general reaction: enthalpy change consider the reaction a a + b b c c + d d = c (c) + d (d) - a (a)...
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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
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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)
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
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We can find (CO2) if (C(graphite)) and
(O2) can be determined.
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
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We can find (CO2) if (C(graphite)) and
(O2) can be determined.
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
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Therefore, (C(graphite)) = 0
(O2) = 0
(both results at 1 bar, 25oC).
Hence, we may write:
= (CO2)
0f
H
0f
H
0H 0f
H
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Therefore, (C(graphite)) = 0
(O2) = 0
(both results at 1 bar, 25oC).
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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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. 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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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. 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 }
= {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 . 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
Suppose we wish to determine . In principle, this quantity can be obtained by measuring the heat given off by the following reaction
C(graphite) + ½ O2(g) CO(g)
)(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
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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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
If we reverse reaction (a) and add (b)
CO2(g) CO(g) + ½ O2(g) = 283.0 kJ
0H
0H
0H
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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
If we reverse reaction (a) and add (b)
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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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
If we reverse reaction (a) and add (b)
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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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
If we reverse reaction (a) and add (b)
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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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
If we reverse reaction (a) and add (b)
CO2(g) CO(g) + ½ O2(g) = 283.0 kJ
C(graphite) + O2(g) CO2(g) = -393.5 kJ
C(graphite) + ½ O2(g) CO(g) = -110.5 kJ
0H
0H
0H0H0rxnH
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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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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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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.
Tm
qc
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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.
or (The symbol s is also used in place of c).
Tmqc
Tmcq
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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.
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
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
Heat changes are measured in calorimetry. There are two principal types of heat that can be
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
Heat changes are measured in calorimetry. There are two principal types of heat that can be
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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Which quantity is measured, qp or qV, will depend on the type of calorimeter used to make the experimental measurements.
A device which is open to the atmosphere will give qp, i.e. the enthalpy change.
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Which quantity is measured, qp or qV, will depend on the type of calorimeter used to make the experimental measurements.
A device which is open to the atmosphere will give qp, i.e. the enthalpy change.
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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Which quantity is measured, qp or qV, will depend on the type of calorimeter used to make the experimental measurements.
A device which is open to the atmosphere will give qp, i.e. the enthalpy change.
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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Calorimetry Heats of combustion are usually measured in a
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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Calorimetry Heats of combustion are usually measured in a
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
Therefore: heat gained = 2.000 x 104 J - 0.836 X 104 J by calorimeter = 1.164 x 104 J
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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 JThe heat gained by the calorimeter:
= 3.36 x 103 JoC-1 x 2.05 oC = 6.89 x 103 J
Tcmq
TCq
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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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The total heat gained by the calorimeter + water contents = 4.28 x 103 J + 6.89 x 103 J
= 1.117 x 104 J So we can write the following equivalence statement: 0.569 g benzoic acid 1.117 x 104 J