Entropy, Free Energy, and Nuclear Chemistry Notes

Thermodynamics Review

• Thermodynamics: Study of heat and motion.
• Thermochemistry: Subset of thermodynamics (studied in Chpt 5).

Definitions

• System: Part of the universe under study.
• Surroundings: Rest of the universe (not under study).
• Exothermic Reaction: Transfers thermal energy from the system to the surroundings.
• Endothermic Reaction: Transfers thermal energy from the surroundings to the system.
• First Law of Thermodynamics: Law of conservation of energy.
• Enthalpy Change: Thermal energy transferred at constant pressure.
• State Function: Quantity whose value is determined only by the initial and final state of a system.

Spontaneous Processes

• Many naturally occurring processes tend to be exothermic, but enthalpy is only part of the story.
• Examples of Spontaneous Processes:
  • Water flows downhill.
  • Ice melts on a warm day.
  • Eating an apple and excreting $CO<em>2$ and $H</em>2O$. Ingesting $CO<em>2$ and $H</em>2O$ does not spontaneously produce an apple.
  • Zinc metal dissolving in a strong acid solution: $Zn(s) + 2 H<em>3O^+(aq) \rightarrow Zn^{2+}(aq) + H</em>2(g) + 2H_2O(l)$

Entropy and Spontaneity

• Two factors are important: Enthalpy and Entropy.
• Entropy (S): Measure of disorder (more disorder, more entropy).
  • $AS = S<em>2 - S</em>1$
  • Balls naturally roll down hills, not up. Negative $\Delta H$ favors spontaneity.
  • Rooms naturally get messy, more disordered. Positive $\Delta S$ favors spontaneity.

Generalizations About Entropy Changes

• As the temperature of a system increases, entropy increases ($\Delta S$ is positive).
• As the volume in which gas particles can move increases, entropy increases ($\Delta S$ is positive).
• As the number of gas particles in the system increases, entropy increases ($\Delta S$ is positive).
• As the size and complexity of the molecules increase, entropy increases.
• Examples:
  • $2C<em>2H</em>4(g) \rightarrow C<em>4H</em>8(g)$: $\Delta S$ is negative because there are fewer molecules (less disorder).
  • $C<em>4H</em>8(g; 298K; 1 atm) \rightarrow C<em>4H</em>8(g; 298K; 0.5 atm)$: $\Delta S$ is positive because volume increased.
  • $C<em>4H</em>8(g; 298K; 1 atm) \rightarrow C<em>4H</em>8(g; 398K; 1 atm)$: $\Delta S$ is positive because temperature increased.
  • Ranking entropy: H2O(s) < H2O(l) < H_2O(g)

Spontaneity and $\Delta H$ and $\Delta S$

• $2C(graphite) + O_2(g) \rightarrow 2CO(g)$
  • $\Delta H$ is negative (favors).
  • $\Delta S$ is positive (favors).
• $N<em>2(g) + 2O</em>2(g) \rightarrow 2NO_2(g)$
  • $\Delta H$ is positive (disfavors).
  • $\Delta S$ is negative (disfavors).

Second Law of Thermodynamics

• In a spontaneous process, the entropy of the universe increases (\Delta S_{univ} > 0).
• $\Delta S<em>{univ} = \Delta S</em>{sys} + \Delta S_{surr}$
• $\Delta S<em>{surr}$ relates to the heat change of the system ($\Delta H</em>{sys}$) and temperature.
• $\Delta S<em>{surr} = -\frac{\Delta H</em>{sys}}{T}$
• The 2nd Law: \Delta S{univ} = \Delta S{sys} - \frac{\Delta H_{sys}}{T} > 0
• T \cdot (\Delta S{sys} - \frac{\Delta H{sys}}{T}) > 0
• \Delta H{sys} - T\Delta S{sys} < 0

Gibb's Free Energy

• $AG = AH - TAS$
• If $\Delta G$ is negative, the reaction is spontaneous.
• If $\Delta G$ is positive, the reaction is non-spontaneous.
• If $\Delta G$ is zero, the reaction is at equilibrium.
• $\Delta G<em>{sys} = \Delta H</em>{sys} - T\Delta S_{sys}$

Calculating $\Delta H$, $\Delta S$, and $\Delta G$

• $\Delta H<em>{rxn}^\circ = \Sigma n \Delta H</em>f^\circ (products) - \Sigma n \Delta H_f^\circ (reactants)$

  • Standard state (1 atm and 1 M soln)
  • Formation reaction
  • $n$ is the number of moles in the balanced reaction

• Consider the reaction: $2Al(s) + Fe<em>2O</em>3(s) \rightarrow Al<em>2O</em>3(s) + 2Fe(s)$

• $\Delta S_{rxn}^\circ = \Sigma nS^\circ (products) - \Sigma nS^\circ (reactants)$

  • Note: $S^\circ$ is used, not $\Delta S^\circ$. The zero of entropy has been assigned (no zero of enthalpy has been assigned, so only changes can be measured).

• 3rd Law of Thermodynamics: The entropy of a perfect crystal is zero at 0 K.

• For the previous reaction:

  • $\Delta S^\circ = (1 mol)(51.0 J/mol \cdot K) + (2 mol)(27.2 J/mol \cdot K) - (2 mol)(28.33 J/mol \cdot K) - (1 mol)(90.0 J/mol \cdot K) = -41.2 J/K$

• Using $\Delta G = \Delta H^\circ - T\Delta S^\circ$, at 25°C:

  • $\Delta G^\circ = -848.0 kJ - 298K (-0.0412 kJ/K) = -835.7 kJ$

• Another way to calculate $\Delta G^\circ$ is:

  • $\Delta G<em>{rxn}^\circ = \Sigma n \Delta G</em>f^\circ (products) - \Sigma n \Delta G_f^\circ (reactants)$

• Examples:

  • a) $\Delta G^\circ = -, \Delta H^\circ = +, \Delta S^\circ = ?$
    • $(-) = (+) - (?)$
    • $\Delta S^\circ$ = positive
  • b) $\Delta G^\circ = +, \Delta H^\circ = +, \Delta S^\circ = ?$
    • $(+) = (+) - (?)$
    • $\Delta S^\circ$ = negative or undetermined
  • c) $\Delta S^\circ = +, \Delta H^\circ = -, \Delta G^\circ = ?$
    • $(?) = (-) - (+)$
    • $\Delta G^\circ$ = negative
  • d) $\Delta S^\circ = -, \Delta H^\circ = -, \Delta G^\circ = ?$
    • $(?) = (-) - (-)$
    • $\Delta G^\circ$ = positive or negative

• Cyanopyridine Example:

  • $C<em>4H</em>6(g) + C<em>2N</em>2(g) \rightarrow C<em>6H</em>4N<em>2 + H</em>2(g)$
  • Values provided in table in the original document.
  • $\Delta H = 257.92 kJ - 111.27 kJ - 296.52 kJ = -149.87 kJ$
  • $\Delta S = 320.69 J/K + 129.83 J/K - 276.3 J/K - 239.78 J/K = -65.57 J/K$
  • $\Delta G = \Delta H - T\Delta S = (-149.87 kJ) - (298 K)(-0.06557 kJ/K) = -130.33 kJ$
  • Conclusion: Yes, the reaction is worthwhile to attempt in the lab because it is spontaneous (negative $\Delta G$).
  • Methanol Example:
  • Calculate $\Delta G^\circ$ for the reaction:
  • $2CH<em>3OH(g) + 3O</em>2(g) \rightarrow 2CO<em>2(g) + 4H</em>2O(g)$
  • $\Delta G^\circ = (2 mol)(-394 kJ/mol) + (4 mol)(-229 kJ/mol) - (2 mol)(-163 kJ/mol) = -1378 kJ$

• Temperature and Spontaneity Example:

  • $\Delta H^\circ = 72.4 kJ/mol$ and $\Delta S^\circ = 142 J/mol \cdot K$
  • $T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{72.4 kJ/mol}{0.142 kJ/mol \cdot K} = 510 K$

Oxidation Numbers

• Each atom in a pure element has an oxidation number of zero.
• For monatomic ions, the oxidation number is equal to the charge of the ion.
• The following elements always have the same oxidation number in compounds:
  • F is -1
  • H is +1, except when with a metal
  • O is -2, except when in a peroxide ion ($O_2^{2-}$
• The sum of the oxidation numbers in a molecule or ion must add up to the charge of the molecule or ion.

Examples For Oxidation States

• $O_2$: 0
• $HNO_3$: +1 +5 -2
• $CO$: +2 -2
• $CO_2$: +4 -2
• $SO_4^{2-}$: +6 -2
• $S_8$: 0
• $Fe<em>2S</em>3$: +3 -2
• $Cr<em>2O</em>7^{2-}$: +6 -2

Oxidation-Reduction (Redox)

• Oxidation: Loss of electrons
• Reduction: Gain of electrons
• LEO the lion says GER (Loss of Electrons is Oxidation, Gain of Electrons is Reduction)
• In a reaction, if one thing is oxidized, another thing must be reduced.

Redox Example

• $Cu^{2+}(aq) + Zn(s) \rightarrow Cu(s) + Zn^{2+}(aq)$
  • What is being oxidized? Zn
  • What is being reduced? Cu
• Oxidizing Agent: Helps the other thing get oxidized (oxidizing agent is reduced).
• Reducing Agent: Helps the other thing get reduced (reducing agent is oxidized).
• $Cl<em>2 + H</em>2S \rightarrow S_8 + HCl$
  • Oxidized: $H_2S$
  • Reduced: $Cl_2$
  • Oxidizing agent: $Cl_2$
  • Reducing agent: $H_2S$

Voltaic Cells

• Voltaic (or Galvanic) Cell generates electricity spontaneously.

• Electrons are produced in one ½-cell (oxidation ½-cell).

• Electrons are gained in the other ½-cell (reduction ½-cell).

• Anode: Electrode where oxidation is occurring.

• Cathode: Electrode where reduction is occurring.

• Electrons flow from negative electrode (anode) to positive electrode (cathode).

• Negative ions in the salt bridge go in the opposite direction of the electrons.

• Voltage of the cell results from the two ½-cell reactions. $E<em>{cell} = E</em>{ox} + E_{red}$

• Reverse a ½-reaction Change the sign of the potential.

• Multiply a ½-reaction by a number DO NOT multiply the potential by a number.

• Add two ½-reactions Add their potentials: $E^\circ<em>{cell} = E^\circ</em>{red} + E^\circ_{ox}$

• A positive cell voltage means the reaction is spontaneous in the direction written.

Voltaic Potential Examples

• A voltaic cell has a Mg electrode in 1 M $Mg(NO<em>3)</em>2(aq)$ and a Ag electrode in 1 M $AgNO_3(aq)$.
  • $Mg^{2+}(aq) + 2e^- \rightarrow Mg(s) \quad E^\circ = -2.37 V$ (anode)
  • $Ag^+(aq) + e^- \rightarrow Ag(s) \quad E^\circ = +0.80 V$ (cathode)
  • $E^\circ<em>{cell} = E^\circ</em>{cathode} - E^\circ_{anode} = 0.80 V - (-2.37 V) = +3.17 V$
• A voltaic cell has a Cu electrode in 1 M $CuSO<em>4(aq)$ and a Fe electrode in 1 M $FeSO</em>4(aq)$.
  • $Cu^{2+} + 2e^- \rightarrow Cu(s) \quad E^\circ = +0.337 V$ (cathode)
  • $Fe^{2+} + 2e^- \rightarrow Fe(s) \quad E^\circ = -0.44 V$ (anode)
    • $E^\circ_{cell} = (0.337 V) - (-0.44 V) = 0.777 V$
• Will the following reaction happen spontaneously? $2Al(s) + 3Sn^{4+}(aq) \rightarrow 2Al^{3+}(aq) + 3Sn^{2+}(aq)$
  • $Al^{3+} + 3e^- \rightarrow Al(s) \quad E^\circ = -1.66 V$ (anode)
  • $Sn^{4+} + 2e^- \rightarrow Sn^{2+}(aq) \quad E^\circ = +0.15 V$ (cathode)
  • $E^\circ_{cell} = 0.15 - (-1.66) = +1.81 V$

Thermodynamics and Equilibrium

• The equation $E^\circ = \frac{RT}{nF}lnK$ relates the standard state cell potential for a chemical reaction to the equilibrium constant.

• $R = 8.314 J/mol \cdot K$, $F = 96485 C/mol$

• Using $\Delta G = -RTlnK$ and $E^\circ = \frac{RT}{nF}lnK$

  • $\Delta G = -nFE^\circ$

• Relating $E^\circ$ to K:

  • $E = E^\circ - \frac{RT}{nF}lnQ$
  • Nernst Equation: $E = E^\circ - \frac{0.0257 V}{n}lnQ$ at 25°C
  • At equilibrium: Q = K and E = 0
  • $E^\circ = \frac{0.0257 V}{n}lnK$

• The redox reaction $Co(s) + 2H^+ \rightarrow Co^{2+}(aq) + H_2(g)$. E° has a potential of 0.277 V.

  • What is $\Delta G^\circ$ for this reaction?
  • $\Delta G^\circ = -nFE^\circ = -(2)(96485 C/mol)(0.277 V) = -53.5 kJ/mol$
  • What is K for this reaction?
  • $E^\circ = \frac{0.0257}{n}lnK$
  • lnK = 21.56
  • K = 2.1 x 10^9

Nuclear Chemistry

• Radioactive decay: The disintegration of an unstable atomic nucleus by spontaneous emission of radiation.
• Radioisotopes: Nuclei that undergo radioactive decay.

Differences Between Chemical and Nuclear Reactions:

Types of Radiation and Decay

• The more mass the particle has, the less penetrating it is.
• The faster the particle is, the more penetrating it is.

Nuclear Equations and Symbols

• Elements may change in nuclear reactions.
• The sum of mass numbers of reactants must equal the sum of mass numbers of products.
• The sum of atomic numbers of reactants must equal the sum of atomic numbers of products.

Nuclear Symbols

Types of Decay

• Alpha Decay (also called alpha emission): Nucleus loses an alpha particle ($\^4_2He$).
  • Example: $\^{222}<em>{86}Rn \rightarrow \^{218}</em>{84}Po + \^4_2He$
• Beta Decay (also called beta emission): Nucleus loses a beta particle ($\^0_{-1}e$).
  • Example: $\^3<em>1H \rightarrow \^3</em>2He + \^0_{-1}e$
• Gamma Radiation: A photon of high-energy gamma ray ($\gamma$) is emitted (typically occurs after another radioactive decay).
• Positron Emission: A positron ($\^0_{+1}e$) is emitted (equal mass but opposite charge of an electron).
  • Example: $\^{18}<em>9F \rightarrow \^{18}</em>8O + \^0_{+1}e$
• Electron Capture: Nucleus absorbs an electron ($\^0_{-1}e$) and then releases an X-ray.
  • Example: $\^{125}<em>{53}I + \^0</em>{-1}e \rightarrow \^{125}_{52}Te$

Radioactive Decay Series

• Parent: Unstable in a radioactive decay step.
• Daughter: Stable in a radioactive decay step.

Energy from Nucleus

• $E = mc^2$
• For nuclear reactions, mass changes may be measurable.
• For chemical reactions, mass changes are not measurable.

Half-Life

• Half-life: Time required for one-half of the sample to undergo radioactive decay.

• Radioactive decay is a first-order process.

• $ln[A] = -kt + ln[A]_0$ or with activity (defined as disintegrations/time).

• $t_{1/2} = \frac{0.693}{k}$

Examples

• You obtain a new sample of cobalt-60, half-life 5.25 years, with a mass of 400 mg. How much cobalt-60 remains after 19.5 years?
  • $k = \frac{0.693}{5.25 years} = 0.132$
  • $ln[A] = -(0.132)(19.5) + ln[400]$
  • $ln[A] = -2.574 + 5.991$
  • $ln[A] = 3.417$
  • $A = e^{3.417} = 30.48 mg$
• A sample of radon-222 has an initial activity of $7.0 × 10^4$ disintegrations per second (dps). After 6.6 days, its activity is $2.1 × 10^4$ dps. What is the half-life of radon-222?
  • $ln[2.1 × 10^4] = -k(6.6 days) + ln[7.0 × 10^4]$
  • $11.16 = -6.6k + 11.16$
  • $k = 0.183$
  • $t_{1/2} = \frac{0.693}{0.183} = 3.8 days$
• You obtain a 20.0 mg sample of mercury-190 which has a half-life of 20 minutes. How long will it take for 15.0 mg of your sample to decay?
  • $k = \frac{0.693}{20 min} = 0.0347 min^{-1}$
  • $ln[15] = -0.0347 min^{-1} t + ln[20]$
  • $t = 8.29 min$

Radioisotopic Dating

• Use certain isotopes to estimate the age of various items.
  • $^{235}U$ [half-life = 4.5 billion years] - Determines age of rock.
  • $^3H$ [half-life 12.3 years] - Used to date aged.
• Carbon-14 Dating [Half-life of 5730 years] - Used to date things that were living things.

Nuclear Fission

• Fission: When a large nucleus breaks into smaller nuclei.
  • Happens through a nuclear chain reaction.
  • Only certain isotopes undergo nuclear chain reactions.
• Disasters:
  • Three-mile Island, Pennsylvania, 1979
  • Chernobyl, former Soviet Union, 1986
  • Fukushima-Daiichi, Japan, 2011

Nuclear Fusion

• Reaction takes smaller nuclei and builds larger ones.
• Releases tremendous amounts of energy.
  • 1 g of H would release same as 20 tons of coal.