chemistry-140 lecture 13

Chapter 6: Enthalpy Changes for Chemical Reactions

Chapter Highlights

energy transfer, specific heat and heat transfer

heat of fusion & heat vapourization

exothermic & endothermic reactions

first law of thermodynamics

enthalpy changes

calorimetry

Hess’s Law

Chemistry-140 Lecture 13

Thermodynamics

Thermodynamics: energy and its transformations.

Thermochemistry: energy changes and chemical reactions

Force: "push" or "pull" exerted on an object.

Work: energy required to overcome a force:

Work = force x distance.

Heat: energy transferred from one object to another because of a temperature difference.


Kinetic and Potential Energy

Kinetic energy (EK): energy of an object due to its motion.

EK = mv2 (m = mass, v = velocity)

Potential energy (EP): energy stored by an object

due to its relative position.

1

2


Energy

Energy can be expressed in a wide variety of units.

The joule (J) is the metric unit of energy and is the energy

possessed by a 2 kg object moving at a velocity of 1 m/s.

1 J (joule) =1 kg-m2/s2.

The calorie (cal) is the energy required to raise the

temperature of 1 g of water by 1°C.

1 cal = 4.184 J

1 Cal = 1 kcal = 1000 cal = 4.184 kJ


Energy Transfer

The environment of a chemical reaction is separated into system and surroundings.

System: reactants, products, solvents, etc. in the reaction.

Surroundings: the universe, including the vessel.

2 H2(g) + O2(g) 2 H2O(l) + energy

Processes that lower a system's internal energy are spontaneous. Processes that increase a system's internal energy are nonspontaneous.


Heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C,

The specific heat of a substance, C, is the amount of heat required to raise the temperature of a 1 g sample of the substance by 1°C,

Energy Transfer & Specific Heat

quantity of heat supplied

mass of object temperature change

C =

heat transferred = q = C x Theat transferred = q = C x T


Example 6.2:

A lake has a surface area of 2.6 x 106 m2 and an

average depth of 10 m. What quantity of heat (kJ) must be

transferred to the lake to raise the temperature by 1 oC?

(Assume a density of 1.0 g/cm3 the lake water)

Energy Transfer & Specific Heat


Answer:

Determine the mass of water and calculate the energy

required using the concept of specific heat.

Volume(H2O) = (2.6 x 106 m2)(10 m) = 2.6 x 107 m3 =

Mass(H2O) = (2.6 x 1013 cm3)(1.0 g/cm3) =

Since:

q = m x C x T = (2.6 x 1013 g)(4.184 J/g-K)(1.0 K)

=

2.6 x 1013 cm32.6 x 1013 cm3

2.6 x 1013 g2.6 x 1013 g

quantity of heat supplied

mass of object temperature change

C =

1.1 x 1011 kJ1.1 x 1011 kJ


Energy and Changes of State

Heat transfer: heat is

transferred to a

substance with no

structural change

Phase changes or phase

transitions: a

substance's structure is

altered as in melting,

freezing, vaporizing,

and condensing.


500 kJ

500 kJ

Ice, 2.0 kg, 0oC

Iron, 1.0 kg, 0oC

0oC

0oC

1100oC


Heat of fusion: the enthalpy change associated with melting

a substance, Hfus (kJ/mol).

Heat of vapourization; the enthalpy associated with

vapourizing a substance, Hvap (kJ/mol).

Heating a substance is endothermic. BUT... during a phase

transition the temperature remains constant



The energy required (q) for the five possible processes is determined by the amount of sample.

For warming:

q = C x mass x T,

(specific heat, C, is different for each physical state)

For the phase transitions: q = (# of moles) x H.



Question (similar to example 6.4):

Calculate the enthalpy change upon converting 1.00

mol of ice at -25 oC to water vapour (steam) at 125 oC

under a constant pressure of 1 atm. The specific heat of

ice, water and steam are 2.09 J/g-K, 4.18 J/g-K and 1.84

J/g-K, respectively. Also for water; Hfus = 6.01 kJ/mol

and Hvap = 40.67 kJ/mol.



Answer:

Divide the total change into calculable segments and

calculate the enthalpy change for each segment. Sum to get

the total enthalpy change (Hess's Law).

Segment 1: Heat ice from -25 oC to 0 oC.

Segment 2: Convert ice to water at 0 oC.

Segment 3: Heat water from 0 oC to 100 oC.

Segment 4: Convert water to steam at 100 oC.

Segment 5: Heat steam from 100 oC to 125 oC.


3: Water

2: Melting

1: Ice

5: Steam4: Boiling

Heat absorbed

Heat liberated

Heat added (kJ)

Energy and Changes of StateT

empe

ratu

re o C



Segment 1: Heating the ice from -25 oC to 0 oC.

H1 = (1.00 mol)(18.0 g/mol)(2.09 J/g-K)(25 K) =

Segment 2: Convert ice to water at 0 oC.

H2 = (1.00 mol)(6.01 kJ/mol) =

Segment 3: Heating the water from 0 oC to 100 oC.

H3 = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) =

940 J 940 J

6.01 kJ6.01 kJ

7520 J7520 J



Segment 4: Convert water to steam at 100oC.

H4 = (1.00 mol)(40.67 kJ/mol) =

Segment 5: Heating the steam from 100oC to 125oC.

H5 = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) =

Total Enthalpy Change:

H = H1 + H2 + H3 + H4+ H5

= (0.94 kJ) + (6.01 kJ) + (7.52 kJ) + (40.67 kJ) + (0.83 kJ)

=

40.67 kJ40.67 kJ

830 J830 J

55.97 kJ55.97 kJ


Energy is neither created nor destroyed; only exchanged between system and surroundings.

Internal energy is the total energy of a system. Only changes in internal energy, E, can be measured.

E = Efinal - Einitial

(+ve = gain by system & -ve = loss to surroundings)

E = q + w

(heat added is positive & heat withdrawn is negative)

(work done is positive & work done by is negative)

The First Law of Thermodynamics

heatadded

internalenergy

workdone



Energy transferredfrom surroundings

to system

Energy transferredfrom system

to surroundings


Heat Transfer

Question:

The H2(g) and O2(g) in a cylinder are ignited and the

system loses 550 J to its surroundings. The expanding gas

does 240 J of work on its surroundings. What is the change

in internal energy of the system?


Heat Transfer

Answer:

Energy flows from the system, so q = -550 J.

Work is done by the system, so w = -240 J

Therefore:

E = q + w = (-550 J) + (-240 J) = -790 J-790 J



Term Test #1

Friday October 12th, 2001

6:30 P.M.

Term Test #1


6:30 P.M.


A to H Education Gym

I to Z Rm 104 Odette Bldg




Covers Material From Chapters 4 - 6Covers Material From Chapters 4 - 6

Duration: 75 minutes

Contents: SIX “Problems”!!



A Periodic Table & ALL Required

Constants will be Supplied

A Periodic Table & ALL Required

Constants will be Supplied


Chapter Highlights





enthalpy changes

calorimetry

Hess’s Law


Energy is neither created nor destroyed; only exchanged between system and surroundings.

Internal energy is the total energy of a system. Only changes in internal energy, E, can be measured.

E = Efinal - Einitial

(+ve = gain by system & -ve = loss to surroundings)

E = q + w

(heat added is positive & heat withdrawn is negative)

(work done is positive & work done by is negative)


heatadded

internalenergy

workdone



Energy transferredfrom surroundings

to system

Energy transferredfrom system

to surroundings


Heat & Enthalpy Changes

Enthalpy: amount of heat energy possessed by a substance. Enthalpy change corresponds to the heat change of the

system at constant pressure:

Endothermic reaction: heat is added (positive value).

Exothermic reaction: heat is withdrawn (negative value).

H = qp

H = Hfinal - Hinitial

H = qp

H = Hfinal - Hinitial


Enthalpies of Reaction

Enthalpy change: the sum of the absolute enthalpies of the products minus the absolute enthalpies of the reactants.

2 H2(g) + O2(g) 2 H2O(l) + energy

2 H2(g) + O2(g) 2 H2O(l) H = -483.6 kJ

H = H(products) - H(reactants)H = H(products) - H(reactants)



Enthalpy is an extensive property. The magnitude of H is proportional to the amount of reactant consumed.

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

H = -802 kJ

2 CH4(g) + 4 O2(g) 2 CO2(g) + 4 H2O(g)

H = -1604 kJ



The enthalpy change is equal in magnitude but opposite in sign to H for the reverse reaction.

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

H = -802 kJ

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



The enthalpy change depends on the states of the reactants and the products.

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

H = -802 kJ

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

2 H2O(g) 2 H2O(l)

H = -88 kJ



Question:

Ammonium nitrate can decompose by the reaction:

NH4NO3(s) N2O(g) + 2 H2O(g) H = -37.0 kJ

Calculate the quantity of heat produced when 2.50 g of NH4NO3 decomposes at constant pressure.



Answer:

We know the heat produced from 1 mole of NH4NO3.Calculate moles in 2.50 g and determine the heat produced by that amount.

heat = (2.50 g)

=

1 mol NH NO

80.06 g NH NO 4 3

4 3

- 37.0 kJ

1 mol NH NO 4 3

-1.16 kJ-1.16 kJ


Calorimetry

Calorimetry: The measurement of heat flow. Measurements are made using a calorimeter.

Recall that: The heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C,

In constant-pressure calorimetry, the pressure remains constant because the apparatus is open to the atmosphere. At constant pressure, the heat change of the reaction is the enthalpy change, H.

heat transferred = q = C x Theat transferred = q = C x T


Calorimetry

Question:

When 50 mL of 1.0 M HCl(aq) and 50 mL of 1.0 M

NaOH(aq) are mixed in a constant-pressure calorimeter,

the temperature increases from 21.0 to 27.5 oC. Calculate

H for this reaction assuming a specific heat of 4.18 J/g-oC

and a density of 1.0 g/mL for the final solution.


Calorimetry

Answer:

Recall, C =

q = C x m x T

Since the total solution is 100 mL and the density is 1.0 g/mL,

then m = (100 mL) (1.0 g/mL) = 100 g

T = (27.5 - 21.0 oC) = 6.5 oC

and C = 4.18 J/g-oC

q

m x T


Calorimetry

Answer:

then qp = (4.18 J/g-oC) (100 g) (6.5 oC)

= -2700 J =

Since moles in solution were (0.050 L)(1.0 M ) = 0.050 mol

H = =

-2.7 kJ-2.7 kJ

- 2.7 kJ

0.050 L

-54 kJ/mol-54 kJ/mol


Bomb Calorimetry

In bomb calorimetry, the apparatus is sealed and the

experiment is a constant-volume process. The heat

change of the reaction is the internal energy change.

E = qevolved = -Ccalorimeter x TE = qevolved = -Ccalorimeter x T


Bomb Calorimetry


A Student “Bomb” Calorimetry

Thermometer

Cardboard orStyrofoam Lid

NestedStyrofoam Cups

Exothermic ReactionOccurs in Solution


Bomb Calorimetry

Question:

Hydrazine, N2H4, and its derivatives are widely used

as rocket fuels.

N2H4(l) + O2(g) N2(g) + 2 H2O(g)

When 1.00 g of hydrazine is burned in a bomb

calorimeter, the temperature of the calorimeter increases

by 3.51 oC. If the calorimeter has a heat capacity of

5.510 kJ/oC, what is the quantity of heat evolved?


Bomb Calorimetry

Answer:

Recall that:

qevolved = (-5.510 kJ/oC) x (3.51 oC)

qevolved = -19.3 kJ

Calculate this on a per mole basis:

qevolved = =

E = qevolved = -Ccalorimeter x TE = qevolved = -Ccalorimeter x T

-618 kJ/mol-618 kJ/mol- 19.3 kJ

1.00 g N H2 4

32.0 g N H

1 mol N H2 4

2 4


Term Test #1


6:30 P.M.

Term Test #1


6:30 P.M.


Covers Material From Chapters 4 - 6Covers Material From Chapters 4 - 6





A Simple Periodic Table & ALL

Required Constants will be Supplied

A Simple Periodic Table & ALL

Required Constants will be Supplied



Chapter Highlights





enthalpy changes

calorimetry

Hess’s Law


Hess’s Law

If a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes

for each step.

For example:

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

(2) 2 H2O(g) 2 H2O(l)

H = -88 kJ


Hess’s Law

Add: (1) + (2)

(1) + (2) CH4(g) + 2 O2(g) + 2 H2O(g)

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

H = (-802 - 88) kJ = -890 kJ

Net Equation:

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

H = -890 kJ


Hess’s Law

Question:

Calculate H for the reaction:

2 C(s) + H2(g) C2H2(g)


Hess’s LawQuestion:

Given the following:

(1) C2H2(g) + O2(g) 2 CO2(g) + H2O(l)

H = -1299.6 kJ

(2) C(s) + O2(g) CO2(g)

H = -393.5 kJ

(3) H2(g) + O2(g) H2O(l)

H = -285.9 kJ

1

2

5

2


Hess’s Law

Answer:

Step 1: Target equation has C2H2(g) as a product so we reverse the first equation and change the sign on H:

-(1) 2 CO2(g) + H2O(l) C2H2(g) + O2(g)

H = +1299.6 kJ

5

2


Hess’s Law

Step 2: Target equation has 2 C(s) as a reactant so multiply the second equation x 2:

2 x (2) 2 C(s) + 2 O2(g) 2 CO2(g)

H = -787.0 kJ


Hess’s Law

Step 3: Target equation has H2(g) as a reactant so we keep the third equation as is:

(3) H2(g) + O2(g) H2O(l)

H = -285.9 kJ

1

2


Step 4: Add the three equations from steps 1,2 and 3:

-(1) 2 CO2(g) + H2O(l) C2H2(g) + O2(g)

2 x (2) 2 C(s) + 2 O2(g) 2 CO2(g) (3) H2(g) + O2(g) H2O(l)

Total

2 CO2(g) + H2O(l) + 2 C(s) + 2 O2(g) + H2(g) + O2(g)

C2H2(g) + O2(g) + 2 CO2(g) + H2O(l)

5

2

1

2

1

25

2

Hess’s Law


Hess’s Law

Net

2 C(s) + H2(g) C2H2(g)

-(1) H = +1299.6 kJ

2 x (2)H = -787.0 kJ

(3)H = -285.9 kJ

Total H = +226.7 kJH = +226.7 kJ


Enthalpies of Formation

We can calculate enthalpy associated with many changes.

Vapourization (Hvap for converting liquids to gas)

Fusion (Hfus for melting solids)

Combustion (Hcom for combusting in oxygen)

Enthalpy of the reaction that forms a substance from its

constituent elements is called

the enthalpy of formation, Hf, or the heat of formation.2 C(s) + H2(g) C2H2(g)

Hf = +226.7 kJ



Standard state: the state that is most stable for the substance at the temperature of interest, usually 298 K and 1 atm.

C(s) diamond graphite fullerene

Standard enthalpy of formation, H°f, is the enthalpy of the reaction that forms the substance from its constituent

elements in their standard states.

2 C(s) + H2(g) C2H2(g)

Hof = +226.7 kJ



Enthalpies of reaction can be calculated by applying Hess's law to the enthalpies of formation of the participants:

H°rxn = [nH°f(products)] - [mH°f(reactants)]



The standard enthalpy of formation for ethanol, C2H5OH, is:

2 C(graphite) + 3 H2(g) + O2(g) C2H5OH(l)

H°f = -277.7 kJ

By definition, H°f = 0 for the standard state of an element.

Thus, H°f = 0 for C(graphite)

H°f = 0 for H2(g)

H°f = 0 for O2(g)

1

2


Enthalpies of FormationQuestion:

Calculate H for the combustion of propane, C3H8(g) using tables of standard enthalpies of formation.

TABLE 6.2 Selected StandardEnthalpies of Formation

Substance Hof (kJ/mol)

C3H8(g) -103.8

CO2(g) -393.5

H2O(l) -285.8


Enthalpies of Formation Answer:

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


H°rxn = [3H°f(CO2) + 4H°f(H2O)] - [H°f(C3H8) + 5H°f(O2)]

= [3(-393.5 kJ/mol) + 4(-285.8 kJ/mol)] -

[(-103.8 kJ/mol) + 5(0.0 kJ/mol)]

= (-2324 kJ) - (-103.8) =

-2220 kJ/mol -2220 kJ/mol



Question:

The standard enthalpy change for the reaction

CaCO3(s) CaO(s) + CO2(g) is 178.1

kJ.

Calculate the standard enthalpy of formation of CaCO3(s) from the standard enthalpies of formation of CaO(s)

and CO2(g); -635.1 kJ and 393.5 kJ respectively.


Enthalpies of Formation Answer:


H°rxn = [H°f(CaO) + H°f(CO2)] - [H°f(CaCO3)]

178.1 kJ = [(-635.1 kJ/mol) + (-393.5 kJ/mol)] -

[H°f(CaCO3)]

[H°f(CaCO3)] = -635.1 kJ - 393.5 kJ - 178.1 kJ

[H°f(CaCO3)] =

-1206.7 kJ/mol-1206.7 kJ/mol


Textbook Questions From Chapter #6Specific Heat: 15, 18, 24

Changes of State: 30

Enthalpy: 34, 38

Hess’s Law: 40, 42, 44, 48, 54

Calorimetry: 62

General & Conceptual 70, 78

Textbook Questions From Chapter #6Specific Heat: 15, 18, 24

Changes of State: 30

Enthalpy: 34, 38

Hess’s Law: 40, 42, 44, 48, 54

Calorimetry: 62

General & Conceptual 70, 78


chemistry-140 lecture 13

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energy energy

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energy changes

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systems internal energy

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