Energy is the ability to do work or transfer...

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Thermochemistry

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Energy

· Energy is the ability to do work or transfer heat.· Energy used to cause an object that has mass to move is called work.· Energy used to cause the temperature of an object to rise is called heat.

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Units of Energy

· The SI unit of energy is the joule (J).

· An older, non-SI unit is still in widespread use: the calorie (cal). 1 cal = 4.184 J

1J = 1kg m

2

s2

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Definitions:System and Surroundings

· The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).· The surroundings are everything else (here, the cylinder and piston).

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Definitions: Work

· Energy used to move an object over some distance is work.· w = F ´ d where w is work, F is the force, and d is the distance over which the force is exerted.

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Heat

· Energy can also be transferred as heat.· Heat flows from warmer objects to cooler objects.

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Conversion of Energy

· Energy can be converted from one type to another.· For example, the cyclist above has potential energy as she sits on top of the hill.

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Conversion of Energy

· As she coasts down the hill, her potential energy is converted to kinetic energy.· At the bottom, all the potential energy she had at the top of the hill is now kinetic energy.

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First Law of Thermodynamics

· Energy is neither created nor destroyed.· In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Initialstate

Finalstate

E of system decreases

Inte

rnal

ene

rgy,

E

Energy lost tosurroundings

EFinal < Einitial

DE < 0 ( negative)

E Final

E Initial

Initialstate

Finalstate

E of system increases

Inte

rnal

ene

rgy,

E

Energy gainedfrom surroundings

E final > E initial

DE > 0 (positive)

E initial

E final

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Internal Energy

The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

Initialstate

Finalstate


Inte

rnal

ene

rgy,

E


EFinal < Einitial

DE < 0 ( negative)

E Final

E Initial

Initialstate

Finalstate

E of system increasesIn

tern

al e

nerg

y, E


E final > E initial

DE > 0 (positive)

E initial

E final

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Internal Energy By definition, the change in internal energy, DE, is the final energy of the system minus the initial energy of the system:

DE = Efinal − Einitial

Initialstate

Finalstate


Inte

rnal

ene

rgy,

E


EFinal < Einitial

DE < 0 ( negative)

E Final

E Initial

Initialstate

Finalstate

E of system increases

Inte

rnal

ene

rgy,

E


E final > E initial

DE > 0 (positive)

E initial

E final

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Changes in Internal Energy· If DE > 0, Efinal > Einitial· Therefore, the system absorbed energy from the surroundings.· This energy change is called endergonic.

· If DE < 0, Efinal < Einitial · Therefore, the system released energy to the surroundings.· This energy change is called exergonic.

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Changes in Internal Energy

· When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).· That is, DE = q + w.

System

ΔE>0

Heat q > 0

Work w > 0

Surroundings

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DE, q, w, and Their Signs

For q + means system gains heat - means system loses heat

For w + means work done on system - means work done by system

For DE + means net gain of energy by system - means net loss of energy by system

Sign Conventions for q, w and DE

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Exchange of Heat between System and Surroundings

System

ΔH>0Endothermic

Surroundings

Heat

System

ΔH<0Exothermic

Surroundings

Heat

When heat is absorbed by the system from the surroundings, the process is endothermic.

When heat is released by the system into the surroundings, the process is exothermic.

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State Functions

The internal energy of a system is independent of the path by which the system achieved that state.

Below, the water could have reached room temperature from either direction...it makes no difference to the final energy of the system if it reached its final temperature by heating up or cooling down.

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State Functions

· Therefore, internal energy is a state function.· It depends only on the present state of the system, not on the path by which the system arrived at that state.· And so, DE depends only on Einitial and Efinal.

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State Functions

· However, q and w are not state functions.· Whether the battery is shorted out or is discharged by running the fan, its DE is the same.· But q and w are different in the two cases.

Charged batteryHeat

Work

Energy lost bybattery

Discharged battery

Heat ΔE

A B

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Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).

P= F/A P= F/A

Initial state Final state

h

#h ΔV

A = cross sectional area

When a gas expands it does work on its surroundings: W = Fd.

In this case: F = PA

and d = #h.

So

W = Fd

W = (PA)#h

W = P#V

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Work done by a gas

Since the gas expands, and does work on its surroundings, the energy of the gas decreases, this is considered negative work.

W = - P#V

If the surroundings compress the gas, decreasing its volume, this increases the energy of the gas, it is considered positive work.

W = + P#V

P= F/A P= F/A

Initial state Final state

h

#h ΔV

A = cross sectional area

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EnthalpyThe energy to

· form a substance from its basic elements,

· to change the phase of a substance,

· or to have a reaction occur

is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance (PV) to occupy.

H = E + PV

The change in enthalpy, #H, is usually the quantity of interest.

#H = #E + #(PV)

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Enthalpy

· If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.· Enthalpy is the internal energy plus the product of pressure and volume:

H = E + PV

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Enthalpy

· When the system changes at constant pressure, the change in enthalpy, DH, is DH = D(E + PV)· This can be written DH = DE + PDV

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Enthalpy

· Since DE = q + w and w = -PDV, we can substitute these into the enthalpy expression: DH = DE + PDV DH = (q+w) − w DH = q· So, at constant pressure, the change in enthalpy is the heat gained or lost.

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Endothermicity and Exothermicity

System

ΔH>0Endothermic

Surroundings

Heat

System

ΔH<0Exothermic

Surroundings

Heat

A process is endothermic when DH is positive.

A process is exothermic when DH is negative.

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Enthalpy of Reaction

The change in enthalpy, DH, is the enthalpy of the products minus the enthalpy of the reactants:

DH = Hproducts − Hreactants

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Enthalpy of Reaction

This quantity, DH, is called the enthalpy of reaction, or the heat of reaction.

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The Truth about Enthalpy

· Enthalpy is an extensive property.· DH for a reaction in the forward direction is equal in size, but opposite in sign, to DH for the reverse reaction.· DH for a reaction depends on the state of the products and the state of the reactants.

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Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure DH through calorimetry, the measurement of heat flow.

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Heat Capacity and Specific Heat

The amount of energy required to raise the temperature of a substance by 1 K (1°C) is its heat capacity.

The amount of energy required to raise the temperature of one gram of a substance by 1 K (1°C) is its specific heat (c).

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Heat Capacity and Specific Heat

Specific heat, then, is

Specific Heat = heat transferredmass x temperature change

c =q

m x ΔT

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Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

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Constant Pressure Calorimetry

Because the specific heat for water is well known (4.184 J/g-K), we can measure DH for the reaction with this equation: q = m x C x DT

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Bomb Calorimetry

· Reactions can be carried out in a sealed “bomb” such as this one.· The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.

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Bomb Calorimetry

· Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, DE, not DH.· For most reactions, the difference is very small.

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Phase Changes

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Energy Changes Associated with Changes of State

Chemical and physical changes are usually accompanied by changes in energy. Recall the following terms:

· When energy is put into the system, the process is called endothermic.

· When energy is released by the system, the process is called exothermic.

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Endothermic processes Exothermic processes

energy is taken into the system from surroundings

energy is released from the system to surroundings

melting a solid freezing a liquid

boiling or evaporating a liquid

condensing a gas

sublimation of a solid deposition of a gas

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1 Which of the following is/are exothermic? I. boiling II. melting III. freezing

A I only

B I and II onlyC III only

D I, II and III

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The heat of fusion (DHfus) is the energy required to change a solid at its melting point to a liquid.

The heat of vaporization (DHvap) is defined as the energy required to change a liquid at its boiling point to a gas.

Note that these quantities are usually per 1.00 mole, whereas q = mCDT involves mass in grams.

But

ane

Die

thyl

eth

er

Wat

er

Mer

cury

10

40

80

Hea

t of f

usio

n an

d va

poriz

atio

n k

J/m

ol

Heat of vaporizationHeat of fusion

24

5

2941

58

2367

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2The heat of vaporization for butane is 24 kJ/mol. How much energy is required to vaporize 2 mol of butane?

A 2kJ

B 12 kJ

C 22 kJ

D 48 kJ

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3The heat of vaporization for butane is 24 kJ/mol. How much energy is required to vaporize 0.33 mol of butane?

A 8 kJ

B 12 kJ

C 16 kJ

D 72 kJ

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4 How much energy is required to melt 0.5 mol of ice?

A 2kJ

B 3 kJ

C 6 kJ

D 12 kJ

Heat of fusion for H2O (s)

Heat of vaporization for H2O (l)

6 kJ/mol 41 kJ/mol

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5 How much energy is released when 3.0 mol of water freezes?

A 2kJ

B 3 kJ

C 18 kJ

D 123 kJ

Heat of fusion for H2O (s)

Heat of vaporization for H2O (l)

6 kJ/mol 41 kJ/mol

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6How much energy is needed to melt 10.0 mol solid Hg?

A 2.3 kJ

B 5.8 kJ

C 23 kJ

D 230 kJ

Heat of fusion for Hg (s)

Heat of vaporization for Hg (l)

23 kJ/mol 58 kJ/mol E 580 kJ

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7How much energy is needed to vaporize 2.0 mol Hg (l)?

A 11.5 kJ

B 29 kJ

C 42 kJ

D 116 kJ

Heat of fusion for Hg (s)

Heat of vaporization for Hg (l)

23 kJ/mol 58 kJ/mol

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Let's examine this heating curve for water one segment at a time.

This graph is called a heating curve. It illustrates how temperature changes over time as constant heat is applied to a pure solid substance.

Tem

pera

ture

(0 C)

Heat added (each division corresponds to 4 kJ)

Liquid water

Ice and liquid water ( melting)

Liquid water and vapor ( vaporization)

Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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From A to B, ice is heating up from -25oC to 0oC.


Liquid water



Water vapor

-25

25

75

0

50

100

Ice

A

B C

D E

F125

Tem

pera

ture

(0 C)

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From B to C, ice is melting.

The added heat is breaking the hydrogen bonds of the solid, so the temperature is constant.

That's why the slope is zero.

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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From C to D, liquid water is heating up from 0oC to 100oC.

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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From D to E, liquid water is boiling into vapor.

The added heat is breaking the IM forces of the liquid, so the temperature is constant.

That's why the slope is zero.

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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From E to F water vapor, steam, is heating up from 100 oC to 125 oC.

Once again, the slope is 1/(mc). But "c" is different for all the phases of a substance, so the slope is different for for solid, liquid and gaseous H2O.

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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Recall that

1 where m = mass mC C = specific heat of the substance

Since slope and specific heat are inversely proportional, a steeper (greater) slope means that the specific heat is low; in other words, a small amount of heat added (small change in x) causes a high rise in temperature.

Conversely, a small or low sloped segment means that the specific heat is high; in other words, a large amount of heat (large change in x) is required to cause a rise in temperature.

slope =

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8 Which segment(s) contain solid sodium chloride?

A AB only

B AB and BC

C AB, BC and CD

D BC, CD and DE

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9 Which segment(s) contain liquid sodium chloride?

A AB only

B AB and BC

C AB, BC and CD

D BC, CD and DE

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10 What is the melting point(in oC) of sodium chloride?

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11 What is the freezing point(in oC) of sodium chloride?

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12 Which is greater:the specific heat of solid NaCl, or the specific heat of molten (liquid) NaCl?

A solid

B liquid

C Cannot be determined.

D They are equal.

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Here is a summary of some of the main features of a heating curve.

Segments AB, CD, EF

Segments BC & DE

· · slope # 0· · T increasing· · KE increasing· · PE is constant

· · slope = 0· · DT = 0· · KE is constant· · PE is increasing

apply q = mC DT apply DHfus or DHvap

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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Specific heat of ice 2.09 J/g-oC

Specific heat of water 4.18 J/g-oC

Specific heat of steam 1.84 J/g-oC

Heat of fusion (DHfus) of water at 0oC 6.01 kJ/mol

Heat of vaporization (DHvap) of water at 100oC 40.7 kJ/mol


Recall that any given substance has a different value for specific heat as a solid, as a liquid and as a gas. Additionally, melting one mole of a substance (DHfus) and vaporizing one mole of the same substance (DHvap) require different amounts of energy. Here is some useful information regarding H2O.

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Calculating Energy Changes from a Heating Curve

Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at -15.0°C to water at 25.0 °C.

It is a good idea to sketch out the segments first on a phase diagram.

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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Segment 3Warming the liquid from 0°C to25°C.

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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q1 = mCDTq1 = (10.0g) (2.09 J/g-oC) (15.0oC)q1 = 313.5 Jq1 = 0.314 kJ or 313.5J


We are now ready to apply either the calorimetry equation or DHfus or DHvap.

Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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q2 = (DHfus ) (# moles)q2 = (6.01 kJ/mol) [(10.0 g)/ 18.0 g/mol)]q2 = 3.34 kJ

or

6.01/18 J/g x 10g = 3.34 kJ= 3340J


Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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q3 = mCDTq3 = (10.0g) (4.18 J/g-oC) (25.0oC)q3 = 1045 Jq3 = 1.05 kJ


Tem

pera

ture

(0 C)


Liquid water



Water vapor

-25

25

75

125

0

50

100

Ice

A

B C

D E

F

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Now, we add the heat changes for all 3 segments. First, make sure that all quantities are in kilojoules.

Total DH = (q1 + q2 + q3) kJ DH = (0.314 + 3.34 + 1.05) kJ DH = 4.6 kJ or 4698.5J


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0 5 10 3520 2515 30 40 45 50t (min)

T (oC)

20

0

40

60

80

100

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13 How many grams of methane must burn to heat 1kg of water from 250C to 900C?

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Hess’s Law

· DH is well known for many reactions, and it is inconvenient to measure DH for every reaction in which we are interested.

· However, we can estimate DH using published DH values and the properties of enthalpy.

Slide 71 / 87

Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps, DH for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Because DH is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

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Reverse the second equation, so diamond is on the right side. This changes the sign of the DH for this process.

1) C(graphite) + O2 ¾® CO2(g) DH = -394 kJ/mol

2) CO2(g) ¾® C(diamond) + O2(g) DH = +396 kJ/mol

Combustion of graphite1) C(graphite) + O2(g) --> CO2(g)

-394 kJ

Combustion of diamond2) C(diamond) + O2(g) --> CO2(g)

-396 kJ

Sample Problem:Using the information below, calculate the DH for the reaction

C(graphite) ¾® C(diamond)

Calculation of DH using Hess's Law

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1) C(graphite) + O2 ¾® CO2(g) DH = -394 kJ/mol

2) CO2(g) ¾® C(diamond) + O2(g) DH = +396 kJ/mol

Now, adding the two equations and their enthalpies together yields

C(graphite) ¾® C(diamond) DH = + 2 kJ/mol

Calculation of DH using Hess's Law

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14 Use Hess's Law to calculate the heat of reaction, DH, for the reaction A ➝ C. A ➝ B #H = +20 kJ B ➝ C #H = +60 kJ

A 20 kJ

B 40 kJ

C 60 kJ

D 80 kJ

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15 Use Hess's Law to calculate the heat of reaction, DH, for the reaction A ¾® C. A ¾® B DH = +25 kJ B ¾® C DH = -100 kJ

A -75 kJ

B + 75 kJ

C +125 kJ

D -125 kJ

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16 Use Hess's Law to calculate the heat of reaction, DH, for the reaction A ¾® C. B ¾® A DH = -40 kJ B ¾® C DH = -95 kJ

A +55 kJ

B -55 kJ

C +135 kJ

D -135 kJ

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Calculation of DH

We can use Hess’s law in this way:

DH = S n DHf°products – S m DHf° reactants

where n and m are the stoichiometric coefficients.

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Enthalpies of Formation

An enthalpy of formation, DHf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

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Standard Enthalpies of Formation

Standard enthalpies of formation, DHf °, are

measured under standard conditions (25°C and 1.00 atm pressure).

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Enthalpies of Formation· Enthalpy of formation, DHf , is a specific kind of DH, just like DHvap or DHfus.

· Enthalpy of formation, DHf , is defined as the enthalpy change for the reaction in which

ONE MOLE OF A COMPOUND IS MADE FROM ITS ELEMENTS IN THEIR PURE FORM.

· Standard enthalpy of formation , DHf°, has the o symbol which indicates the following conditions:

- temperature of 25oC = 298 K - pressure of 1.0 atm - solutions are 1.0 M (molar concentration)

Not STP!

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Standard Enthalpies of Formation for Selected Elements (kJ/mol)

Li (s) 0

Li (g) 159.3

O2 (g) 0

O (g) 247.5

I2 (s) 0

I2 (g) 62.25

Br2 (l) 0

Br2 (g) 30.71

· Note that phase is important for DHf° values; for example, the enthalpy of formation is different for H2O (g) than it is for H2O (l).

· Also note that all pure elements in their normal physical state have enthalpies of formation of zero, DHf° = 0.

Standard Enthalpies of Formation

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17 Which of the following will not have a standard heat of formation of zero?

A gold, Au (s)B potassium, K (s)C mercury, Hg (s)D iodine, I2 (s)

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18 Which of the following will not have a standard heat of formation of zero?

A neon, Ne (g)B carbon, C (g)C chlorine, Cl2 (g)D nitrogen, N2 (g)

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Calculation of DHSample Problem

Calculate the DH for the reaction C3H8(g) + 5O2(g) ¾® 3CO2(g) + 4H2O(l)

DH = S n DHf°products – S m DHf° reactants

where n and m are coefficients

DH = (3 mol x DHf°CO2 + 4 mol x DHf°H2O) - ( DHf°C3H8 + 5 mol x DHf° O2)

DH = (3 x -393.5 + 4 x -285.8 ) - (-103.85 + 5 x 0)

DH = - 2220 kJ

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Calculate the standard enthalpy change for the reactionCaCO3(s) ¾® CaO(s) + CO2(g)

Given the standard enthalpies of formation of CaO(s) = -635.5k kJ/molCO2(g) = -393.5 kJ/molCaCO3(s) = -1207.1 kJ/mol

DH = S n DHf°products - S m DHf° reactants

= [DHf°CaO + DHf°CO2] - [DHf°CaCO3]

=[(1 mol)(-635.5 kJ/mol) + (1 mol)(-393.5 kJ/mol)]- [(1 mol)(-1207.1 kJ/mol)]

= 178.1 kJ

Calculation of DHSlide 86 / 87

Energy in Foods

Most of the fuel in the food we eat comes from carbohydrates and fats.

Beer typically contains 3.5% alcohol

Slide 87 / 87

Energy in Fuels

The vast majority of the energy consumed in this country comes from fossil fuels.

Energy is the ability to do work or transfer...

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