Chapter 12: Thermodynamics - Why Chemical Reactions Happen

Chemical Thermodynamics

• Chemical thermodynamics studies the energetics of chemical reactions.

12.1 Spontaneous Processes

• Spontaneous process: Occurs without continuous external intervention.
• Nonspontaneous process: Occurs only with continuous energy input.
• Spontaneity depends on the dispersion of energy during a process.
• Thermodynamics predicts whether a process occurs under given conditions.
• Spontaneous processes occur naturally.
• Nonspontaneous processes require energy input.
• Spontaneity is determined by comparing the chemical potential energy before and after the reaction.
• A thermodynamically favorable reaction results in the system having less potential energy after the reaction than before.
• Spontaneity does not indicate the speed of the reaction (fast or slow).

Reversibility of Process

• Spontaneous processes are irreversible due to net energy release, proceeding in one direction only.
• Reversible processes proceed back and forth between end conditions.
  • Reversible processes are at equilibrium.
  • This yields no change in free energy.
• If a process is spontaneous in one direction, it's nonspontaneous in the opposite direction.

Spontaneous Processes and Energy

• Spontaneous processes release energy from the system, typically moving from higher to lower potential energy (exothermic).
• Some spontaneous processes proceed from lower to higher potential energy (endothermic).
• Two factors determine whether a reaction is spontaneous:
  • Enthalpy change ($\Delta H$).
  • Entropy change ($\Delta S$).

Enthalpy Change ($\Delta H$)

• Enthalpy change is measured in kJ/mol.
• Exothermic reactions occur when the bonds in the products are stronger than in the reactants.
  • Stronger bonds indicate more stable molecules.
  • Exothermic reactions release energy; $\Delta H$ is negative.
• Endothermic reactions occur when the bonds in the products are weaker than in the reactants.
  • Endothermic reactions absorb energy; $\Delta H$ is positive.
• Enthalpy change is favorable for exothermic reactions and unfavorable for endothermic reactions.

Spontaneity and Enthalpy

• Many spontaneous processes are exothermic (e.g., combustion), but not all.
• Endothermic reactions (\Delta H_{rxn} > 0) can be spontaneous.

12.2 Entropy and Second Law of Thermodynamics

• Second Law of Thermodynamics: The universe's total entropy increases in any spontaneous process.
• Entropy (S): A measure of disorder in a system.
• Entropy increases as the number of energetically equivalent ways of arranging components increases.
• Entropy has units of J/mol.
• $S = k \ln W$
  • k = Boltzmann constant = $1.38 × 10^{-23}$ J/K
  • W is the number of energetically equivalent ways a system can exist.
• Random systems require less energy than ordered systems.

Examples of Entropy

• Salt dissolving in water: increases entropy.
  • $NaCl(s) \rightarrow NaCl(aq)$
• Evaporation of $H<em>2O(l)$ to $H</em>2O(g)$: increases entropy.

Entropy and the Second Law of Thermodynamics

• The first law of thermodynamics: energy cannot be created or destroyed. $\Delta E<em>{univ} = 0 = \Delta E</em>{sys} + \Delta E_{surr}$
• Energy can be transferred, but the total energy of the universe remains constant.
• For a spontaneous process, the total entropy change of the universe must be positive: $\Delta S<em>{univ} = \Delta S</em>{sys} + \Delta S_{surr}$
  • For a reversible process, $\Delta S_{univ} = 0$
  • For an irreversible (spontaneous) process, \Delta S_{univ} > 0
• If the system's entropy decreases, the surroundings' entropy must increase by a larger amount.
• When $\Delta S<em>{sys}$ is negative, $\Delta S</em>{surr}$ must be positive and large for a spontaneous process.

Entropy: Expansion of Gas in a Vacuum

• When a stopcock is opened, gas atoms distribute themselves in multiple arrangements.
• These arrangements are energetically equivalent states for the expansion of a gas.
  • Potential energy remains the same whether molecules are in one flask or evenly distributed.
  • Certain states are more probable than others.

Macrostates and Probability

• There is only one possible arrangement for state A and one for state B.

Macrostates and Microstates

• Macrostate C can be achieved through several different arrangements of particles (microstates).
• All microstates have the same macrostate.
• There are six different particle arrangements that result in the same macrostate.

Macrostates and Probability

• States A and B have only one possible arrangement each.
• State C has six possible arrangements.
• The macrostate with the highest entropy also has the greatest dispersal of energy.
  • State C has higher entropy than states A or B.

Changes in Entropy ($\Delta S$)

• $\Delta S = S<em>{final} – S</em>{initial}$
• Entropy change is favorable when the result is a more random system ($\Delta S$ is positive).
• Changes that increase entropy:
  • Reactions whose products are in a more random state.
    • Solid < liquid < gas.
  • Reactions with a larger number of product molecules than reactant molecules.
  • Increase in temperature.
  • Solids dissociating into ions upon dissolving.

Entropy Change in State Change

• When materials change state, the number of macrostates they can have changes.
• The more degrees of freedom the molecules have, the more macrostates are possible.
• Solids have fewer macrostates than liquids, which have fewer macrostates than gases.
  • S{solid} < S{liquid} < S_{gas}

12.3 Absolute Entropy and Molecular Structure

• Standard molar entropy ($S^o$): The absolute entropy of one mole of a substance in its standard state.
  • Solid: Pure solid, most stable allotrope of an element (1 bar, 25°C).
  • Liquid: Pure liquid (1 bar, 25°C).
  • Gas: Pure gas (1 bar, 25°C).
  • Solution: 1 M (1 bar, 25°C).

Trends in Entropies

• Example: $H_2O$ at 298 K
  • $H_2O(l) = 69.9$ J/(mol·K)
  • $H_2O(g) = 188.8$ J/(mol·K)
  • $\Delta S_{vap} = 119$ J/(mol·K)

Factors Affecting Entropy Change

• Entropy increases when temperature increases.
• Entropy increases when volume increases.
• Entropy increases when the number of independent particles increases.

Entropy and Structure

• Entropy is affected by mass, molecular structure, mass distribution, and rigidity.

Practice Problem: Predicting the Sign of $\Delta S$

• a. $H<em>2O(g) \rightarrow H</em>2O(l)$: $\Delta S$ is negative (gas to liquid).
• b. Solid carbon dioxide sublimes: $\Delta S$ is positive (solid to gas).
• c. $2 N<em>2O(g) \rightarrow 2 N</em>2(g) + O_2(g)$: $\Delta S$ is positive (increase in moles of gas).

Heat Transfer and Changes in Entropy of the Surroundings

• The second law requires the entropy of the universe to increase for a spontaneous process.
• Processes like condensation of water vapor are spontaneous, even though water vapor is more random than liquid water.
• If a process is spontaneous but the entropy change of the process is unfavorable, there must have been a large increase in the entropy of the surroundings.
• The entropy increase must come from heat released by the system; the process must be exothermic!

Heat Exchange and $\Delta S_{surr}$

• Exothermic system processes add heat to the surroundings, increasing the entropy of the surroundings.
• Endothermic system processes take heat from the surroundings, decreasing the entropy of the surroundings.
• The amount the entropy of the surroundings changes depends on the original temperature.
• The higher the original temperature, the less effect addition or removal of heat has.

12.4 Applications of the Second Law

• Universe = system + surroundings: $\Delta S<em>{univ} = \Delta S</em>{sys} + \Delta S_{surr}$
  • If \Delta S_{univ} > 0, the process is spontaneous.
  • If \Delta S_{univ} < 0, the process is nonspontaneous.

Entropy and Heat

• Relationship between entropy gain and temperature: $\Delta S = \frac{q_{rev}}{T}$
  • $q_{rev}$ = flow of heat for reversible process
• Reversible process: Happens so slowly that an incremental change can be reversed by another tiny change, restoring the original state of the system with no net flow of energy between the system and its surroundings.

Quantifying Entropy Changes in the Surroundings

• The entropy change in the surroundings is proportional to the amount of heat gained or lost.
  • $q<em>{surr} = –q</em>{sys}$
• The entropy change in the surroundings is also inversely proportional to the temperature.
• At constant pressure and temperature, the overall relationship is:
  • $\Delta S<em>{surr} = - \frac{\Delta H</em>{sys}}{T}$

Entropy of the Universe

• $\Delta S_{sys} = \frac{q}{T}$
• It takes $6.01 × 10^3$ J of energy to melt 1.00 mol of ice at 0°C. Assuming the process occurs reversibly:
  • $\Delta S_{sys} = \frac{(1.00 \text{ mol})(6.01 × 10^3 \text{ J/mol})}{273 \text{ K}} = 22.0 \text{ J/K}$
• Suppose 1.00 mol of ice melts at room temperature on a countertop. Heat flows out of the countertop:
  • $\Delta S_{surr} = \frac{(1.00 \text{ mol})(-6.01 × 10^3 \text{ J/mol})}{295 \text{ K}} = -20.4 \text{ J/K}$

Calculating Changes in Entropy in the Surroundings

• Example: Combustion of propane gas:
  • $C<em>3H</em>8(g) + 5 O<em>2(g) \rightarrow 3 CO</em>2(g) + 4 H_2O(g)$
  • $\Delta H_{rxn} = –2044$ kJ
  • a. Calculate the entropy change in the surroundings at 25 °C.
  • b. Determine the sign of the entropy change for the system.
  • c. Determine the sign of the entropy change for the universe. Is the reaction spontaneous?
• Solution:
  • a. $\Delta S<em>{surr} = \frac{-\Delta H</em>{rxn}}{T} = \frac{-(-2044 \text{ kJ})}{298 \text{ K}} = \frac{2044000 \text{ J}}{298 \text{ K}} = 6859 \text{ J/K}$
  • b. The number of moles of gas increases, implying a positive $\Delta S_{sys}$.
  • c. Since both $\Delta S<em>{sys}$ and $\Delta S</em>{surr}$ are positive, $\Delta S_{univ}$ is positive, and the reaction is spontaneous.

Spontaneity and Entropy

• $\Delta S_{sys} = +22.0 \text{ J/K}$
• $\Delta S_{surr} = –20.4 \text{ J/K}$
• $\Delta S<em>{univ} = \Delta S</em>{sys} + \Delta S_{surr} = +1.6 \text{ J/K}$, spontaneous

Temperature Dependence of $\Delta S_{surr}$

• When heat is added to cool surroundings, it has more of an effect on entropy than if the surroundings were already hot.
• Water freezes spontaneously below 0 °C because the heat released on freezing increases the entropy of the surroundings enough to make $\Delta S$ positive.
• Above 0 °C, the increase in entropy of the surroundings is insufficient to make $\Delta S$ positive.

Entropy Change in the System and Surroundings

• When the entropy change in a system is unfavorable (negative), the entropy change in the surroundings must be favorable (positive) and large in order to allow the process to be spontaneous.

12.5 Calculating Entropy Changes

• Entropy change for the system:
  • $\Delta S<em>{sys} = S</em>{final} – S_{initial}$
  • $\Delta S^o<em>{rxn} = \sum n</em>{products}S^o<em>{products} – \sum n</em>{reactants}S^o_{reactants}$
  • n = coefficients of the products/reactants in the balanced equation

Sample Exercise 12.3: Calculating Entropy Changes

• Methanol reforming reaction:
  • $CH<em>3OH(g) + H</em>2O(g) \rightarrow CO<em>2(g) + 3 H</em>2(g)$
  • Calculate the value of $\Delta S^o_{rxn}$ at 298 K.
  • $\Delta S^o<em>{rxn} = \sum n</em>{products} S^o<em>{products} – \sum n</em>{reactants} S^o_{reactants}$
  • $= [1 \text{ mol } CO<em>2 (213.8 \frac{\text{J}}{\text{mol} \cdot \text{K}}) + 3 \text{ mol } H</em>2 (130.6 \frac{\text{J}}{\text{mol} \cdot \text{K}})] - [1 \text{ mol } CH<em>3OH (239.9 \frac{\text{J}}{\text{mol} \cdot \text{K}}) + 1 \text{ mol } H</em>2O (188.8 \frac{\text{J}}{\text{mol} \cdot \text{K}})]$
  • $= (605.6 – 428.7) \text{ J/K} = 176.9 \text{ J/K}$

12.6 Gibbs Free Energy

• Free energy: A measure of the maximum amount of work that a thermodynamic system can perform.
• Gibbs free energy (G): The maximum amount of energy released by a process occurring at constant temperature and pressure that is available to do useful work.
• $\Delta S<em>{univ} = \Delta S</em>{sys} - \frac{\Delta H_{sys}}{T}$
• Multiply by -T and rearrange:
• $-T \Delta S<em>{univ} = \Delta H</em>{sys} - T \Delta S_{sys}$
• Define Gibbs Free Energy:
• $\Delta G = \Delta H - T \Delta S$

Spontaneous Chemical Processes

• Driving forces that contribute to spontaneity:
  • The system experiences an increase in entropy: \Delta S > 0
  • The process is exothermic: \Delta H < 0
• Free energy (G) relates enthalpy, entropy, and temperature for a process:
  • $\Delta G^o<em>{rxn} = \Delta H^o</em>{rxn} - T \Delta S^o_{rxn}$
  • $\Delta H^o<em>{rxn} = \sum n</em>{products} \Delta H^o<em>{f, products} - \sum n</em>{reactants} \Delta H^o_{f, reactants}$
  • $\Delta S^o<em>{rxn} = \sum n</em>{products} S^o<em>{products} - \sum n</em>{reactants} S^o_{reactants}$

Third Law of Thermodynamics

• The third law of thermodynamics states that the entropy of a perfect crystal of a substance at absolute zero is equal to 0.
• There is a minimum randomness in a perfect crystal at 0 K.
• Unlike enthalpy and internal energy, absolute values of entropy can be determined.

Calculating $\Delta G^o_{rxn}$

• The standard free energy of formation is the free energy change to form one mole of a compound from its elements in their standard states.
• $\Delta G^o<em>{rxn} = \sum n \Delta G^o</em>f [prods] - \sum m \Delta G^o_f [reacts]$, where n and m are the coefficients of the products and reactants in the reaction.

Temperature Dependence of $\Delta G$

• The dependence of $\Delta G$ on temperature arises mainly from the $T \Delta S$ term in the definition of $\Delta G$.
• The sign of $\Delta G$ is dominated by the sign of $\Delta H$ at low temperatures and by the sign of $\Delta S$ at high temperatures.
• Negative values of $\Delta H$ and positive values of $\Delta S$ favor spontaneity.

Example: Temperature Dependence

• Assuming that $\Delta H^o$ and $\Delta S^o$ do not change with temperature, calculate the temperature for which $\Delta G^o$ is 0 for the reaction:
  • $CS<em>2(l) \rightleftharpoons CS</em>2(g)$
  • At 298 K, $\Delta H^o = 27.66$ kJ and $\Delta S^o = 86.39$ J/K.
  • $\Delta G^o = \Delta H^o - T \Delta S^o$
  • $0 = 27660 \text{ J} - T (86.39 \text{ J/K})$
  • $T = \frac{27660 \text{ J}}{86.39 \text{ J/K}} = 320.2 \text{ K}$

Example: Calculate $\Delta G^o$

• Given the following chemical reaction and data at 298 K:
  • $N<em>2(g) + 3 H</em>2(g) \rightleftharpoons 2 NH_3(g)$
  • $\Delta H^o_f = 0 \text{ kJ/mol} \quad 0 \text{ kJ/mol} \quad -46.1 \text{ kJ/mol}$
  • $S^o = 191.5 \frac{\text{J}}{\text{mol} \cdot \text{K}} \quad 130.6 \frac{\text{J}}{\text{mol} \cdot \text{K}} \quad 192.3 \frac{\text{J}}{\text{mol} \cdot \text{K}}$
• Assuming that $\Delta H^o$ and $\Delta S^o$ do not change with temperature, calculate $\Delta G^o$ at 1000 K.
  • $\Delta S^o_{rxn} = [(2 </li> <li>= -198.7 \frac{\text{J}}{\text{mol} \cdot \text{K}}$
  • $\Delta H^o_{rxn} = [2(-46.1)] - [0] = -92.2 \text{ kJ/mol}$
• $\Delta G^o<em>{rxn} = \Delta H^o</em>{rxn} - T \Delta S^o_{rxn}$
  • $\Delta G^o_{rxn} = -92200 \frac{\text{J}}{\text{mol}} - (1000 \text{ K})(-198.7 \frac{\text{J}}{\text{mol} \cdot \text{K}})$
  • $\Delta G^o_{rxn} = 106500 \frac{\text{J}}{\text{mol}} = 106.5 \text{ kJ/mol}$

Calculating Standard Free Energy of Formation

• Standard free energy formation ($\Delta G^o_f$):
• A compound’s $\Delta G^o_f$ value is the change in free energy associated with the formation of one mole of a compound in its standard state from its elements in their standard states.
  • $\Delta G^o<em>{rxn} = \sum n</em>{products} \Delta G^o<em>{f, products} - \sum n</em>{reactants} \Delta G^o_{f, reactants}$
  • n = coefficients of the products/reactants in the balanced equation

Free Energy Change

• Internal energy change of system can be used to perform work:
  • $\Delta E = q + w$
  • $\Delta H = \Delta G + T \Delta S$
  • $\Delta G = \Delta H - T \Delta S$ (energy that does work)

12.7 Temperature and Spontaneity

• $\Delta H$ = positive
• $\Delta S$ = positive
• $\Delta G = \Delta H – T \Delta S$
  • $=$ negative above a certain temperature, when T \Delta S > \Delta H

Spontaneity and Entropy- Effects of ΔH, ΔS, and T on G and Spontaneity

• Negative $\Delta H$ and positive $\Delta S$: $\Delta G$ is always negative; always spontaneous.
• Negative $\Delta H$ and negative $\Delta S$: $\Delta G$ is negative at lower temperatures; spontaneous at lower temperatures.
• Positive $\Delta H$ and positive $\Delta S$: $\Delta G$ is negative at higher temperatures; spontaneous at higher temperatures.
• Positive $\Delta H$ and negative $\Delta S$: $\Delta G$ is always positive; never spontaneous.