Unit 6 Chp 5 B and L Notes.ppt

IMPORTANT NUMBERS:

Guess what each represents

•

120

IMPORTANT NUMBERS:

MATCHING GAME

1. 120

a. “A” days until Exam

2. 35

b. “B” Day labs remain

3. 27

c. days instruction remain

4. 6

d. days until AP Chem

IMPORTANT NUMBERS:

1. 120

a. days until AP Chem

Exam

2. 35

b. days instruction remain

3. 27

c. “A” days until Exam

Unit 6

Unit 6

Chapter 5 (Mostly) THERMOCHEMISTRY (Part I)

TWO FORMS OF ENERGY:

Work: Energy used to

Work

a force acting over a distance.

w = F x d

Heat

energy transferred between objects because

of temperature difference.

The Universe

is divided into two halves.

• The system is the part you are concerned with. • The surroundings are the rest.

Transferring Energy

Transferring Energy

Thermochemistry is the study of heat change in chemical reactions or physical changes.

The system is the specific part of the universe that is of interest in the study. Everything else is called

surroundings.

open

mass & energy Human Body Exchange:

closed Energy Light Bulb

Potential and Kinetic Energy

Potential and Kinetic Energy

Energy

: The capacity to do work or produce heat

Potential Energy (PE)

: Energy due to

position or composition

Kinetic Energy (KE)

: Energy due to the

motion of the object

KE =

1

2

mv

2

ELECTROSTATIC POTENTIAL ENERGY

E

_EL

Most Important Potential Energy is: ELECTROSTATIC POETENTIAL ENERGY E_el

*This describes how

particles interact with one another

Coulomb’s Law:

E_el = k Q₁*Q₂ d

UNITS USED TO MEASURE E?

Joules: J or kJ (1000J = 1 kJ)

Calorie: cal

Conversion: 1 Cal = 4.184 J

Calorie (food): 1 Cal = 1000 cal = 4.184 J

*Capital C for food calories (used to be called kCal)

First Law

First Law

Law of Conservation of Energy

: Energy

can neither be created nor destroyed, but

can be converted between forms

The First Law of Thermodynamics

: The

total energy content of the universe is

constant. In other words, energy is

neither created or destroyed but

converted into other forms of energy.

EVIDENCE OF ENERGY EXCHANGE?

• _{Changes in a substance’s properties or change into a}

different substance requires an exchange of energy.

• _{How can you identify whether or not energy is being}

exchanged?

• _{Changes in temperature!}

• _{Temperature itself is not heat but it does allow us to}

determine what is happening with heat

• _{Heat is exchanged until thermal equilibrium is}

Energy Change in Processes

Energy Change in Processes

Endothermic:

Exothermic:

Systems in which

energy flows into

the system as the

reaction proceeds.

(enter) +q

Systems in which

A system transferring energy as heat.

Exothermic (hot water) Endothermic (ice water)

• These are similar to

WHAT TYPES OF PROCESSES INVOLVE

HEAT TRANSFERS?

• _{Heating or cooling of a substance (heating =}

endo, cooling = exo)

• _{Phase Changes (depends on type of change)}

• _{Chemical Transformations (depends on overall}

energy exchange)

• _{Solution Formation (depends on IMFs of both}

solute and solvent)

• _{How can we measure or calculate these heat}

HEAT TRANSFERS

• _{Must obey the law of conservation of energy.}

• _{Heat flows from the warmer body to the cooler body} • _{Heat is exchanged until thermal equilibrium is}

established

• _{Unfortunately, there is no instrument available that}

directly measures heat. Instead, we measure

temperature changes, and then use them to calculate changes in heat energy.

• _{ΔT, is directly proportional to the amount of heat}

absorbed, which we will identify by the symbol q.

q = mcΔT

The specific heat capacity (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

Heat (q) absorbed or released:

q

=

m x c x



T

+q = endothermic, heat goes in



T

=

T

_final

-

T

_initial

-q = exothermic, heat goes out

Finding the Quantity of Heat

from Specific Heat Capacity

A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 250C to 300.0C? The specific heat capacity (c) of Cu

is 0.387 J/g*K.

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

c of Fe = 0.444 J/g • 0C

t = t_final – t_initial = 50C – 940C = -890C

q = mct = -34,000 J

Other items to solve for

…

1. It requires 129 kJ to heat 466 g of a liquid

substance from 8.50°C to 74.60°C. What is the specific heat capacity of this substance? What is the identity of this substance?

CALORIMETRY

•

Method for measuring heat

exchanges of a system.

• _{Uses include mixing two}

chemicals together, finding an unknown metal, or

calculating enthalpy (ΔH) of chemical reaction.

•

Typically done in a coffee

ENTHALPY? Δ

H

°

• _{Enthalpy (ΔH) is the specific amount of heat}

exchanged by a system under standard conditions,

expressed as + or – quantity, expressed as a magnitude

• _{°indicates standard conditions}

• _{Standard conditions: 1 atm and 25 °C or 298 K} • _{Units of ΔH °= kJ/mol}

• _{Can be related to q in the q = mcΔT equation}

• _{How can we design an experiment to calculate the}

Calculating Enthalpy

Δ

H

°

No heat enters or leaves!

*System and surroundings are opposites, key idea:

*When using formulas, always use water data!!!

q

_sys

= -q

_surr

q_cal = C_cal x t

Reaction at Constant P: H = q_rxn

Formulas:

Formulas:

*During experiment, some heat is lost to the calorimeter, usually heat of calorimeter is negligible

Calorimetry Example:

(Integrating mC

Δ

T and stoich)

Magnesium metal reacts with hydrochloric acid: Mg(s) + 2HCl(aq)  MgCl₂(aq) + H₂(g)

In an experiment, 0.158 g Mg metal is combined with 100.0 mL HCl in a coffee cup calorimeter. The Mg fully reacts. The temperature rises from 25.6°C to 32.8°C. Find ΔH in kJ/mol. (Use 1.00 g/mL as density of soln and C_soln = 4.18J/g°C as the specific heat.)

AP SAMPLE QUESTION

A 100 g sample of a metal was heated to 100oC and then

quickly transferred to an insulated container holding 100 g of water at 22oC. The temperature of the water rose to reach

a final temperature of 35oC. Which of the following can be

concluded?

A.The metal temperature changed more than the water

temperature did; therefore the metal lost more thermal energy than the water gained.

B.The metal temperature changed more than the water

temperature did, but the metal lost the same amount of thermal energy as the water gained.

C.The metal temperature changed more than the water

temperature did; therefore the heat capacity of the metal must be greater than the heat capacity of the water.

D.The final temperature is less than the average starting

AP SAMPLE QUESTION

A piece of Fe(s) at 25°C is placed into H2O(l) at 75°C in an insulated

container. A student predicts that when thermal equilibrium is reached, the Fe atoms, being more massive than the H₂O molecules, will have a higher average kinetic energy than the  H₂O  molecules. Which of the following best explains why the student’s prediction is incorrect?

A.At thermal equilibrium, the less massive  H₂O  molecules would have a higher average kinetic energy than the Fe atoms because they are more free to move than are the Fe atoms.

B.At thermal equilibrium, the collisions between the Fe atoms and

the H₂O molecules would cease because the average kinetic energies of their particles would have become the same.

C.At thermal equilibrium, the movement of both the Fe atoms and

the H₂O molecules would cease; thus, the average kinetic energy of their particles would have to be the same.

SAMPLE QUESTION

For an experiment, 50.0g of H₂O was added to a coffee-cup calorimeter, as shown in the diagram above. The

initial temperature of

the H₂O was 22.0°C, and it

absorbed 300.J of heat from an object that was carefully placed inside the calorimeter. Assuming no heat is transferred to the surroundings, which of the following was the approximate temperature of

the H₂O after thermal equilibrium was reached? Assume that the specific

heat capacity of H₂O is 4.2J/(g⋅K).

A student mixes 50mL of 1.0M HCl and 50mL of 1.0M NaOH in a

coffee-cup calorimeter and observes the change in temperature until the mixture reaches thermal equilibrium. The initial and final

temperatures (°C) of the mixture are shown in the diagram above of the laboratory setup. Based on the results, what is the change in

temperature reported with the correct number of significant figures?

Enthalpy (



H

) Heat of

Reactions or Processes

• The heat of a chemical reaction

 

H = H

_(products)

– H

_(reactants)

• At constant pressure:



H = q

(



H can be substituted for q in q = mcΔt)

 

H + = endothermic reaction

Endothermic

Endothermic

Reaction

Reaction

Exothermic

Exothermic

Reaction

Both Enthalpy diagrams for EXOTHERMIC Processes

E

nt

ha

lp

y,

H

CH₄ + 2O₂

CO₂ + 2H₂O

H_initial

H_final heat out H < 0

Both Enthalpy diagrams for ENDOTHERMIC Processes

E

nt

ha

lp

y,

H

H_initial H_final

H₂O(l) H₂O(g)

heat in H > 0

Water condensing in a cold pipe.

• What

’

s Happening in the above process?

ADDING NUMBERS TO THE

GRAPHS:

Endo or Exothermic?

Calculate _H.

Calculate E_a, both catalyzed and uncatalyzed.

SPECIFIC TYPES OF SPECIAL

ENTHALPIES 

H

1. Heat of Fusion for water H_fus (Endo)

2. Heat of Vaporization for water



H

_vap

(Endo)

3. Heat of Combustions H_c (Exo)

*Must look up heat value depending on hydrocarbon 4. Heat of Formations H_f0 (Both)

*Must look up heat value depending on substance 5. Heat of Solutions (dissolutions) H_soln

H₂O (s) H₂O (l) H = 6.01 kJ

Some Important Types of Enthalpy Change

heat of combustion (Hcomb)

heat of formation (H_f)

heat of fusion (H_fus)

heat of vaporization (H_vap)

C₄H₁₀(l) + 13/2O₂(g) 4CO₂(g) + 5H₂O(g)

K(s) + 1/2Br₂(l) KBr(s)

NaCl(s) NaCl(l)

C₆H₆(l) C₆H₆(g)

Heat of fusion



H

_fus

(melting)

H₂O (s) H₂O (l) H = 6.01 kJ



H

_fus

H

₂

O =

6.01 kJ

Example:

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

6.4

Melting = endo +



H

Freezing = exo -



H

•

You can represent any

substance as a fusion.

It is melting (s)



(l)

Heat of vaporization



H

_vap

(boiling)

H₂O (l) H₂O (g) H = 41.6 kJ



H

_vap

H

₂

O =

41.6 kJ

Example:

41.6 kJ are absorbed for every 1 mole of water that boils at 1000C and 1 atm. How much heat required for 2 moles

water?

Boiling = endo +



H

Condensing = exo -



H

•

You can represent any

substance as a

vaporization. It is

boiling (l)



(g) and is

always a +



H

USEFUL PROCESSES TO NOTE:

•

Phase changes for

different states of

matter.

•

S



L

fusion (



H

_fus

)

•

L



g

vaporization

(



H

_vap

)

•

Can also determine

SAMPLE CALCULATION

A 2.00 mol sample

of C

₂

H

₅

OH undergoes the

phase transition illustrated in

the diagram. Δ

H

_vap

, of 

C

₂

H

₅

OH  is +38.6kJ/mol. What

is the change in enthalpy in

the phase transition shown in

the diagram?

The diagram above represents the melting of H₂O(s). A 2.00 mol sample

of H₂O(s) at 0°C melted, producing H₂O(l) at 0°C. Based on the diagram, which of the following best describes the amount of heat required for this process and the

changes that took place at the molecular level?

A.3.01kJ of heat was absorbed to decrease the average speed of the water molecules in the liquid, which decreases the distance between molecules.

B.6.02kJ of heat was absorbed to increase the number of hydrogen bonds between water molecules in the liquid compared to the solid.

C.12.0kJ of heat was absorbed to decrease the polarity of the water molecules, which increases the density of the liquid compared to the solid.

AP SAMPLE QUESTION

A molecular solid coexists with its liquid phase at its melting point. The solid-liquid mixture is heated, but the

temperature does not change while the solid is melting. The best explanation for this phenomenon is that the heat

absorbed by the mixture

A.is lost to the surroundings very quickly

B.is used in overcoming the intermolecular attractions in the solid

C.is used in breaking the bonds within the molecules of the solid

D.causes the nonbonding electrons in the molecules to move to lower energy levels

A sample of CHCl₃(s) was exposed to a constant source of heat for a period of time. The graph above shows the change in the

ANSWER CHOICES:

A. The specific heat capacity of the liquid is significantly higher than that of the solid, because the particles in the liquid state need to absorb more thermal energy to increase their average speed.

B. The specific heat capacity of the solid is significantly higher than that of the gas, because the particles in the solid state need to absorb more thermal energy to increase their average speed. C. The enthalpy of fusion is greater than the enthalpy of

vaporization, because separating molecules from their bound

crystalline state requires more energy than separating molecules completely from the liquid state.

D. The enthalpy of vaporization is greater than the enthalpy of

INTEGRATED PROBLEM

Calculate the Enthalpy Change, ΔH, upon converting 1.00 mol of ice at -25°C to 125°C under a constant pressure of 1 atm. The specific heats of ice, liquid water and steam are: 2.03 J/g*K, 4.18 J/g*K, 1.84 J/g*K respectively.

For H₂O, ΔH_fus = 6.01 kJ/mol and ΔH_vap = 40.67 kJ/mol

First, analyze the problem.

What are we asked to find?

Can you represent it graphically?

Where?

SPECIFIC TYPES OF SPECIAL

ENTHALPIES 

H

1. Heat of Fusion for water H_fus (Endo)

2. Heat of Vaporization for water



H

_vap

(Endo)

3. Heat of Combustions Hc (Exo)

*Must look up heat value depending on hydrocarbon

4. Heat of Formations H_f0 (Both)

*Must look up heat value depending on substance

5. Heat of Solutions (dissolutions) H_soln

H₂O (s) H₂O (l) H = 6.01 kJ

Some Important Types of Enthalpy Change

heat of combustion (Hcomb)

heat of formation (H_f)

heat of fusion (H_fus)

heat of vaporization (H_vap)

C₄H₁₀(l) + 13/2O₂(g) 4CO₂(g) + 5H₂O(g)

K(s) + 1/2Br₂(l) KBr(s)

NaCl(s) NaCl(l)

Heat of Combustion



H

_comb

CH₄ (g) + 2O₂ (g) CO₂ (g) + 2H₂O (l) H = -890.4 kJ

Is H negative or positive for this process in the diagram?

Heat of Combustion

*Very Exothermic H_comb

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

Standard Enthalpy of Formation

(



H

0

)

f

Standard enthalpy of formation (H0) is the heat change

that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

*The standard enthalpy of formation

of any element in its most stable form

is zero.

H0 (O

2) = 0

f

H0 (O

3) = 142 kJ/mol

f

H0 (C, graphite) = 0

f

H0 (C, diamond) = 1.90 kJ/mol

f

6.6

WRITING EQ. EXAMPLES:

Write an equation to represent the formation of CO₂(g)

from the elements in their standard states. C(s) + O₂(g)  CO₂(g)

Write an equation to represent the formation of NH₃(g)

from the elements in their standard states.

N₂(g) + H₂(g) NH₃(g)

Write an equation to represent the heat of combustion for one mole of butane, C₄H₁₀.

C₄H₁₀ (g) + O₂ (g)  4CO₂ (g) + 5H₂O (g)

H₂O (s) H₂O (l) H = 6.01 kJ

1. The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations Rules using



H

2. If you reverse a reaction, the sign of H changes

H₂O (l) H₂O (s) H =

-

6.01 kJ

3. If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

H₂O (s) H₂O (l) H = 6.01 kJ

4. The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

H₂O (l) H₂O (g) H = 44.0 kJ

Example: How much heat is evolved when 266 g of white phosphorus (P₄) burn in air?

P₄ (s) + 5O₂ (g) P₄O₁₀ (s) H = -3013 kJ

266 g P₄ 1 mol P4 123.9 g P₄

x -3013 kJ 1 mol P₄

ENTHALPY OF REACTION



H

_RXN

•

For a chemical reaction, the 

enthalpy of

reaction 

(Δ

H

_rxn

) is the difference in

enthalpy between products and reactants.

•

The units of Δ

H

_rxn

 are kilojoules

per mole

of each

reactant or product.

•

Must take into consideration the limiting

SOLVING FOR ENTHALPY PROBS

N_2(g) + 3H_2(g)  2NH_3(g) H=-150 kJ

1. How much heat was released if 34 grams of ammonia reacted?

2. The chemical equation shown above represents the reaction between Mg(s) and HCl(aq). When 12.15g of Mg(s) is added to 500.0mL of 4.0MHCl(aq), 95kJ of heat is released.

The experiment is repeated with 24.30 g of Mg(s) and 500.0 mLof 4.0M HCl(aq). What is the amount of heat released by the reaction?

3. The chemical equation above represents the reaction between HCl and NaOH. When equal volumes of 1.0 M HCl and 1.0 M NaOH are mixed, 57.1 kJ of heat is released. If the experiment is repeated with 2.0 M of HCl, how much heat would be released?

*Always base heat on Limiting Reactant

•

-190 kJ

SPECIFIC TYPES OF SPECIAL

ENTHALPIES 

H

1. Heat of Fusion for water H_fus (Endo)

2. Heat of Vaporization for water



H

_vap

(Endo)

3. Heat of Combustions H_c (Exo)

*Must look up heat value depending on hydrocarbon

4. Heat of Formations H_f0 (Both)

*Must look up heat value depending on substance

5. Heat of Solutions (dissolutions) Hsoln

H₂O (s) H₂O (l) H = 6.01 kJ

ENTHALPY OF SOLUTION,



H

_SOLN

• _{The change in enthalpy that occurs when a}

specified amount of solute dissolves in a given quantity of solvent. 

• _{Use total mass of both solute and solvent in the}

problems involving mcaT and only use the mol of solute in calculating H_soln

• H_soln can be endo or exothermic.

IM FORCES IN SOLN FORMATION

1. Solute-solute

interactions between solute particles must

be overcome in order to disperse the solute particles thru

the solvent.

2. Solvent-solvent

interactions between solvent particles

must be overcome to make room for the solute particles in

the solvent.

HYDRATION PROCESS OF NACL IN WATER

Heat of Solution

Heat of Solution

The

Heat of Solution

is the amount of heat

energy absorbed (endothermic) or released

(exothermic) when a specific amount of

solute dissolves in a solvent. It depends on

three factors: (next slide)

Some Examples:

Substance

Substance Heat of Solution Heat of Solution (kJ/mol)

(kJ/mol)

NaOH

NaOH -44.51-44.51

NH

NH44NONO33 +25.69+25.69

KNO

KNO₃₃ +34.89+34.89

HCl

Steps in Solution Formation

Steps in Solution Formation



H

₁

Expanding the solute



H

₂

Expanding the solvent



H

₃

Interaction of solute and solvent to

form the solution

•Have to overcome attractive forces.



H

₁

> 0

•Always endothermic, requires energy

•If polar,



H

₁

is large +

•If nonpolar,



H

₁

is small +

1. Break apart solute.

2. Break apart Solvent

• Have to overcome attractive forces.

 

H

₂

> 0

• Always endothermic, requires energy

• If polar,



H

₂

is large +

• If nonpolar,



H

₂

is small +



H

₃

depends on what you are

mixing.

•If Molecules attract each other:



H

₃

is large and negative.

•(polar and polar or non and non)

•If Molecules don

’

t attract:



H

₃

is small and negative.

•(polar and non)

*This explains the rule

“

Like

dissolves Like

”

“

“

Like Dissolves Like

Like Dissolves Like

”

”

Fats BenzeneBenzene

SteroidsSteroids HexaneHexane

WaxesWaxes TolueneToluene

Polar and ionic solutes dissolve best in polar solvents Nonpolar solutes dissolve best in nonpolar solvents

Salts WaterWater

SugarsSugars Small alcoholsSmall alcohols

Acetic acidAcetic acid

Predicting Solution Formation

Predicting Solution Formation

Solute/

Solute/

Solvent

Solvent





H

H

₁₁

(solute)

(solute)





H

H

₂₂

(solvent)

(solvent)





H

H

₃₃

(mix)

(mix)





H

H

_{solsol’’nn}

Outcome

Outcome

Polar/

Polar/

Polar

Polar

+ large

+ large + large+ large - large- large

+/-small

+/-small

Solution Solution forms forms Nonpolar/ Nonpolar/ Polar/ Polar/ + small

+ small + large+ large - small- small + large+ large

No solution

No solution

forms

forms

Nonpolar/

Nonpolar/

Nonpolar

Nonpolar

+ small

+ small + small+ small - small- small

+/-small

+/-small

Solution Solution forms forms Polar Polar

/

Nonpolar Nonpolar + large

+ large + small+ small - small- small + large+ large

No solution

No solution

MISCIBLE OR IMMISCIBLE?

Miscible

: Pairs of

liquids that mix well

(ethanol and water)

Immiscible

: pairs of

liquids that do not

dissolve in one another

(hexane and water)

NATURAL TENDENCY: SOLNS

Mixing is spontaneous: occurs of its own accord w/out input of energy from outside.

Entropy ΔS: measure of randomness or disorder of a system

ORDER OF DISORDER

•

Phases of matter and entropy:

•

Least

Greatest

Solids < Liquids < Gases

Signs of



S:

More disorder = +



S

*favored by nature

MORE DISORDER

• _{Only Temperature when entropy is 0 is at 0 Kelvin}

(absolute zero, no motion)

• _{Predicting Entropy Change:}

• _{1. Gases have most entropy}

• _{2. Mixtures have more entropy than pure substances}

or elements

• _{3. Side of the chemical equation with more moles}

DOES ENTROPY INCREASE OR DECREASE?

(AS WRITTEN FORWARD DIRECTION)

H₂(g) + Cl₂(g)  2HCl(aq)

NaCl(s) + H₂O(l)  H₂O(l) + NaCl(aq)

AP SAMPLE QUESTION

CALORIMETRY PRAC PROB

100. mL of 0.500 M HCl was mixed with 100. mL of 0.500 M NaOH in a calorimeter with negligible heat capacity. The initial temperature of the HCl and

NaOH solutions was the same, 22.50°C and the final temperature of the mixed solution was 25.86°C.

Calculate the heat change for the neutralization reaction in kJ/mol. (Use 1.0 g/mL as the density of the soln and 4.18 J/g°C for specific heat capacity.)

HCl(aq)+NaOH(aq)  NaCl(aq) + H₂O(l)

YOU TRY:

In a typical experiment, 100 mL of water is placed in the polystyrene foam cup and the initial temperature of the water recorded. 5.05 g potassium nitrate, KNO₃(s), is

added to the water while stirring. The temperature of the solution falls and the minimum temperature achieved is recorded as the final temperature. Determine the value of the molar heat of solution of potassium nitrate in kJ mol-1.

(density of water is 1.00 g mL-1 and specific heat capacity

of water is 4.18 J°C-1g-1)The results of the experiment are

shown in the table below.

The standard enthalpy of reaction (H0 ) is the enthalpy of

a reaction carried out at 1 atm. rxn

aA + bB cC + dD

H0

rxn = [cH0f (C) + dH0f (D)] - [aH0f (A) + bH0f (B)]

H0

rxn = nH0f (products) - mHf0 (reactants)

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

2 Ways to Utilize Hess’ Law

• _{Reaction rearrangement of Heats of formations}

method if given a series of reactions.

• _{Summation of heats of formation if given the}

Hess

Hess

’

’

s

s

Law

Hess

Hess’’s Law Example Problem (Rearrangement of s Law Example Problem (Rearrangement of

Rxns)

Rxns)

Calculate _{H for the combustion of methane, CH4:}

CH₄ + 2O₂  CO₂ + 2H₂O

    Reaction _Ho  

C + 2H₂  CH_{4 -74.80 kJ}

C + O₂  CO_{2 -393.50 kJ}

H₂ + ½ O₂  H₂O _{-285.83 kJ}

CH₄  C + 2H₂ +74.80 kJ

C + O₂  CO₂ -393.50 kJ

2H₂ + O₂  2 H₂O -571.66 kJ

Sum up reaction and



H

Calculation of Heat of Reaction

Calculation of Heat of Reaction

(Summation of Heats)

(Summation of Heats)

Calculate H for the combustion of methane, CH4:

CH₄ + 2O₂  CO₂ + 2H₂O

    Substance _H

f  

CH_{4 -74.80 kJ}

O_{2 0 kJ}

CO_{2 -393.50 kJ}

H₂O _{-285.83 kJ}



H

_rxn

=





H

_f

(products) -





H

_f

(reactants)



H

_rxn

= [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]

Using Hess’s Law to Calculate an Unknown H

Given the following information, calculate the unknown H:

Equation A: CO(g) + 1/2O₂(g) CO₂(g) H_A = -283.0 kJ

Equation B: N₂(g) + O₂(g) 2NO(g) H_B = 180.6 kJ CO(g) + NO(g) CO₂(g) + 1/2N₂(g) H = ?

Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less

SOLUTION:

Equations A and B have to be manipulated by reversal and/or multiplication by factors in order to sum to the first, or target, equation.

Multiply Equation B by 1/2 and reverse it.

H_B = -90.3 kJ CO(g) + 1/2O₂(g) CO₂(g) H_A = -283.0 kJ

NO(g) 1/2N₂(g) + 1/2O₂(g)

H_rxn = -373.3 kJ CO(g) + NO(g) CO₂(g) + 1/2N₂(g)

CALCULATE Δ

H

_F

• _{You can also use Hess’ Law to solve for a heat of}

formation if the heat of rxn is known. (It is just algebra!)

EX: Find the heat of formation for water, given the following:

Bond Energy and Enthalpy

Bond Energy and Enthalpy

D

= Bond energy per mole of bonds

Energy required

Energy released

Breaking bonds always requires energy

Breaking = endothermic

Forming bonds always releases energy

Forming = exothermic

The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.

H_{2 (g)} H _(g) + H _(g) _H0 = 436.4 kJ

Cl_{2 (g)} Cl _(g) + Cl _(g) _H0 = 242.7 kJ

HCl _(g) H _(g) + Cl _(g) _H0 = 431.9 kJ

O_{2 (g)} O _(g) + O _(g) H0 = 498.7 kJ O O

N_{2 (g)} N _(g) + N _(g) H0 = 941.4 kJ N N

Bond Energy

Bond Energies

Use bond energies to calculate the enthalpy change for: CH_4(g) + 2O_2(g) 2CO_2(g) + 2H₂O_(g)



H

0

=



D(

broken

) –



D(

formed

)

H0 = -3156 kJ/mol

Type of Bond Enthalpy (kJ/mol)

C – H 413 O = O 498 C = O 805 H – O 646

Steps:

1.Draw Lewis structures for each of the reactants and products.

2.Look up bond energy and multiply energy to break or form each bond by the amount of bonds broken or formed.

3.Sum together total energy to break bonds. Sum total energy to form bonds.

Use bond energies to calculate the enthalpy change for: H_{2 (g)} + F_{2 (g)} 2HF _(g)

H0 = BE(reactants) – BE(products)

Type of

bonds broken bonds brokenNumber of Bond energy (kJ/mol) change (kJ)Energy

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds

formed bonds formedNumber of Bond energy (kJ/mol) Energy change (kJ)

H F 2 568.2 1136.4

Estimate



H

_f

for Sodium Chloride

(Using Hess’ Law)

Na(s) + ½ Cl

₂

(g)



NaCl(s)

Lattice Energy 786 kJ/mol

Ionization Energy for Na 495 kJ/mol

Electron Affinity for Cl -349 kJ/mol

Bond energy of Cl₂ 239 kJ/mol

Enthalpy of sublimation for Na 109 kJ/mol

Na(s)  Na(g) + 109 kJ

Na(g)  Na+(g) + e- + 495 kJ

½ Cl₂(g)  Cl(g) + ½(239) kJ

kJ)

Cl(g) + e- _ Cl-(g) - 349 kJ

Na+(g) + Cl-(g) _ NaCl(s) -786 kJ

AP SAMPLE QUESTION

What is the standard enthalpy change Δ

H

o

, for the reaction

represented above?

Δ

H

o

f

of C

2

H

2

(

g

) is 230 kJ mol

-1

Δ

H

o

f

of C

6

H

6

(

g

) is 83 kJ mol

-1

Assume that the bond enthalpies of the oxygen and

AP SAMPLE QUESTION

The enthalpy change for the reaction: 2Al(s) +Fe₂O₃(s)→2Fe(s)+Al₂O₃(s) is −860kJ/mol.

Based on the standard enthalpies of formation ΔH∘ f provided in the table, what is the

approximate ΔH∘

f for Fe2O3(s)  ?