General reaction: Enthalpy change Consider the reaction a A + b B c C + d D = c (C) + d (D) - a (A) - b (B) 61.

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General reaction: Enthalpy change

Consider the reaction

a A + b B c C + d D

= c (C) + d (D) - a (A) - b (B)

0H 0f

H0f

H0f

H0f

H

2

General reaction: Enthalpy change

Consider the reaction

a A + b B c C + d D

= c (C) + d (D) - a (A) - b (B)

For the case of CO2, the result would read

= (CO2) - { (C(graphite)) + (O2)}

can be measured by experiment.

0H 0f

H0f

H0f

H0f

H

0H 0f

H0f

H0f

H0H

3

We can find (CO2) if (C(graphite)) and

(O2) can be determined.

0f

H

0f

H0f

H

4



Unfortunately, there is no way to determine (C(graphite)) and (O2). In fact, there is no

way to determine for any element.

0f

H

0f

H0f

H

0f

H0f

H0f

H

5



Unfortunately, there is no way to determine (C(graphite)) and (O2). In fact, there is no

way to determine for any element.

A simple way out of the difficulty is to arbitrarily define the standard enthalpy of formation of any element in its stable form as zero.

0f

H

0f

H0f

H

0f

H0f

H0f

H

6

Therefore, (C(graphite)) = 0

(O2) = 0

(both results at 1 bar, 25oC).

0f

H

0f

H

7


(O2) = 0


Hence, we may write:

= (CO2)

0f

H

0f

H

0H 0f

H

8


(O2) = 0


Hence, we may write:

= (CO2)

Tables of are put together in this way.

0f

H

0f

H

0H 0f

H

0f

H

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Example: The metabolism of glucose in our bodies can be represented by the equation:

C6H12O6 + 6 O2(g) 6 CO2(g) + 6 H2O(l).

Calculate the standard enthalpy for the reaction.

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C6H12O6 + 6 O2(g) 6 CO2(g) + 6 H2O(l).

Calculate the standard enthalpy for the reaction. The standard enthalpies of formation of C6H12O6(s), CO2(g), and H2O(l), are -1274.5 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.

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C6H12O6 + 6 O2(g) 6 CO2(g) + 6 H2O(l).

Calculate the standard enthalpy for the reaction. The standard enthalpies of formation of C6H12O6(s), CO2(g), and H2O(l), are -1274.5 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.

Note, by definition (O2) = 0.0f

H

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Hence: = {6 + 6 } -{ + 6 }

0H )(COH 2(g)0f

)O(HH (l)20f

)OH(CH 6(s)126

0f

)(OH 2(g)0f

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Hence: = {6 + 6 } -{ + 6 }

= {6 mol x (-393.5 kJ/mol) + 6 mol x (-285.84 kJ/mol)} - { 1 mol (-1273.3 kJ/mol) }

= -2803 kJ

0H )(COH 2(g)0f

)O(HH (l)20f

)OH(CH 6(s)126

0f

)(OH 2(g)0f

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Hess’s Law

Suppose we wish to determine . )(COH (g)0f

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Hess’s Law

Suppose we wish to determine . In principle, this quantity can be obtained by measuring the heat given off by the following reaction

)(COH (g)0f

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Hess’s Law


C(graphite) + ½ O2(g) CO(g)

)(COH (g)0f

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Hess’s Law


C(graphite) + ½ O2(g) CO(g)

What is the major problem with this approach?

)(COH (g)0f

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Can avoid the problem in the following manner: (a) CO(g) + ½ O2(g) CO2(g) = -283.0 kJ

(b) C(graphite) + O2(g) CO2(g) = -393.5 kJ

0H

0H

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If we reverse reaction (a) and add (b)

CO2(g) CO(g) + ½ O2(g) = 283.0 kJ

0H

0H

0H

20




CO2(g) CO(g) + ½ O2(g) = 283.0 kJ

C(graphite) + O2(g) CO2(g) = -393.5 kJ

0H

0H

0H0H

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CO2(g) CO(g) + ½ O2(g) = 283.0 kJ


0H

0H

0H0H

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CO2(g) CO(g) + ½ O2(g) = 283.0 kJ

C(graphite) + ½ O2(g) CO2(g) = -393.5 kJ

0H

0H

0H0H

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CO2(g) CO(g) + ½ O2(g) = 283.0 kJ


C(graphite) + ½ O2(g) CO(g) = -110.5 kJ

0H

0H

0H0H0rxnH

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Hence

= = -110.5 kJ/mol)(COH (g)0f

0rxnH

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The preceding is an illustration of Hess’s Law.

Hess’s Law: The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps.

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Hess’s Law energy diagram

kJ393ΔHtotal

kJ283ΔH

kJ110ΔH

enth

alpy

2CO

2OCO 2/1

2OC

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Specific Heat (Specific Heat Capacity)

Specific heat: The specific heat of a substance is the amount of heat required to raise the temperature of 1 g of that substance by 1 oC.

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Tm

qc

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or (The symbol s is also used in place of c).

Tmqc

Tmcq

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or (The symbol s is also used in place of c). A common (non SI) unit is . At 25 oC the value is

0.998 cal g-1 oC-1 for water. (i.e. approximately 1 cal g-1 oC-1).

Tmqc

Tmcq

Cgcal

o

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Heat Capacity Heat Capacity: This is defined by

TqC

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Water has a relatively high heat capacity.

TqC

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Water has a relatively high heat capacity. The units of C are cal oC-1.

The conversion to SI units is

1 cal oC-1 = 4.184 J K-1

TqC

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The human body has a high percentage (by mass) of water, and thus, a relatively high heat capacity.

CqT

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The human body has a high percentage (by mass) of water, and thus, a relatively high heat capacity.

If C is high, then for modest changes in heat q, the change in temperature of the body is small. Humans are sensitive to even small changes in body temperature.

CqT

T

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Calorimetry

Heat changes are measured in calorimetry. There are two principal types of heat that can be

measured.

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Calorimetry


measured.(1) Heat changes measured under constant pressure

conditions. Symbol qp This quantity is equal to the enthalpy change: pqΔH

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Calorimetry


measured.(1) Heat changes measured under constant pressure

conditions. Symbol qp This quantity is equal to the enthalpy change:

(2) Heat changes measured under constant volume conditions. Symbol qV This quantity is equal to the change in the internal energy:

pqΔH

VqΔE

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Which quantity is measured, qp or qV, will depend on the type of calorimeter used to make the experimental measurements.

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A device which is open to the atmosphere will give qp, i.e. the enthalpy change.

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A device that is a sealed container, that is one having a fixed volume, will yield qV, i.e. the internal energy change.

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A device that is a sealed container, that is one having a fixed volume, will yield qV, i.e. the internal energy change.

The enthalpy and the internal energy are connected by the formula:

H = E + p V

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44

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Calorimetry Heats of combustion are usually measured in a

calorimeter.

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calorimeter. A known mass of the sample is placed in a steel

container which is filled under pressure with oxygen gas. The sample is ignited electrically, and the temperature rise in the surroundings indicates the amount of heat given off.

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calorimeter. A known mass of the sample is placed in a steel

container which is filled under pressure with oxygen gas. The sample is ignited electrically, and the temperature rise in the surroundings indicates the amount of heat given off.

The surroundings are a known mass of water plus the actual calorimeter device. The assumption is usually made that no heat is lost to the outside world.

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The heat capacity of the calorimeter is determined using a compound for which the heat of combustion is accurately known.

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Example: If a given amount of a standard compound releases 2.000 x 104 J, and that raises the temperature of 1000.0 g of water in the calorimeter by 2.00 oC, determine the heat taken up by the calorimeter. For water, c = 4.18 Jg-1 oC-1.

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Example: If a given amount of a standard compound releases 2.000 x 104 J, and that raises the temperature of 1000.0 g of water in the calorimeter by 2.00 oC, determine the heat taken up by the calorimeter. For water, c = 4.18 Jg-1 oC-1.

= (1000.0 g) (4.18 Jg-1 oC-1)(2.00 oC) = 8.36 kJ

Tcmq OH2

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Now: heat released = heat gained + heat gained by sample by water by calorimeter

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Now: heat released = heat gained + heat gained by sample by water by calorimeter

Therefore: heat gained = 2.000 x 104 J - 0.836 X 104 J by calorimeter = 1.164 x 104 J

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The heat capacity of the calorimeter is:

1.164 x 104 J = 2.00 oC

= 5.82 x 103 J oC-1

TqC

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Sample Calculation A 0.569 g sample of benzoic acid is burned in a

constant volume bomb calorimeter. The heat capacity of the calorimeter is 3.36 x 103 J/oC. If the temperature rise registered in the 500.0 g of water in the calorimeter is 2.05 oC, calculate the heat given off by the benzoic acid in kJ/mol.

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Sample Calculation A 0.569 g sample of benzoic acid is burned in a

constant volume bomb calorimeter. The heat capacity of the calorimeter is 3.36 x 103 J/oC. If the temperature rise registered in the 500.0 g of water in the calorimeter is 2.05 oC, calculate the heat given off by the benzoic acid in kJ/mol.

The chemical reaction is :

C6H5COOH + 15/2 O2 7 CO2 + 3 H2O

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The heat gained by the water: = 500.0 g x 4.18 Jg-1 oC-1 x 2.05 oC = 4.28 x 103 J

Tcmq

57

The heat gained by the water: = 500.0 g x 4.18 Jg-1 oC-1 x 2.05 oC = 4.28 x 103 JThe heat gained by the calorimeter:

= 3.36 x 103 JoC-1 x 2.05 oC = 6.89 x 103 J

Tcmq

TCq

58

The total heat gained by the calorimeter + water contents = 4.28 x 103 J + 6.89 x 103 J

= 1.117 x 104 J

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= 1.117 x 104 J So we can write the following equivalence statement: 0.569 g benzoic acid 1.117 x 104 J

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= 1.117 x 104 J So we can write the following equivalence statement: 0.569 g benzoic acid 1.117 x 104 JThe molar mass of benzoic acid = 122 g/mol

General reaction: Enthalpy change Consider the reaction a A + b B c C + d D = c (C) + d (D) - a (A) - b (B) 61.

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