Thermodynamic Stability As A Function Of Enthalpy Of Formation

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Thermodynamic Stability As A Function Of Enthalpy Of Formation

*Synthesis* and *decomposition* reactions are two important chemical reactions. For example, the combustion of carbon (as coal) and the decomposition of the carbon dioxide that is the product of the combustion are synthesis reaction and decomposition reaction respectively (figure 11.2).

In general, a systhesis reaction has the following form:
*element or compound* + *element or compound* -------> *compound*

In general, a decomposition reaction has the following form:
*compound* -------> *two or more elements or compounds*
*Thermodynamic stability* is the non-spontaneity of the decomposition of the product of a synthesis chemical reaction.

(a)What is *enthalpy of formation*?

(b) Show that the carbon dioxide that is the product of the combustion of carbon (as coal) is thermodynamically stable.

**The strings**:

S_{7}P_{7}A_{71} (Static Equilibrium)
**The math**:

Pj Problem of Interest is of type *equilibrium*. In Chemical reactions, there is a distinction between, *thermodynamic stability*, *kinetic stability* (characterized by imperceptible rate of change eventhough change is occuring), and *dynamic stability associated with reversible reactions*. *Thermodynamic stability* refers to the non-spontaneity of the decomposition of a substance at room temperature. It is in this sense that it is *static equilibrium*.

(a)Every system has some internal energy. In chemical reactions, the internal energy is the sum of the bond energy associated with chemical bonds and the kinetic energy of molecular species.

Let *U* denote the *internal energy* of an arbitrary system. Then, a change in the internal energy of the system occurs only if:

(i) heat flows out or into the system

(ii) work is done by the system or on the system.

Let ΔU denote this change then,

ΔU = q + w ----(1)

Where q represents heat absorbed by the system and w represents work done on the system.

In equation (1), *q* has a positive sign if heat flows into the system and a negative sign if heat flows out of the system. Similarly, *w* has a positive sign if work is done on the system and a negative sign if work is done by the system. There are various types of work that can be done on the system or by the system. For example, electric, mechanical, magnetic, pressure-volume, etc.

Assuming pressure-volume work, equation (1) becomes:

ΔU = q + PΔV ; where P is pressure and ΔV is change in volume.

So, q_{p} = ΔU - PΔV-------(2)

The subscript *p* indicates a constant pressure process.

In the pressure-volume type of work, work done by the system = -PΔV. When we substitute this expression in (2), we have:

q_{p} = ΔU + PΔV = Δ(U + PV)----(3).

The quantity (U + PV) is called *Enthalpy* and is denoted by *H*.

Thus, q_{p} = ΔH = H_{2} - H_{1} ----(4)

When ΔH is < 0, reaction is *exothermic* (heat is produced). Exothermic reactions are more likely to be spontaneous.

When ΔH is > 0, reaction is *endothermic* (heat is absorbed).

The standard of reference for enthalpy measurement is 298 Kelvin (25^{o}C) and 100.000 KPascal. *Free elements* (elements not in a compound) are assumed to have zero enthalpy.

The net amount of energy produced or absorbed when a mole of a compound is formed from its *free elements*, is called *Enthalpy of Formation*.

The symbol for *enthalpy of formation* is ΔH_{f}^{o}. The superscript "o" indicates that the value given are at 25^{0}C and 100.000Kpa. The subscript "f" indicates that the enthalpy is *enthalpy of formation*

(b) The synthesis reaction that produces carbon dioxide from the combustion of coal is as follows:

C_{(cr)} + O_{(g)} ---------> CO_{2(g)} + energy (393.5 KJ) ----(5)

Equation 5 says that one mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide and 393.5 kJoule of energy.

So, the enthalpy of formation of CO_{2(g)} = -393.5 kJ.

So, the decomposition of CO_{2(g)} will require 393.5 kJ of energy

Therefore, CO_{2(g)} cannot decompose spontaneously

So, CO_{2(g)} is thermodynamically stable.

The *enthalpy of reaction* is the enthalpy change for a reaction and is the sum of enthalpy of formation of products minus the sum of enthalpy of formation of reactancts at 298 Kelvin (25^{0}C) and 100.000 KPascal. In mathematical terms:

ΔH_{r}^{o} = ΣΔH_{f(products)}^{o} - ΣΔH_{f(reactants)}^{o} --------(6).

So, if the *enthalpy of formation* of each reactant and product is known (this information is available for many reactants and products), one can calculate the amount of energy produced or absorbed and as a result, the *thermodynamic stability* of the product of a synthesis reaction and in general, whether a reaction will be exothermic or endothermic. For example, the decomposition of one mole of mecury fulminate produces 268 kJ, hence it is explosive.

Some chemical reactions consists of series of intermediate reactions. In such cases, *the enthalpy change for the reaction is the sum of the enthalpy changes for the series of reactions*. This is known as *Hess's Law*.

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