Hess’s Law and Standard Enthalpies of Formation - PowerPoint Presentation

Slide1

Hess’s Law and Standard Enthalpies of Formation

Thermochemistry

Pages 242 - 252Slide2

Hess’s Law

Enthalpy changes are state functions.

It does not matter if ΔH for a reaction is calculated in one step or a series of steps = Hess’s Law

By using values of ΔH of known reactions, we can use Hess’s law to solve for enthalpies of reactions whose values we do not know. Slide3

Hess’s Laws Problem-Solving

Manipulate equations so that they add up to the desired equation.

There are 3 ways we can manipulate equations:

1. We can reverse the entire equation (switch reactants and products).

2. We can multiply the entire equation by a factor (such as 3, 2, 1/2 )

3. We can do both #1 and #2.

**When you manipulate an equation you must manipulate the ΔH value in the EXACT same way!!**Slide4

Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal.

Reverse any reactions as needed to give the required reactants and products

Multiply reactions to give the correct numbers of reactants and products

(Trial and error- allow the final reaction to guide you)

**Start by finding a substance that only appears once in the reactants.Slide5

Hess’s Law Example

Calculate

ΔH

for

N

2

(g) + O

2

(g) 2NO(g)Given:N2(g) + 2O2 (g) 2NO2(g) ΔH = 66.4 kJ/mol2NO(g) + O2(g) 2NO2(g) ΔH = -114.1kJ/molA: ΔH = 180.5 kJ/molSlide6

Hess’s Law Practice Together

Given

2H

2

(g) + C(s) CH

4

(g)

ΔH =

-74.81

kJ/mol2H2(g) + O2(g) 2H2O (l) ΔH = -571.66 kJ/molC(s) + O2(g) CO2(g) ΔH = -393.52 kJ/molCalculate ΔH for CH4(g) + 2O

2(g) CO2(g) + 2H2O(l)A:

ΔH = -890.37 kJ/molSlide7

Hess’s Law Practice Problem

Given the following thermochemical data, calculate the ΔH° for:

Ca

(s) + 2H

2

O(l)

Ca

(OH)

2

(s) + H2(g)H2(g) + ½ O2(g) H2O(l) ΔH° = -285kJCaO(s) + H2O(l) Ca(OH)2(s) ΔH° = -64 kJCa(s) + ½ O2(g) CaO(s) ΔH°

= -635 kJA: -414 kJSlide8

Hess’s Law Practice Problem

Calculate ΔH for

4NH

3

(g) + 3O

2

(g) 2N

2

(g) + 6H

2O(l) Given the following reactions and ΔH values,a. 2N2O(g) O2(g) + 2N2(g) ΔH= -164 kJb. 2NH2(g) + 3N2O(g) 4N2(g) + 3H2O(l) ΔH= -1012 kJA: ΔH= -1532 kJSlide9

Hess’s Law Practice Problem

For the reaction:

H

2

O (l) H

2

O(g) ΔH = +44kJ

How much heat is evolved when 9.0 grams of water vapor is condensed to liquid water?

A: 22kJ evolved or -22kJ

Slide10

Standard Enthalpy of Formation

ΔH°

f

The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.

ΔH°

f

The degree symbol (°) on a thermodynamic function indicates the process has been carried out under standard conditions.

Standard state is

not the same as standard temperature and pressure (STP)Slide11

ΔH°f

Key Ideas

ΔH

°

f

is always given per mole of compound formed.

ΔH°

f

involves formation of a compound from its elements with the substances in their standard states.For an element:It is the form which the element exists in at 25°C and 1 atm.For a compound:For a gas it is a pressure of exactly 1 atm (IUPAC – 1bar )For a substance in solution, it is a concentration of exactly 1M.For a pure solid or liquid, is the pure solid or liquid. ΔH°f for an element in its standard state, such as Ba(s) or N2(g), equals 0.Slide12

ΔH°

reaction

=

Σn

p

ΔH°

f

(products) -

Σnr ΔH°f (reactants) The enthalpy change for a given reaction can be calculated by subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products.When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes.When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer.Elements in their standard states are not included in the ΔHreaction calculations. That is, ΔH°f for an element in its standard state is zero.Slide13

Working with Standard Enthalpy of Formation…

Consulting your textbook Appendix (4) and knowledge on standard states, list the standard enthalpy of formation for each of the following substances.

Al

2

O

3

(s)

Ti(s)

P

4(g)SO42-(aq)F2 (g)Slide14

Using the standard heats of formation, calculate ΔH for the following reactions.

HCl

(g) H

+

(

aq

) +

Cl

-

(aq)2NO2(g) N2O4 (g)C2H2(g) + H2(g) C2H4(g)2NaOH(s) + CO2(g) Na2

CO3(s) + H2O(g)Slide15

Standard Enthalpy of Formation Practice

The heat released when HNO

3

reacts with

NaOH

is 56 kJ/mole of water produced. How much energy is released when 400.0mL of 0.200M HNO

3

is mixed with 500.0mL of 0.150 M

NaOH

?The enthalpy of neutralization for the reaction of a strong acid with a strong base is -56kJ/mol of water produced. How much energy will be released when 200.0mL of 0.400M HCl is mixed with 150.0mL of 0.500M NaOH? How does this compare with the answer in the first part? Why?Slide16

Bond Energies – Calculating Heat of Reaction

All substances must be in the gaseous form.

Breaking bonds

requires energy (Endothermic)

Forming bonds

releases energy (Exothermic)

ΔH =

ΣE

bonds

broken - ΣEbonds formed Slide17

Bond Energies and Heat of Reaction Example

Calculate the value of ΔH for the reaction below using the average bond energies.

H

2

(g)

+ F

2

(g) 2HF(g)

BondAverage Bond Energy (kJ/mol)H-H432

H-F565F-F

160Slide18

Bond Energies and Heat of Reaction Practice

Calculate the value of ΔH for the reaction below using the average bond energies.

H

2

(g)

+ ½ O

2

(g) H

2O(g)BondAverage Bond Energy (kJ/mol)H-H

432 O=O496

H-O463Slide19

Individual Practice

57, 58, 59, 60, 61, 63, 65, 67, 71Slide20

Reminders

Hand-warmer Lab Report Due

___________

!

Practice and Review Ch. 6 Concepts and Problems

Begin reading Ch. 7