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