Example 5.5. Calculating bond energies from thermodynamic data

Problem:The standard heat of formation of methane, CH₄ is
–74.9 kJ×mol^-1. The H-–H bond energy is 436 kJ×mol^-1; standard enthalpy of atomization for carbon is given in the table 5.1. Calculate the C–H bond energy.

Solution: The equation of methane formation is

We can envision an alternative path from elementary substances to the compound that follows a series of successive reactions:

The heat of methane formation is equal to the sum of the heats reactions (1), (2) and (3).

ΔH_f° = ΔH₁° + ΔH₂° + ΔH₃°

The only unknown quantity is ΔH₃°, which is negative of the atomization energy (negative, because step 3 involves the formation of chemical bonds). Substituting and solving for the atomization energy we obtain

ΔH_at° = ΔH₁° + ΔH₂° – ΔH_f° = 715 + 872 – (–74.9) = 1662 kJ

This quantity is the total amount of energy needed to break all four C–H bonds in 1 mol of CH₄. Division by 4 gives average bond energy of 415 kJ×mol^-1.

Answer: The average C–H bond energy is 415 kJ×mol^-1.

Enthalpy of ionization.To calculate energy of crystalline lattice for ionic solid, enthalpies of ionizationandelectron affinitiesare used.The enthalpy of ionizationis the enthalpy change that accompanies formation of 1 mole of cations from isolated atoms or other cations in a ground state. The electron affinity is the energy that is realized or adsorbed when electron(s) are added to 1 mole of isolated atoms or anions in a ground state.

5.1.3Spontaneous and Nonspontaneous Reactions. Entropy and Gibbs Energy

A spontaneous reactionis a reaction that, once begun, takes place without outside assistance. An example is a reaction between hydrogen and oxygen. Once the mixture is ignited, the reaction to form water continues until all of one reactant is consumed. The reverse reaction, decomposition of water to hydrogen and oxygen, is nonspontaneous. This reaction will occur when an electric current is passed through aqueous solution (containing an electrolyte such as NaOH). Thus decomposition continues as long as the electric current is supplied.

Predicting, whether a given reaction will occur spontaneously or not under a given set of conditions is important for chemists. The reaction between hydrogen and oxygen and many other spontaneous reactions are exothermic. So we can assume that one of the driving forces that determine whether a reaction is spontaneous is a tendency to give off energy. However, there are also rare endothermic spontaneous reactions. Dinitrogen tetroxide decomposes spontaneously at room temperature (the reaction is not complete):

Dissolution is another spontaneous process, which can be endothermic. For example, ammonium nitrate dissolves spontaneously in water, even though energy is absorbed when this reaction takes place.

Therefore, there is another factor that influences whether reaction is spontaneous or not. This factor is entropy.

Entropy.Thermodynamic characteristic, called entropy (denoted as S) is a measure of the disorder of the system[5]. Entropy of a substance depends on its aggregate state:

S_solid < S_liqud << S_gas

For example, for water S_ice < S(H₂O)_liqud (70.1)<< S(H₂O)_gas(188.7)

The entropy of a substance (mixture) increases with increasing temperature. For a gas the entropy increases with volume increase (greater space is favorable for disorder).

The standard entropy change for a reaction can be calculated, using sums of the standard entropies of all products and reactants. Each S° should be multiplied by the corresponding coefficient in the equation:

DS°_r = ΣS°(products) – ΣS°(reactants).

Predicting the sign of entropy change for chemical reaction. For some chemical reactions, it is possible to predict the sign of DS. For example, let us consider the reaction

2 Cu(NO₃)_2(s) ® 2 CuO_(s) + 4 NO_2(g) + O_2(g)

which occur when copper nitrate is heated. The reactant is a solid, and, as we know, solids have low entropies. Among the products there is one solid, but there are also two gases. The gases have much larger entropies that any of the solids. Therefore, for this reaction DS > 0.

When gases are present among both the reactants and products, we can compare amounts of matter of gases before and after the reaction. If the total amount of matter of gases increased after the reaction, DS > 0. For example, DS is positive for decomposition of N₂O_4(g) (NO_2(g) is formed). Generally, decrease of the complexity of the particles in the system (for example, dissociation of diatomic molecule like H₂, Cl₂ or HI) lead to increase of entropy. Dissolution of solids also results in great increase of entropy.

Date: 2015-12-11; view: 937