of Constant Heat Summation

If a reaction takes place in many steps then its standard reaction enthalpy is the sum of the standard enthalpies of intermediate reactions into which the overall reaction may be divided at the same temperature.

If Δ_{r}H_{1}, Δ_{r}H_{2}, Δ_{r}H_{3} …………….represent enthalpies of reactions which result in same products as Δ_{r}H

Let us assume that a reaction A → B along one route has enthalpy Δ_{r}H

Formation of same product may happen along another route which is comprised of several reactions with enthalpies Δ_{r}H_{1}, Δ_{r}H_{2}, Δ_{r}H_{3} ……………. Then, we have

Δ_{r}H = Δ_{r}H_{1} + Δ_{r}H_{2} + Δ_{r}H_{3} …………….

It can be represented by following figure:

**Enthalpy of Combustion:** Enthalpy change per mole (or per unit amount) of a substance, when the substance undergoes combustion and all the reactants and products are in their standard states at the specified temperature, is called enthalpy of combustion.

**Enthalpy of Atomization:** The enthalpy change on breaking one mole of bonds completely to obtain atoms in gas phase is called enthalpy of atomization. In case, of diatomic molecules, enthalpy of atomization is also the bond dissociation enthalpy.

**Bond Enthalpy:** Enthalpy during making or breaking of a chemical bond is called bond enthalpy. There are two different terms related to enthalpy changes associated with chemical bonds: (a) Bond dissociation enthalpy and (b) Mean bond enthalpy.

**Diatomic Molecules:** In this case, bond dissociation enthalpy is about the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase. It is important to note that bond dissociation enthalpy is same as enthalpy of atomization; in case of diatomic molecule. But in case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule.

**Polyatomic Molecules:** Let us take example of methane to understand bond dissociation enthalpy. All four C-H bonds in methane are identical in bond length and energy. But, energies required to break individual C-H bonds in each successive step differ.

CH_{4}(g) → CH_{3}(g) + H(g);Δ_{bond}H^{Θ} = =427 kJ mol^{-1}

CH_{3}(g) → CH_{2}(g) + H(g);Δ_{bond}H^{Θ} = =439 kJ mol^{-1}

CH_{2}(g) → CH (g) + H(g);Δ_{bond}H^{Θ} = =452 kJ mol^{-1}

CH (g) → C (g) + H(g);Δ_{bond}H^{Θ} = =374 kJ mol^{-1}

Enthalpy of reaction = 1665 kJ mol^{-1}

In such cases, we use mean bond enthalpy of C-H bond.

Mean = 1665 ÷ 4 = 416 kJ mol^{-1}

**Lattice Enthalpy:** The enthalpy change associated with one mole of an ionic compound’s dissociation into its ions in gaseous state is called lattice enthalpy. It is impossible to directly determine lattice enthalpies so indirect method is used for the purpose. For this, Born-Haber Cycle is constructed which shows various steps of dissociation of lattice into ions. Based on this cycle, enthalpies of different steps are found and added to get lattice enthalpy.

**Enthalpy of Solution:** Enthalpy change accompanying the dissolution of one mole of a substance in a specified amount of solvent is called enthalpy of solution.

A process which has the ability to proceed on its own, without any external assistance is called spontaneous process. Spontaneous process is irreversible and may only be reversed by an external agency.

Decrease in enthalpy may be a contributory factor to spontaneity but is not true for all cases.

In an isolated system, there is always a tendency for the system's energy to become more disordered or chaotic and this could be a criterion for spontaneous change.

Entropy of an isolated system increases with increase in disorder in the system. With reference to a chemical reaction, this entropy change can be attributed to rearrangement of atoms or ions from one pattern in the reactants to another in the products. Change in entropy during a chemical reaction may be estimated qualitatively by considering the increased randomness of structures of species.

Entropy can be measured quantitatively through statistical methods. But in this chapter, we will quantify entropy by relating it to the heat evolved. When heat is added to the system, there is an increase in molecular motions; which eventually increases randomness in the system. Heat added to a system at lower temperature causes greater randomness than same heat added to it at higher temperature. So, we can say that entropy change is inversely proportional to temperature.

Following equation gives the relation between ΔS (entropy change), q and T.

`ΔS=(q_(re\v))/T`

When a system is in equilibrium, its entropy is maximum and change in entropy ΔS = 0.

Gibbs energy is given by following equation:

G = H - TS

The change in Gibbs energy for the system can be written as follows:

ΔG_{sys} = ΔH_{sys} - TΔS_{sys} - S_{sys}ΔT

ΔT = 0 at constant temperature.

Or, ΔG_{sys} = ΔH_{sys} - TΔS_{sys}

A simple way to write the above equation is as follows:

ΔG = ΔH – TΔS

Thus, Gibbs energy change = enthalpy change – temperature × entropy change

We know: ΔS_{total} = ΔS_{sys} + ΔS_{surr}

So, increase in enthalpy of surrounding is equal to decrease in enthalpy of system. So, entropy change of surroundings can be given by following equation.

ΔS_{surr} `=(ΔH_(su\rr))/T=-(ΔH_(sy\s))/T`

ΔS_{total} = ΔS_{sys} + `(-(ΔH_(sy\s))/T)`

Or, TΔS_{sys} = TΔS_{sys} - ΔH_{sys}

For spontaneous process ΔS_{total} > 0, so

TΔS_{sys} - ΔH_{sys} > 0

Or, -(ΔH_{sys} - TΔS_{sys}) > 0

We have earlier seen that ΔG = ΔH – TδS

So, we can write -ΔG > 0

Or, ΔG = ΔH – TδS = 0

If ΔG < 0, the process is spontaneous

If ΔG > 0, the process is non-spontaneous

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. Isolated systems spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy.

The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero.

A + B ⇌ C + D is Δ_{r} = 0

For a reaction in which all reactants and products are in standard state, Gibbs energy is related to the equilibrium constant as follows:

0 = Δ_{r}G^{Θ} + RT ln

Or, Δ_{r}G^{Θ} = - RT ln K

Or, Δ_{r}G^{Θ} = - 2.303 RT log K

We also know that

Δ_{r}G^{Θ} = Δ_{r}H^{Θ} - TΔ_{r}S^{Θ} = - RT ln K

For strongly endothermic reactions, value of ΔH may be large and positive but value of K will be much smaller than 1. Reaction is unlikely to form much product in this case.

In case of exothermic reactions, ΔH is negative and ΔG is likely to be large and negative. In this case, K will be much larger than 1. There is likelihood of strongly exothermic reaction.

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