\( \newcommand{\xrightleftharpoons}[2]{\overset{#1}{\underset{#2}{\rightleftharpoons}}} \) \( \newcommand{\conc}[1]{\left[\mathrm{#1}\right]} \) \( \newcommand{\chem}[1]{\mathrm{#1}} \)

Chapter 16

Thermodynamics

Shaun Williams, PhD

Thermodynamics vs. Kinetics

  • Domain of Kinetics
    • Rate of a reaction depends on the pathway from reactants to products.
  • Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products.

The energies of the reactants and products constitutes the thermodynamics domain. The energy profile from reactant to product constitutes the kinetic domain.

Spontaneous Processes and Entropy

Concept Check

Consider \(2.4\,\mathrm{moles}\) of a gas contained in a \(4.0\,\mathrm{L}\) bulb at a constant temperature of \(32^\circ \mathrm{C}\). This bulb is connected by a valve to an evacuated \(20.0\,\mathrm{L}\) bulb. Assume the temperature is constant.

What should happen to the gas when you open the value?

The gas should spread evenly throughout the two bulbs.

The Expansion of an Ideal Gas Into an Evacuated Bulb

Two bulbs separated by a closed valve. Initally all the gas in contained in the left bulb and the right bulb is evacuated. When the valve is opened, the gas evenly distributes throughout both bulbs.

Entropy

The Microstates That Give a Particular Arrangement (State)

The arrangement of 4 particles (each labeled) in the two bulbs with opened valve. There is one way to arrange them so that they are all in one bulb (one microstate). There are four ways to arrangement them so that 3 are in one bulb and one in the other (4 micrstate). Ther are 6 ways to arrange them so that two are in one bulb and two are in the other.

Positional Entropy

The solid is highly ordered so it has a very low entropy. The liquid which is less ordered but still compact has a higher entropy. The gas is very disordered and spread out so it has a much higher entropy.

Concept Check

Predict the sign of \(\Delta S\) for each of the following:

Second Law of Thermodynamics

\(\Delta S_{surr}\)

Interplay of \(\Delta S_{sys}\) and \(\Delta S_{surr}\) in Determining the Sign of \(\Delta S_{univ}\)

Signs of Entropy Changes
\(\Delta S_{sys}\) \(\Delta S_{surr}\) \(\Delta S_{univ}\) Process Spontaneous?
+ + + Yes
- - - No (reaction will occur in opposite direction)
+ - ? Yes, if \(\Delta S_{sys}\) has a larger magnitude than \(\Delta S_{surr}\)
- + ? Yes, if \(\Delta S_{surr}\) has a larger magnitude than \(\Delta S_{sys}\)

Free Energy (\(G\))

$$ \Delta S_{univ} = -\frac{\Delta G}{T}\,\left( \text{at constant }T\text{ and }P \right) $$

Effect of \(\Delta H\) and \(\Delta S\) on Spontaneity

Case Result
\(\Delta S > 0\), \(\Delta H <0\) Spontaneous at all temperatures
\(\Delta S > 0\), \(\Delta H > 0\) Spontaneous at high temperatures (where exothermicity is relatively unimportant)
\(\Delta S < 0\), \(\Delta H < 0\) Spontaneous at low temperatures (where exothermicity is dominant)
\(\Delta S < 0\), \(\Delta H > 0\) Process not spontaneous at any temperatures (reverse process is spontaneous at all temperatures)

Third Law of Thermodynamics

Standard State Values

Concept Check

A stable diatomic molecule spontaneously forms from its atoms. Predict the sign of

Free Energy

The Meaning of \(\Delta G\) for a Chemical Reaction

When the energy profile of the reaction entirely downhill between reaction and product. If equilibrium is more stable then the reaction will not go to completion.

Change in Free Energy to Reach Equilibrium

Regardless of what fraction of reactant A you start with, the reaction will progress either forward or backward until an equilibrium is established.

Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction

\(\Delta G^\circ\) \(K\)
\(\Delta G^\circ = 0\) \(K=1\)
\(\Delta G^\circ < 0\) \(K>1\)
\(\Delta G^\circ > 0\) \(K<1\)

Free Energy and Work

/