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<br><br><br> # Chapter 16: <br> Thermodynamics <br><br><br><br> ### Eric Van Dornshuld #### Mississippi State University #### 10/24/2020 .footnote[ [Textbook: Chapter 16](https://openstax.org/books/chemistry-2e/pages/16-introduction) ] --- ### Spontaneity A **spontaneous** process occurs naturally under certain conditions, once started, whereas a **nonspontaneous** process will not take place unless it is driven by a continual input of energy from an external source. <br> .pull-left[ Ice spontaneously melts at *T* > 0 °C <img src="ice-th.png" width="420px" style="display: block; margin: auto;" /> ] .pull-right[ Ice does not spontaneously melt at *T* < 0 °C <img src="ice-tc.png" width="315px" style="display: block; margin: auto;" /> ] What is the *driving force* for spontaneous processes? --- ### Dispersal of Matter Consider an isolated system. The left flask contains a gas. The right flask contains a vacuum. <img src="dispersal-of-matter.jpg" width="800px" style="display: block; margin: auto;" /> When valve is opened, the gas spontaneously uniformly disperses itself. No work is done `$$w = -P\Delta V = 0 \quad (P = 0~\mathrm{in~a~vacuum})$$` and there is no change in internal energy `$$\Delta U = q + w = 0 + 0 = 0$$` since the system is isolated and it does not exchange heat with its surroundings. Since no change in energy occurred, energy is not the driving force. The driving force is related to the *uniform dispersal of matter*. .footnote[ Access for free at https://openstax.org/books/chemistry-2e/pages/16-1-spontaneity ] --- ### Dispersal of energy Heat always spontaneously flows from a hotter object to a colder object. Energy spontaneously disperses itself uniformally. <img src="heat-transfer.jpg" width="800px" style="display: block; margin: auto;" /> The system (X and Y), if isolated, does not experience a temperature change, simply a redistribution of thermal energy. The driving force is related to the *uniform dispersal of energy*. <br> > As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy. .footnote[ Access for free at https://openstax.org/books/chemistry-2e/pages/16-1-spontaneity ] --- ### Spontaneity <br> .largeish[ **Other Examples** * Reactions will always tend toward equilibrium spontaneously and not away * Highly soluble salts will always dissolve in pure water and disperse itself spontaneously and never reform into a solid * Iron will spontaneously rust in the presence of oxygen and water * A ball will always roll down a hill and not up * A hot pan on a stove that is turned off will always cool down to room temperature * The scent of perfume naturally disperses and never congregates back to the source * Dornshuld spontaneously screams when he stubs his toe ] --- class: center, middle [Entropy Overview](https://dornshuld.chemistry.msstate.edu/notes/ch16/entropy/entropy.html) --- ### Entropy Entropy, *S*, is an extensive property and state function (depends on *final* and *initial* states of system) **Entropy change**, Δ*S*, is a measure of how dispersed energy is. For an reversible process: `$$\Delta S = \dfrac{q_p}{T} = \dfrac{\Delta H}{T}$$` Here, a transfer of thermal energy at a certain temperature results in a dispersal of energy and causes an entropy change. The magnitude of Δ*S* depends on the ratio of heat to temperature. -- .pull-left[ **Analogy** Give $1,000 dollars to a needy person. Give $1,000 dollars to a billionaire. Which person was more significantly affected by this transfer of wealth? This "effect" is analogous to `\(\Delta S\)`. ] -- .pull-right[ **Analogy** Give 1 L of ice 334 kJ worth of heat and 0 °C and it becomes a liquid (at 0 °C). Give the sun 334 kJ worth of heat and the effect is trivial. Which object was more significantly affected by this transfer of heat? This "effect" is analogous to `\(\Delta S\)`. ] --- ### Entropy <br> **The *Zero*** Entropy lives on an absolute scale. The entropy of a perfect crystal at absolute zero is defined to be zero. <br> > Since entropy can be thought of as a measure of our lack of precise knowledge of the behavior of the many particles in a complex system (the less knowledge the higher the entropy) it means that we actually have perfect knowledge of the energy behavior of a system at absolute zero. This is equivalent to zero entropy. The practical effect of this is that it means we can assign a zero value to this state, which means complete knowledge. <div class="credit" style="text-align:right;"> — [Ronald L Klaus](https://www.quora.com/What-is-zero-entropy?page_id=1#!n=18), 2016 </div> <br> Entropy is measured in a variety of ways including measuring the heat required to raise the temperature of a reversible process a given amount. --- **A process is spontaneous if it increases the entropy of the universe** `$$\Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}} \qquad \mathrm{where}~~\Delta S = \dfrac{q}{T} = \dfrac{\Delta H}{T}$$` `\begin{align*} \Delta S_{\mathrm{univ}} > 0 &\quad \mathrm{for~a~spontaneous~process}\\ \Delta S_{\mathrm{univ}} < 0 &\quad \mathrm{for~a~nonspontaneous~process}\\ \Delta S_{\mathrm{univ}} = 0 &\quad \mathrm{for~a~system~at~equilibrium} \end{align*}` -- <br> <img src="spontaneous-melting.png" width="400px" style="display: block; margin: auto;" /> --- **Example: Will Ice Spontaneously Melt?** The entropy change for the process below is 22.1 J K<sup>–1</sup> and requires 6.025 kJ of heat from the surroundings. `$$\mathrm{H_2O}(s) \longrightarrow \mathrm{H_2O}(l)\quad \mathrm{at~0~^{\circ}C}$$` Is the process spontaneous at –10.00 °C? Is it spontaneous at 10.00 °C? <br> -- .pull-left[ At –10.00 °C `\begin{align*} \Delta S_{\mathrm{univ}} &= \Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}} \\[1.5ex] &= \dfrac{\Delta H}{T}_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}\\[1.5ex] &= \dfrac{-6.025\times 10^{3}~\mathrm{J}}{263.15~\mathrm{K}} + \dfrac{22.1~\mathrm{J}}{\mathrm{K}} \\[1.5ex] &= -0.8~\mathrm{J~K^{-1}} \end{align*}` No, it is not spontaneous at –10.00 °C. ] -- .pull-right[ At +10.00 °C `\begin{align*} \Delta S_{\mathrm{univ}} &= \Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}} \\[1.5ex] &= \dfrac{\Delta H}{T}_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}\\[1.5ex] &= \dfrac{-6.025\times 10^{3}~\mathrm{J}}{283.15~\mathrm{K}} + \dfrac{22.0~\mathrm{J}}{\mathrm{K}} \\[1.5ex] &= 0.8~\mathrm{J~K^{-1}} \end{align*}` Yes, it is spontaneous at 10.00 °C. ] --- **Example: Will Ice Spontaneously Melt?** The entropy change for the process below is 22.1 J K<sup>–1</sup> and requires 6.025 kJ of heat from the surroundings. `$$\mathrm{H_2O}(s) \longrightarrow \mathrm{H_2O}(l)\quad \mathrm{at~0~^{\circ}C}$$` So where did 22.1 J K<sup>–1</sup> come from? It comes from calorimetry! <br> -- `$$\Delta H_{\mathrm{fus}} = 6.025~\mathrm{kJ~mol^{-1}}$$` -- `\begin{align*} \Delta S &= \dfrac{\Delta H}{T}\\[1.5ex] &= \dfrac{6025~\mathrm{J}}{273.15~\mathrm{K}}\\[1.5ex] &= 22.1~\mathrm{J~K^{-1}} \end{align*}` --- ### Predicting entropy .pull-left[ We can predict if entropy increases or decreases with some straightforward rules of thumb. For example, entropy increases 1. as you go from a solid -> liquid -> gas 2. as you increase the number of moles of gas 3. with molecular weight (due to more electrons) and complexity ] -- .pull-right[ **Example 1:** Entropy is positive for the following process. `$$\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g)$$` **Example 2:** Entropy is negative for the following process `$$3\mathrm{H_2}(g) + \mathrm{N_2}(g) \longrightarrow \mathrm{2NH_3}(g)$$` since we are *reducing* the number of moles of gas from 4 to 2. **Example 3:** Which has larger entropy? * He(*g*) * Xe(*g*) Entropy is larger Xenon gas as it has a larger molecular weight (and, consequently, more electrons). ] --- <br><br><br><br> > .large["[Available energy is] the greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition." <div class="credit" style="text-align:right;"> — [Josh Williard Gibbs](https://upload.wikimedia.org/wikipedia/commons/c/c7/Josiah_Willard_Gibbs_-from_MMS-.jpg), 1873 </div> ] --- ### Standard State The **standard state** of a material (substance, mixture, or solution) specifies a reference point for pressure, concentration, and sometimes temperature * Pressure: 1 atm * Concentration: 1 *M* * Temperature: *varies*; usually 298.15 K (25 °C) but can be anything and is used as a reference point to calculate properties under standard and nonstandard conditions. Standard state values are denoted with a ° symbol: Δ*H*°, Δ*S*°, Δ*G*° -- **Examples: The standard state (with *T* = 25 °C) for** * water = H<sub>2</sub>O (*l*) * oxygen = O<sub>2</sub> (*g*) * carbon = C(*s*, graphite) -- It is more engergetically stable than C(*s*, diamond) .footnote[ See Thermodynamic values in [Appendix G](https://openstax.org/books/chemistry-2e/pages/g-standard-thermodynamic-properties-for-selected-substances) ] --- ## Gibbs Free Energy **Gibbs free energy**, Δ*G*, for a process (at constant *T* and *P*) is an alternative way to predict the spontaneity of a process since measuring the entropy change of the universe is difficult. It can be expressed as `$$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$` where every term is related to the system. Free energy is a measure of the *maximum* amount of work that can be extracted from a system based on the * *energy produced/consumed* by a process, Δ*H* * *energy gained from/lost to* the surroundings, *T*Δ*S* Free energy is another good definition for spontaneity. * Δ*G* < 0; spontaneous * Δ*G* > 0; nonspontaneous * Δ*G* = 0; at equilibrium --- ## Determining state function quantities Δ*G*, Δ*H*, and Δ*S*, are state functions; they only depend on the final (or products) and initial (or reactants) states of the system. $$\Delta X^{\circ} = \sum \nu \Delta X^{\circ} (\mathrm{products}) - \sum \nu \Delta X^{\circ} (\mathrm{reactants}) $$ -- **Example: Determine the standard enthalpy and entropy for the following process** `$$\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g)$$` Looking at the standard thermodynamic quantities given in [Appendix G](https://openstax.org/books/chemistry-2e/pages/g-standard-thermodynamic-properties-for-selected-substances), we see .pull-left[ `\begin{align*} \mathrm{H_2O}(l) \quad \Delta H^{\circ} &= -285.83~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 70.0~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}` `\begin{align*} \mathrm{H_2O}(g) \quad \Delta H^{\circ} &= -241.82~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 188.8~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}` ] -- .pull-right[ `\begin{align*} \Delta H^{\circ} &= (-241.82) - (-285.83)\\ &= 44.01~\mathrm{kJ}\\ \Delta S^{\circ} &= (188.8) - (70.0)\\ &= 118.8~\mathrm{J~K^{-1}} \end{align*}` ] --- **Example: Determine the standard enthalpy, entropy, and free energy changes (at 25 °C)** `$$\require{color}$$` `$$\mathrm{\textcolor{red}{3}H_2}(g) + \mathrm{N_2}(g) \longrightarrow \mathrm{\textcolor{blue}{2}NH_3}(g)$$` `\begin{align*} & \qquad \Delta H^{\circ} (\mathrm{kJ~mol^{-1}}) && S^{\circ} (\mathrm{J~mol^{-1}~K^{-1}})\\ \mathrm{H_2}(g) & \qquad 0 && 130.7 \\ \mathrm{N_2}(g) & \qquad 0 && 191.6 \\ \mathrm{NH_3}(g) & \qquad -45.9 && 192.8 \end{align*}` <br> -- `$$\Delta H^{\circ} = [\textcolor{blue}{2}(-45.9)] - [\textcolor{red}{3}(0) + (0)] = -91.8~\mathrm{kJ}$$` -- `$$\Delta S^{\circ} = [\textcolor{blue}{2}(192.8)] - [\textcolor{red}{3}(130.7) + (191.6)] = -198.1~\mathrm{J~K^{-1}}$$` -- `\begin{align*} \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta S^{\circ}\\ &= (-91.8~\mathrm{kJ}) - (298.15~\mathrm{K})(-0.1981~\mathrm{kJ~K^{-1}}) \\ &= -32.7~\mathrm{kJ~mol^{-1}} \end{align*}` --- ### Spontaneity and Temperature Dependence Changing the temperature of a process can sometimes affect its spontaneity (i.e. cause a sign change in Δ*G*). `$$\Delta G = \Delta H - T\Delta S$$` <br> <img src="s-h-g.png" width="550px" style="display: block; margin: auto;" /> For processes that are non/spontaneous at low/high temperature, you can determine the temperature at which the process "switches" in spontaneity by setting Δ*G* = 0 and solving for *T*. `$$T = \dfrac{\Delta H}{\Delta S}$$` --- **Example: Temperature Dependence on Spontaneity** What is the temperature (in °C) at which the following process switches between being nonspontaneous and spontaneous? `$$\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g)$$` <br> .pull-left[ `\begin{align*} \mathrm{H_2O}(l) \quad \Delta H^{\circ} &= -285.83~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 70.0~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}` `\begin{align*} \mathrm{H_2O}(g) \quad \Delta H^{\circ} &= -241.82~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 188.8~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}` ] -- .pull-right[ `\begin{align*} \Delta H^{\circ} &= (-241.82) - (-285.83)\\ &= 44.01~\mathrm{kJ}\\ \Delta S^{\circ} &= (188.8) - (70.0)\\ &= 118.8~\mathrm{J~K^{-1}} \end{align*}` `\begin{align*} T &= \dfrac{\Delta H}{\Delta S} \\ &= \dfrac{44.01~\mathrm{kJ}}{0.1188~\mathrm{J~K^{-1}}}\\ &= 370.5~\mathrm{K} = 97.3~\mathrm{^{\circ}C} \end{align*}` ] .footnote[ Recall that the thermodynamic values used are obtained at 298.15 K (25 °C). The answer here is in reasonable agreement with the actual boiling point of water (100 °C). ] --- ### Free Energy and Equilibrium Free energy, *Q*, and *K* are related. Free energy can be considered to be the "driving force" behind why a reaction proceeds in the direction that it does. * Δ*G* < 0 ; *Q* < *K* - reaction proceeds right * Δ*G* > 0 ; *Q* > *K* - reaction proceeds left * Δ*G* = 0 ; *Q* = *K* - reaction is at equilibrium -- Most reactions are carried out under nonstandard conditions. Standard free energy can be related to a nonstandard free energy for a reaction `$$\Delta G = \Delta G^{\circ} + RT \ln Q \qquad R = 8.315~\mathrm{J~mol^{-1}~K^{-1}}$$` -- If the system is at equilibrium, Δ*G* = 0 and *Q* = *K* to give `$$\Delta G^{\circ} = -RT \ln K \qquad R = 8.315~\mathrm{J~mol^{-1}~K^{-1}}$$` -- and Δ*G*° and *K* can be used to predict reactant or product favorability for a reaction * Δ*G*° < 0 ; *K* > 1 - products are more abundant at equilibrium * Δ*G*° > 0 ; *K* < 1 - reactants are more abundant at equilibrium * Δ*G*° = 0 ; *K* = 1 - reactants and products are comparably abundant at equilibrium --- **Example: Δ*G*, *Q*, and *K* ** Determine Δ*G* for the following process at 25 °C. Will the reaction proceed to the right or left? `$$\mathrm{3H_2}(g) + \mathrm{N_2}(g) \rightleftharpoons \mathrm{2NH_3}(g)$$` * Δ*G*° = –32.7 kJ mol<sup>–1</sup> * *Q* = 8.00 × 10<sup>5</sup> -- `\begin{align*} \Delta G &= \Delta G^{\circ} + RT \ln Q \\ &= (-32.7~\mathrm{kJ~mol^{-1}}) + (0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})\ln(8.00 \times 10^{5}) \\ &= 1.00~\mathrm{kJ} \end{align*}` Since Δ*G* is positive, the forward reaction is nonspontaneous and *Q* > *K*. Now determine *K*. -- .pull-left[ `$$\Delta G^{\circ} = -RT \ln K$$` ] .pull-right[ `\begin{align*} K &= e^{-\frac{\Delta G^{\circ}}{RT}}\\ &= e^{-\frac{-32.7~\mathrm{kJ~mol^{-1}}}{(0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}} \\ &= 5.35\times 10^5 \end{align*}` ] --- class: inverse, center, middle # Practice! [Acid/Base Relationships](http://dornshuld.chemistry.msstate.edu/courses/chem2/14-acid-base-relationships/#practice-problems-1) [Titrations](http://dornshuld.chemistry.msstate.edu/courses/chem2/14-titration-calculations/#practice-1) [Chapter 16 Exercises](https://openstax.org/books/chemistry-2e/pages/16-exercises) --- .pull-left[ <img src="spontaneous-melting.png" width="500px" style="display: block; margin: auto;" /> ] .pull-right[ <img src="spontaneous-freezing.png" width="500px" style="display: block; margin: auto;" /> ] --- class: center background-image: url("periodic-table.png") background-position: center background-size: contain