 5 O 8 HC3

3)(22(g)2(g))(125

Changes in Internal Energy (E)

E is positive when energy is absorbed by a system undergoing a chemical or physical change

– Energy can be written as a reactant of the process

kJ 10 3.516 E

O 8 HC kJ 10 3.516 OH 6 CO 53

2(g))(1253

)(22(g)

Enthalpy Changes for Chemical Reactions

Exothermic reactions: release energy in the form of heat to the surroundings (H < 0)

-heat is transferred from a system to the surroundings

Endothermic reactions: gain energy in the form of heat from the surroundings (H > 0)

-heat is transferred from the surroundings to the system

For example, the combustion of propane:

Combustion of butane:

Enthalpy Change (H)Heat content of a substance at constant

pressure- Chemistry is commonly done in open beakers

on a desk top at atmospheric pressure- Therefore enthalpy change (H) is the change

in heat content: H = qp at constant pressure E = qv at constant volume

If E and H < 0:energy is transferred to the surroundings

If E and H > 0: energy is transferred to the system

- Enthalpy and energy differ by the amount of work

E = H + w and w = -PV

Enthalpy Changes for Chemical Reactions

Exothermic reactions generate specific amounts of heat– Because the potential energies of the products are

lower than the potential energies of the reactants

Endothermic reactions consume specific amounts of heat– Potential energies of the reactants are lower than the products

H for the reverse reactionis equal, but has the opposite sign to the forward reaction

Thermochemical Equations

balanced chemical reaction with the H value for the reaction

H < 0 designates an exothermic reaction: heat is a product, the container feels hot

H > 0 designates an endothermic reaction: heat is a reactant, the container feels cold

kJ 3523 - H OH 6 CO 5O 8 HC orxn)(22(g)2(g))12(5

Hess’s Law

If a reaction is the sum of two or more other reactions, H for the overall process is the sum of the H for the component reactions– Hess’s Law is true because H is a state function

If we know the following H values:

kJ 1648H OFe 2 O 3 Fe 4 3

kJ 454H FeO 2O Fe 2 2

kJ 560H OFe 2O FeO 4 1

o3(s)22(g)(s)

o(g)2(g)(s)

o(s)322(g)(s)

1.

2.

Target: ?

Hess’s LawWe can calculate the H for the reaction by

properly adding (or subtracting) the H for reactions 1 and 2– Notice that the target reaction has FeO and O2

as reactants and Fe2O3 as a product– Arrange reactions 1 and 2 so that they also have

FeO and O2 as reactants and Fe2O3 as a product• Each reaction can be doubled, tripled, or multiplied by

a half, etc.• H values are then doubled, tripled, etc.• If a reaction is reversed the H value is changed to the

opposite sign

H

2(2 FeO Fe O kJ

3 4 Fe O Fe O kJ

4 FeO O Fe O kJ

0

s s 2 g

s 2 g 2 3 s

s 2 g 2 3

2 2 2 2 544

3 2 1648

1 2 560

x [ ] ) ( )

1.

2.+

4 FeO 4 Fe 2 O2

+1088 kJ

1

Hess’s Law

Given the following equations and H values

calculate H for the reaction below:

H kJ

2 N O N O 164.1

[2] N + O NO 180.5

[3] N + 2 O NO 66.4

o

2 g 2 g 2 g

2 g 2 g g

2 g 2 g 2 g

[ ]1 2

2

2

You do it!

?H NO 3 NO + ON ogg2g2

Hess’s Law

-The + sign of the H value tells us that the reaction is endothermic.

-The reverse reaction is exothermic, i.e.

3 NO N O + NO H = -155.5 kJg 2 g 2 go

Standard Enthalpy of Formation

Thermochemical standard state conditionsThe thermochemical standard T = 298.15 KThe thermochemical standard P = 1.0000 atm

- Be careful not to confuse these values with STP

Thermochemical standard states of matterFor pure substances in their liquid or solid phase the

standard state is the pure liquid or solid- For aqueous solutions the standard state is 1.00 M

concentrationFor gases the standard state is the gas at 1.00 atm of

pressure- For gaseous mixtures the partial pressure must be

1.00 atm

Standard Molar Enthalpies of Formation, Hf

o

The enthalpy change for the formation of one mole of a substance formed directly from its constituent elements in their standard states– The symbol for standard molar enthalpy of formation is Hf

o (KJ/mol)

The standard molar enthalpy of formation for MgCl2 is:

Standard Molar Enthalpies of Formation (Hf

o)Standard molar enthalpies of formation have been

determined for many substances and are tabulated in Appendix L

Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero.

Example: The standard molar enthalpy of formation for phosphoric acid is -1279 kJ/mol. Write the equation for the reaction for which Ho

f = -1279 kJ

Note: P in standard state is P4 (s)

Phosphoric acid in standard state is H3PO4(s)

Standard Molar Enthalpies of Formation

Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.

Standard Molar Enthalpies of Formation

Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.

You do it!You do it!

Hess’s LawVersion IIFor chemical reaction at standard conditions:

– the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products minus the sum for the reactants

- each enthalpy of formation is multiplied by its coefficient in the balanced chemical equation

tscoefficien tricstoichiomen

Hn Hn H 0reactants f

0products f

0rxn

nn

Final - Initial

Hess’s Law

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Calculate the H°f for the following reaction fromthe data in appendix L:

Hess’s Law

Given the following information, calculate Hf

o for H2S(g)

2 H S + 3 O 2 SO + 2 H O H = -1124 kJ

H ? 0 - 296.8 - 285.8

(kJ / mol)

2 g 2 g 2 g 2 298o

fo

l

You do it!You do it!

CalorimetryAn experimental technique that measures the heat

transfer during a chemical or physical process

Constant pressure calorimetry:A styrofoam coffee-cup calorimeter is

used to measure the amount of heat produced (or absorbed) in a reaction

– This is one method to measure qP (called H) for reactions in solution

qreaction + qsolution = 0

Note: Assuming no heat transfer to the surroundings

CalorimetryIf an exothermic reaction is performed in a

calorimeter, the heat evolved by the reaction is determined from the temperature rise of the solution– This requires a two part calculation

When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947oC. Determine heat of reaction and then calculate the change in enthalpy (as KJ/mol) for the production of NaCH3COO.

CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O(l)

Calorimetry

Constant volume calorimetry:Or Bomb calorimetry measures measure the amount of

heat produced (or absorbed) in a chemical reaction

-this method is used for measuring qv (E)

qreaction + qbomb + qwater = 0