Chapter 19

Part 4: Predicting reactions & the Third Law of Thermodynamics

Is this reaction likely to have a positive or negative change in

entropy?

Entropy and States of Matter

Spontaneity and Changes in State

The melting of ice is disfavored by enthalpy (+ΔH) but favored by entropy (+ ΔS). The freezing of water is favored by enthalpy (- ΔH) but disfavored by entropy (- ΔS). Below 0°C, the enthalpy term ΔH dominates the entropy term TΔS in the Gibbs free-energy equation, so freezing is spontaneous. Above 0°C, the entropy term dominates the enthalpy term, so melting is spontaneous. At 0°C, the entropy and enthalpy terms are in balance.

• ΔH – ΔS + ΔG =?

• ΔH – ΔS – ΔG =?

• ΔH + ΔS + ΔG =?

• ΔH + ΔS – ΔG =?

What are the signs of each of the thermodynamic terms for the non-

spontaneous process represented below?

Answer

• Non-spontaneous

ΔG = +• From the picture

ΔS = +

• Therefore ΔH = +

This is an exothermic reaction. Under what conditions is this

spontaneous?

Answer

• ΔH = —• ΔS = —• ΔG= ΔH - T ΔS

• Therefore when T is small (low temperature) the reaction is spontaneous.

At what temperature is the following spontaneous?

• Br2 (l) Br2 (g)

• ΔH°= 31.0 kJ/mol

• ΔS°= 93.0 J/(Kmol)

• What is the normal boiling point of bromide?

Well?

• At ΔG° = - all rxn’s are spontaneous.

• While ΔS° favors the rxn, ΔH° favors the opposite rxn, (why?) Where these opposite tendencies balance is the is the boiling point of bromide.

• To find this balance and the boiling point, set ΔG° to zero, equilibrium, and solve.

Answer

• ΔG°= ΔH° - T ΔS°

• 0= ΔH° - T ΔS°

• ΔH° = T ΔS° T= ΔH° / ΔS°

• T= 3.10 x 104 J/mol = 333K

93.0 J/(Kmol)

Summary

• At T > 333 K : ΔS° controls the rxn

• At T<333 K : ΔH° controls the rxn. The reaction is spontaneous in the direction that is exothermic.

• At T = 333 K: the reaction is at equilibrium, vapor and liquid coexist.

The Haber-Bosch process

• This is the reaction of nitrogen and hydrogen to produce ammonia.

• The nitrogen (N) and hydrogen (H) are reacted over an Fe catalyst under conditions of 200 atm, 450-500°C; resulting in a yield of 10-20%:

N2(g) +3H2(g) 2NH⇌ 3(g)

ΔH -92.4kJ/mol

Haber Process

• This process reduces the entropy of the system. ΔS = +

• This process is exothermic, but it has a high activation energy.

Third Law of Thermodynamics

• states that the entropy of all crystalline solids approaches zero as their temperature approaches absolute zero. In other words, all substances lose their energy at absolute zero.

From the 3rd Law, we know ΔS is temperature dependent.

• Standard entropy values may be found in tables such as the back of your text.

• Entropy is also a state function and may be calculated for any reaction using the function:

• ΔS°rxn= nΔS°products - nΔS°reactants