11/14 Thermodynamics and Chemical Spontaneity

11/14 Thermodynamics and Chemical Spontaneity

Thermodynamics describes how energy and matter change during chemical reactions and physical processes. It can predict the favored direction under specified conditions, but it does not predict how quickly the process will occur.

A spontaneous process proceeds naturally in a particular direction without continuous external energy input. Examples include heat flowing from a warmer object to a cooler object, gas expansion into an evacuated container, iron corrosion in moist air, and ice melting above its melting point.

Spontaneity is not the same as speed. A process may be thermodynamically favorable but very slow. Kinetics addresses reaction rate, activation energy, and mechanism; thermodynamics addresses direction and equilibrium.

The second law states that a spontaneous process increases the entropy of the universe:

with

At equilibrium, $\Delta S_{\text{univ}}=0$. A process with $\Delta S_{\text{univ}}<0$ is nonspontaneous in the direction written.

Takeaway: Thermodynamic spontaneity identifies a favored direction, while kinetics determines the rate.

Entropy and Its Molecular Meaning

Entropy measures the dispersal of energy and matter. At the molecular level, it is related to the number of accessible microscopic arrangements, or microstates:

Here, $k$ is the Boltzmann constant and $W$ is the number of accessible microstates. More possible arrangements generally mean greater entropy.

For a reversible process,

where $q_{\text{rev}}$ is reversible heat transfer and $T$ is the absolute temperature in kelvins.

Useful qualitative patterns include:

• Melting a solid or vaporizing a liquid generally increases entropy.

• Increasing the number of gas particles generally increases entropy.

• Dissolving a substance often disperses particles and increases entropy.

• Heating a solid or liquid generally increases entropy.

• For a given substance, $S_{\text{solid}}<S_{\text{liquid}}<S_{\text{gas}}$.

The reverse changes generally decrease entropy. These guidelines predict the sign of $\Delta S$, but accurate numerical work requires tabulated entropy values.

For a reaction, use standard molar entropies and the balanced equation:

The coefficient $n$ is the stoichiometric coefficient. For example, in $\mathrm{N_2(g)+3H_2(g)\rightarrow 2NH_3(g)}$, four moles of gas become two moles of gas, so the reaction is expected to have $\Delta S^\circ<0$, although tabulated values are needed for the numerical result.

Takeaway: Use molecular dispersal and the number of gas particles to predict the sign of entropy change, then use tabulated data for precision.

Gibbs Energy, Enthalpy, and Temperature

Enthalpy describes heat absorbed or released by a system at constant pressure:

• $\Delta H<0$: the process is exothermic.

• $\Delta H>0$: the process is endothermic.

Enthalpy alone does not determine spontaneity. An endothermic process can be spontaneous if its entropy increase is sufficiently large, while an exothermic process can be nonspontaneous if it causes a sufficiently large entropy decrease.

The central combination of these effects is Gibbs energy:

At constant temperature and pressure,

Gibbs energy is also related to the entropy change of the universe:

Therefore:

• $\Delta G<0$: the forward process is spontaneous.

• $\Delta G>0$: the forward process is nonspontaneous and the reverse direction is favored.

• $\Delta G=0$: the system is at equilibrium.

A negative $\Delta G$ indicates thermodynamic favorability, not rapid reaction. A process with a large negative $\Delta G$ may still be slow if it has a high activation energy.

The signs of $\Delta H$ and $\Delta S$ help predict temperature dependence:

• If $\Delta H<0$ and $\Delta S>0$, the process is spontaneous at all temperatures, assuming the values remain approximately constant.

• If $\Delta H>0$ and $\Delta S<0$, the process is nonspontaneous at all temperatures.

• If $\Delta H<0$ and $\Delta S<0$, the process is spontaneous at sufficiently low temperature.

• If $\Delta H>0$ and $\Delta S>0$, the process is spontaneous at sufficiently high temperature.

When $\Delta H$ and $\Delta S$ have the same sign, the boundary temperature can be estimated by setting $\Delta G=0$:

Units must be consistent. For example, if $\Delta H=25.0\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S=80.0\ \mathrm{J\,mol^{-1}\,K^{-1}}$, convert entropy to $0.0800\ \mathrm{kJ\,mol^{-1}\,K^{-1}}$. Then:

Above approximately $313\ \mathrm{K}$, $\Delta G<0$; below that temperature, $\Delta G>0$, under the stated assumptions.

Takeaway: Spontaneity reflects the balance between enthalpy and the temperature-weighted entropy change.

Standard Gibbs Energy Change

The standard Gibbs energy change can be calculated from standard Gibbs energies of formation:

The standard Gibbs energy of formation of an element in its standard state is defined as zero. Standard conditions commonly involve pure substances in their standard states, gases at a standard pressure, and dissolved species at a standard concentration; the precise convention may depend on the data table.

An alternative calculation is

Use the same temperature for all quantities and convert units before substituting.

For example, at $298\ \mathrm{K}$, if $\Delta H^\circ=-92.0\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S^\circ=-198\ \mathrm{J\,mol^{-1}\,K^{-1}}$, convert entropy to $-0.198\ \mathrm{kJ\,mol^{-1}\,K^{-1}}$. Then:

Because $\Delta G^\circ<0$, the reaction is spontaneous under standard conditions at $298\ \mathrm{K}$. The negative entropy change does not prevent spontaneity because the favorable enthalpy change is sufficiently large.

Takeaway: Standard-state calculations require balanced stoichiometry, consistent units, and a clearly specified temperature.

Nonstandard Conditions and the Reaction Quotient

Spontaneity can change when concentrations, pressures, or other conditions differ from the standard state. The reaction quotient describes the current composition, and the Gibbs energy under those conditions is

Here, $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$, $T$ is in kelvins, and $Q$ is dimensionless.

For

The symbols represent activities, which measure effective concentration or pressure. In introductory problems, concentrations or partial pressures are often used as approximations. Pure solids and pure liquids are omitted from $Q$ because their activities are approximately constant.

Interpret the result as follows:

• $\Delta G<0$: the system moves spontaneously toward products.

• $\Delta G>0$: the system moves spontaneously toward reactants.

• $\Delta G=0$: the system is at equilibrium.

This equation shows why standard-state spontaneity is not always the same as spontaneity under actual conditions.

Takeaway: To evaluate direction under nonstandard conditions, combine $\Delta G^\circ$ with the current composition through $RT\ln Q$.

Equilibrium, Constants, and Physical Changes

Equilibrium occurs when there is no net thermodynamic driving force for change. At equilibrium,

and

Substituting these conditions into the nonstandard Gibbs equation gives

The signs of $\Delta G^\circ$ and the size of $K$ are related:

• $\Delta G^\circ<0$ corresponds to $K>1$, so products are favored under standard conditions.

• $\Delta G^\circ>0$ corresponds to $K<1$, so reactants are favored under standard conditions.

• $\Delta G^\circ=0$ corresponds to $K=1$, so neither side is favored under standard conditions.

A large equilibrium constant does not mean that the reaction is fast. It means that the equilibrium composition lies toward products. Kinetics determines how long the system takes to approach equilibrium.

The same ideas apply to physical changes. During melting, solid changes to liquid, and both $\Delta H$ and $\Delta S$ are generally positive. Melting becomes spontaneous above the melting temperature because the $T\Delta S$ term makes $\Delta G<0$. At the melting point, $\Delta G=0$, and solid and liquid coexist at equilibrium. Vaporization is similarly increasingly favorable as temperature rises; at the normal boiling point, liquid and gas are in equilibrium at a pressure of one atmosphere under the conventional definition.

Takeaway: Equilibrium identifies the point where the net driving force is zero; it does not describe the time required to get there.

Problem-Solving Checklist and Key Equations

For a quantitative spontaneity problem, use the following sequence:

1. Write and balance the chemical equation, or clearly identify the physical process.

2. Determine which quantities are relevant: $\Delta H$, $\Delta S$, $\Delta G$, $\Delta G^\circ$, $Q$, or $K$.

3. Use kelvins for temperature.

4. Make units consistent, especially when evaluating $T\Delta S$.

5. Substitute into the appropriate equation.

6. Interpret the sign and units of the result.

7. Distinguish thermodynamic favorability from reaction rate.

A final check should ask whether the result answers a direction question or a speed question. A spontaneous process may require a catalyst or may occur very slowly, and a catalyst changes the rate without changing the thermodynamic criterion for spontaneity or the equilibrium position.

The main equations are:

Final takeaway: Entropy describes dispersal, enthalpy describes heat at constant pressure, Gibbs energy combines both for constant-temperature and constant-pressure decisions, and kinetics determines how rapidly the favored change occurs.