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ERT 108/3 PHYSICAL CHEMISTRY FIRST LAW OF THERMODYNAMICS Prepared by: Pn. Hairul Nazirah Abdul...
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Transcript of ERT 108/3 PHYSICAL CHEMISTRY FIRST LAW OF THERMODYNAMICS Prepared by: Pn. Hairul Nazirah Abdul...
ERT 108/3PHYSICAL CHEMISTRY
FIRST LAW OF THERMODYNAMICS
Prepared by:Pn. Hairul Nazirah Abdul Halim
• Thermochemistry is the study of heat produced or required by chemical reactions.
• If we know the ΔU and ΔH for a reaction, we can predict the heat of reaction.
• Enthalpy of a system decreases when heat is releases. ΔH < 0. (Exothermic process).
• Enthalpy of a system increases when heat is absorbed. ΔH > 0. (Endothermic process).
Thermochemistry
• Standard enthalpy change, is the change in enthalpy for a process in which the initial and final substances are in their standard states.
Standard Enthalpy Changes
• Standard states of a substance at a specified temperature is its pure form at 1 bar.
• For example;
- the standard state of liquid ethanol at 298K is pure liquid ethanol at 298K and 1 bar.
• Example of standard enthalpy change, ;
the standard enthalpy of vaporization, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar.
a) Enthalpy of Physical Change
• The standard enthalpy transition, is the standard enthalpy change that accompanies a change of physical state.
• Example, - standard enthalpy of vaporization,
- standard enthalpy of fusion,
• Change of enthalpy is independent of the path between the two states.
• For example, the direct conversion from solid to vapor;
• Or as two steps, first fusion (melting) and then vaporization;
• Conclusion;
• The standard enthalpy changes of a forward process and its reverse must differ only in sign.
• For example, at 298 K;
- enthalpy of vaporization of water is +44 kJ/mol
- enthalpy of condensation is -44 kJ/mol
b) Enthalpy of Chemical Change
Thermochemical equation is a combination of a chemical equation and the corresponding change in standard enthalpy:
= the standard enthalpy change
• Alternatively, we can write the chemical equation with standard reaction enthalpy,
• For combustion of methane, we write;
• The standard reaction enthalpy is;
• Where, v = stoichiometric coefficients
= the standard molar enthalpy of species J at the temp. of interest.
• For the reaction;
• The standard reaction enthalpy is;
• Where is the standard molar enthalpy of species J at the temp. of interest.
• Standard enthalpy of combustion, is the standard reaction enthalpy for the complete oxidation of an organic compound to CO2 gas and liquid H2O.
• Organic compound contains C, H, O and N2 (if N is present).
• Example, the combustion of glucose;
c) Hess’s Law
• Standard enthalpies of individual reactions can be combined to obtain the enthalpy of another reaction.
• This application of the First Law is called Hess’s Law:
The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided.
Example Using Hess’s Law
The standard reaction enthalpy for the hydrogenation of propene is -124 kJ mol-1,
The standard reaction enthalpy for the combustion of
Propane is -2220 kJ mol-1;
Calculate the standard enthalpy of combustion of propene.
Solution
• Add or subtract the reactions given, together with any others needed, so as to reproduce the reaction required.
• Then add or subtract the reaction enthalpies in the same way. Additional data are in Table 2.5.
• The combustion reaction we require is;
• This reaction can be recreated from the following sum:
Standard Enthalpy of Formation
• The standard enthalpy of formation, of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states.
• The reference state of an element is its most stable state at the specified temp. and 1 bar.
• For example, at 298K the reference state of nitrogen is a gas of N2 molecules, mercury is liquid mercury, carbon is graphite.
a) The reaction enthalpy in terms of enthalpies of formation
Example
The standard reaction enthalpy of:
is calculated as follows:
-
The temperature dependence of reaction enthalpies• Standard reaction enthalpies at different
temperatures may be estimated from heat capacities and the reaction enthalpy.
• Kirchhoff’s Law;
is the difference of the molar heat capacities of products and reactants under standard conditions weighted by the stoichiometric numbers;
Example Using Kirchhoff’s Law
The standard enthalpy of formation of gaseous H20 at 298 K is -241.82 kJ mol-1. Estimate its value at 100°C given the following values of the molar heat capacities at constant pressure:
H20(g): 33.58 J K-1mol-1 ;
H2(g): 28.84 J K -1mol-1;
02 (g): 29.37 J K-1 mol-1. Assume that the heat capacities are independent of temperature.
Solution
• When is independent of temperature in the range T1 to T2,. Therefore,
• To proceed, write the chemical equation, identify the stoichiometric coefficients, and calculate
from the data.• The reaction is;
• So;
Example
Calculate for the following reaction at 1450K and 1 bar;
½ H2 (g) + ½ Cl2(g) HCl(g)
Given that (HCl,g) = -92.3 kJ/mol at 298.15K and that;
over this temperature range.
1-1-2
27-30
,
1-1-2
27-3
20
,
1-1-2
27-3
20
,
molJK K
10 x 464.15K
10 x 809.1165.28),HCl (
molJK K
10 x 373.40K
10 x 143.10695.31),Cl (
molJK K
10 x 111.20K
10 x 8363.0064.29),H (
TTgC
TTgC
TTgC
mp
mp
mp
Solution
Use the following equation;
T1 = 298.15 K
T2 = 1450 K
2
27-3
2
27-3
2
27-3
0,
K 10 x 373.40
K 10 x 143.10695.31
2
1
K 10 x 111.20
K 10 x 8363.0064.29
2
1
K 10 x 464.15
K 10 x 809.1165.28
)(
TT
TT
TT
TC mpr
1-1-2
27-30
, molK J K
10 x 595.25K
10 x 844.2215.2)(
TT
TC mpr
1-1450
15.298 2
27-3
-10
mol J K
d K
10 x 595.25K
10 x 844.2215.2
mol kJ 3.92K 1450at
TTT
Hr
3
37-
2
23
-10
K 3
10 x 595.25
K 2
10 x 844.2
215.2mol kJ 3.92K 1450at
TT
THr
1-1-0
337-
223
0
mol kJ 1.95mol kJ )836.23.92(K 1450at
)15.2981450(3
10 x 595.25
)15.2981450(2
10 x 844.2)15.2981450(215.2K 1450at
H
H
r
r