Thermo Diary - Fall 2011
Unit 4 - Thermodynamics - Daily Topics and Equations
As we progress through Thermodynamics we will have more and more equations for your to learn and memorize. I will start putting the equations here as we learn them. Best for you to start memorizing them ASAP. You must have JavaScript enabled in your browser in order to see the formulas correctly.
Wed, 11/2
Heat and work are our two most "tangible" forms of energy. When I say "tangible", I mean that both heat and work are easily measured for a given process. Work (\(w\)) can be defined several ways, but for us chemists, we prefer the idea of expansion work. Expansion work is due to the expansion or compression of a gas. We will define it with the following equation:
\[w=-P\Delta V \]Also remember that on a micro-scale (atoms and molecules) that work is organized molecular motion, and heat is disorganized or random molecular motion. There are various ways to measure heat (\(q\)) - those are shown on Friday, 11/4 (look ahead).
The internal energy (\(E\)) of a system is ALL the types of energies present combined. Internal energy IS a state function. We will usually be most concerned with changes in internal energy. Stated mathematically:
\[\Delta E = E_{\rm f} - E_{\rm i}\]And now, applying the first law we get...
\[\Delta E = q + w \]Internal Energy, \(E\), is also shown as \(U\) in many books and on Quest. Therefore:
\(\Delta U = U_{\rm f} - U_{\rm i}\) \(\Delta U = q + w\)
It is very important to get the sign right on heat and work. Here is the breakdown:
\(+q\) heat is absorbed by the system (into the system)
\(-q\) heat is released by the system (out of the system)
\(+w\) work is done on the system (volume compression)
\(-w\) work is done by the system (volume expansion)
Fri, 11/4
3 Types of Systems used in Thermodynamics...
- open system matter and energy freely exchange with surroundings
- closed system only energy exchanges with surroundings (matter confined to the system)
- isolated system NO exchange of matter or energy with surroundings
Enthalpy is defined: \(H = E + PV\) which means that (at constant pressure) \(\Delta H = \Delta E + P\Delta V\), or rewritten to be \(\Delta E = \Delta H - P\Delta V\).
\(\Delta E = q + w \) at constant pressure becomes,
\(\Delta E = q_P - P\Delta V \), which means that \(\Delta H = q_P\)
\(\Delta E = q + w \) at constant volume becomes, \(\Delta E = q_V\)
Various Ways to Calculate Heat
For any substance not changing phase,
\[q = m\cdot C_{\rm s}\cdot \Delta T\]Note that \(m\) is mass in grams and \(C_{\rm s}\) is specific heat capacity in J/g °C.
This can also be done using mole instead of grams:
\[q = n\cdot C_{\rm m}\cdot \Delta T\]Where \(n\) is the number of moles and \(C_{\rm m}\) is the molar heat capacity in J/mol °C.
You can also calculate heat via a phase change like melting ice to water. For that you will need the enthalpy of the change and the calculation is just:
\[q = m \cdot \Delta H_{\rm change}\]The word "change" above would be the actual type of change: melting, fusion, vaporization, condensation, sublimation.... Can also be written in molar form.
Finally, you can calculate heat via an electric heater. Find the wattage and then multiply by the time to get heat energy:
\[q = P \cdot t\]Where \(P\) is power in watts (1 watt = 1 J/s) and \(t\) is time in seconds (s).
Mon, 11/7
Relating ΔU to ΔH
Remember there are two formulas relating \(\Delta U\) to \(\Delta H\):
\[\Delta U = \Delta H - P\Delta V\] \[\Delta U = \Delta H - \Delta nRT\]The differences above are from the work term which comes from the expansion or compression of gases. Use the top one (\(-P\Delta V\)) for when there is an obvious volume change and a given external pressure. Use the bottom equation (\(-\Delta nRT\)) when you are just looking at a balanced chemical equation. In the equation you HAVE to count gas products and gas reactants in order to calculate \(\Delta n\).
\(\Delta n\) = (mol of gas products) - (mol of gas reactants)
Sometimes the \(\Delta n\) term is explicitly written as \(\Delta n_{\rm gas}\).
Calorimetry
DO check out the Calorimetry Help Sheet that is posted on our Help Sheet pages.
\(q_{\rm cal} = m_{\rm water} \cdot C_{\rm s, water} \cdot \Delta T\) (coffee cup calorimetry)
\(q_{\rm cal} = m_{\rm water} \, C_{\rm s, water} \, \Delta T + C_{\rm hardware}\Delta T\) (bomb calorimetry)
Wed, 11/9
Gas Heat Capacities - Cv and Cp
Gases are unique in that that have distinctly difference heat capacities depending on whether you heat at constant volume or at constant pressure. Solids and liquids do not show this type of differences.
The relationship between \(C_V\) and \(C_P\) for gases is shown in this equation:
\[C_P = C_V + R\]You also get 3 different values for \(C_V\) and for \(C_P\) depending on what form of ideal gas you have. This is explained best via the Equipartition Theorem which says to equally distribute the overall energy across ALL the degrees of freedom. The degrees of freedom are translational modes (3 total for x, y, and z directions), and rotational modes (rotations about the x, y, and z axes. A single atom is too small to have any significant rotational inertia and therefore only has the 3 modes of translational motion. A diatomic (or linear) molecule can have rotational inertia about 2 axes. A polyatomic non-linear molecule can have all 3 rotational modes occupied. Each mode of freedom has \({1\over2}RT\) (J/mol). This is what you sum up to get the overall motional contribution to the internal energy of an ideal gas.
| monatomic | linear or diatomic |
polyatomic non-linear |
|
| modes of freedom trans + rot |
3 + 0 = 3 | 3 + 2 = 5 | 3 + 3 = 6 |
| \(C_V\) | \({3\over2}R\) | \({5\over2}R\) | \({6\over2}R\) |
| \(C_P\) | \({5\over2}R\) | \({7\over2}R\) | \({8\over2}R\) |
FYI: Vibrational modes are only available at very very high temperatures. For this reason we do not quantify the vibrational modes of freedom in these ideal gases.
Also remember that you CAN calculate the internal energy or enthalpy of an ideal gas at any temperature using:
\[E = n\ C_{\rm v}\ T\] \[H = n\ C_{\rm p}\ T\]
Just substitute the appropriate heat capacity given in the above chart.
Introduction to Hess' Law:
\[\Delta H_{\rm rxn} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots \]Fri, 11/11
Hess' Law
You can combine any number of reactions (steps) to equal another "overall" reaction. You also sum the energies involved to get the overall energy. This is show via Hess' Law for enthalpy change:
\[\Delta H_{\rm rxn} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots \]When you "flip" a reaction (switch reactants and products) you must change the sign on \(\Delta H\). If you scale a react up or down (double it, half it, etc...), you must also scale the value of \(\Delta H\).
You can use \(\Delta H_{\rm f}^\circ\) for the rxn steps in Hess' Law.
\[\Delta H_{\rm rxn}^\circ =\sum{n\Delta H^\circ_{\rm f} (\rm products)} - \sum{n\Delta H^\circ_{\rm f} (\rm reactants)}\] Note that the little subscript "f" here means "of formation". A formation reaction is one that produces that compound from just the elements. Please have a look at my Formation Reactions Help Sheet from our Help Sheets page.
You can also get a good approximation of \(\Delta H\) via the summation of bond energies. The concept is the same regardless of what book (or source) you read it from. Here are a few...
From Zumdahl, Chapter 13, section 8. The "D" is for the dissociation energy of the bond. \[\Delta H =\underbrace{\sum{D\:{\rm (bonds\;\;broken)}}}_{\rm energy\;\; required\uparrow} - \underbrace{\sum{D\:{\rm (bonds\;\;formed)}}}_{\rm energy\;\;released\downarrow}\]
From Aktins/Jones, they use "mean bond enthalpies", \(\Delta H_{\rm B}\) \[\Delta H_{\rm rxn}^\circ =\sum{n\Delta H_{\rm B} (\rm reactants)} - \sum{n\Delta H_{\rm B} (\rm products)}\] Of course when using bond energies or enthalpies, you should break only the bonds that need breaking and make the bonds that need making. You don't have to break every bond in the molecule if much of its structure is retained after the reaction.
Mon, 11/14
Entropy (\(S\)) is a measure of energy dispersal. It is a state function. It has units of energy/temp or J/K. When expressed as J/K, entropy is extensive and will scale with the amount of substance. It can also be expressed as J/mol K and that is an intensive property.
Spontaneity: A spontaneous process will proceed forward with no external help. Think self-sufficient in energy (ball rolling down a hill). Non-spontaneous processes are the opposite - they tend to not proceed forward UNLESS there is direct help via energy/work to drive the process forward (ball being pushed back up the hill).
The 2nd Law of Thermodynamics: For any spontaneous process (change), the entropy of the universe must increase. \[\Delta S_{\rm univ} = \Delta S_{\rm sys} + \Delta S_{\rm surr}\]
Note that \(\Delta S_{\rm sys}\) can be negative and the process still be spontaneous as long as \(\Delta S_{\rm surr}\) is positive and a bigger value.
The thermodynamic definition of entropy change is the ratio of the reversible heat flow (\(q_{\rm rev}\)) divided by the absolute temperature:
\( \Delta S = {q_{\rm rev}\over T} \)
A reversible process is one in which the direction of change is always easily redirected (pushed backwards or forwards) by an infintesimal amount of change.
Consider heating a substance such as a gas at constant pressure. The reversible heat is calculated via:
\(q_{\rm rev} = n\ C_{\rm p}\ \Delta T \)
Substituting in for \(q_{\rm rev}\) in the above equation and considering only infintesimal change gives the following equation:
\[{\rm d}S = {n\ C_{\rm p}\ {\rm d}T \over T }\] \[\int_1^2{\rm d}S = n\ C_{\rm p}\ \int_{T_1}^{T_2}{{\rm d}T \over T }\] \[\Delta S = n\ C_{\rm p}\ \ln\left({T_2 \over T_1}\right)\]
So that equation is how to calculate \(\Delta S\) for the change when heating a substance from one temperature to another temperature at constant pressure. Use \(C_{\rm v}\) if you are heating a gas at constant volume. You can even use mass and specific heat if you want:
\[\Delta S = m\ C_{\rm s}\ \ln\left({T_2 \over T_1}\right)\]
Wed, 11/16
Isothermal Expansion of an ideal gas. You stay on the isotherm throughout the entire expansion which means that \(\Delta E = 0\) and \(q = -w\). Also, the external pressure has to constantly match the dropping system pressure - this will result in a reversible process. So this is really isothermal reversible expansion. P is a function of volume (\(P = nRT/V\)) and so one can integrate \({\rm d}w = -nRT\;{\rm d}V/V\) to give:
\(w = -nRT \ln \left({V_2\over V_1}\right)\) \(q = nRT \ln \left({V_2\over V_1}\right)\)
And, Boyle's Law is in effect so that \(P_1/P_2 = V_2/V_1\) and therefore:
\(w = -nRT \ln \left({P_1\over P_2}\right)\) \(q = nRT \ln \left({P_1\over P_2}\right)\)
Entropy (\(S\)) is really a measure of the number of microstates (\(W\)) in the system:
\[S = k \ln W\]Where \(k\) is the Boltzmann Constant (\(1.38\times 10^{-23}\) J/K) which is the same as the universal gas constant, \(R\), divided by Avogadro's number, \(N_{\rm A}\). It is easy to associate the number of available microstates with the available volume (\(V\)). For example, if you double the volume, you will double the number of available microstates. From this, we can consider state1 and then state2 and we get: \[ \Delta S = nR \ln \left({V_2\over V_1}\right) \]
That formula is much like the formula for heat (\(q\)) transfer during isothermal expansion and you can rightfully conclude that: \[ \Delta S = {q_{\rm rev}\over T} \]
Fri, 11/18
The 3rd Law of Thermodynamics: For any perfectly crystalline substance at absolute zero, the entropy is zero.
Note this established the minimum for \(S\) that any substance could have. All absolute entropies are positive. Standard absolute entropy (\(S^\circ\)) is what is put in thermodynamic tables. Please contrast this to the other two state functions shown in tables (\(\Delta H_{\rm f}^\circ\) and \(\Delta G_{\rm f}^\circ\)) which are "formation" reaction values.
Entropy change for a reaction:
\[\Delta S_{\rm rxn}^\circ = \sum n S^\circ({\rm prod}) - \sum n S^\circ({\rm react}) \]
The Gibb's Free Energy (\(G\)) is another state function that is defined from 3 others: \(G = H - TS\). When we hold temperature and pressure constant we get:
\[\Delta G = \Delta H - T\Delta S \]We typically are getting the \(\Delta X\)'s in the above equation from Standard Thermodynamic Data Tables like you have in Appendix 4 in your textbook. If you are at standard state you add the superscript "°"'s to the terms:
\[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]Of course this means that the temperature must be 25°C or 298.15 K in the above equation. You can also show through a simple proof that \(\Delta G\) (a system state function) is equal to \(-T\Delta S_{\rm univ}\) (I showed this in class - see the book sections 10.6 and 10.7). So we can now use \(\Delta G\) as our indicator of spontaneity for a reaction or process.
Definition of Thermodynamic equilibrium: \(\Delta G = 0\)
Also, we can assume that plain ol' \(\Delta G\) (non-standard conditions, no little circle, \(\circ\) ) is approximately equal to the temperature adjusted standard values for \(\Delta H^\circ\) and \(\Delta S^\circ\) via temperature:
\[ \Delta G \approx \Delta H^\circ - T\Delta S^\circ \]This is possible because we assume that \(\Delta H\) and \(\Delta S\) do NOT change (much) with temperature.
Now, if you assume you're at equilibrium, then \(\Delta G = 0\), and you get the following:
\(\displaystyle{\Delta H = T\Delta S}\) \(\displaystyle{\Delta S = {\Delta H \over T}}\) \(\displaystyle{T = {\Delta H \over \Delta S }}\)
In each of these cases, you can substitute the standard values (\(\Delta H^\circ\) and \(\Delta S^\circ\)) in for the "plain" values.
Mon, 11/21
Know the differences in plain ol' \(\Delta G\) and standard \(\Delta G^\circ\). The standard \(\Delta G^\circ\) means that the temperature is 25°C and all gases are 1 atm and all concentrations are 1 M.
Plain \(\Delta G\) is constantly changing during the course of a reaction. Always headed toward ZERO if there is a way to get there. ANY set of conditions is possible for "plain" \(\Delta G\). When the conditions reach the equilibrium state, that is when \(\Delta G = 0\). This is the thermodynamic definition of equilibrium using a system based state function.
\[ \Delta G = \Delta H - T\Delta S \]
Realize there are FOUR possible combinations of sign for \(\Delta H\) and \(\Delta S\) in this equation. Those combo's are (+,+), (-,-), (+,-), and (-,+). Each combination leads to different sign possibilities on \(\Delta G\).
Also remember that \(\Delta G\) and \(\Delta H\) are usually given in kJ/mol while \(\Delta S\) is given in J/mol K. Make sure you convert J to kJ before you combine the entropy term and the enthalpy term in this equation.
Other Odds and Ends...
When heating or cooling a gas, there are two methods: at constant volume and at constant pressure. This leads to TWO versions of calculating entropy change:
\[ \Delta S = n C_{\rm v} \ln \left({T_2\over T_1}\right) \] \[ \Delta S = n C_{\rm p} \ln \left({T_2\over T_1}\right) \]
Note that you use \(C_{\rm v}\) if done at constant volume, or \(C_{\rm p}\) if done at constant pressure.
Also, remember that heat can flow and temperature can remain constant - like for phase changes. For phase changes \(q_{\rm transition} = \Delta H_{\rm transition}\). This means that you can get entropy change via:
\[ \Delta S_{\rm transition} = {\Delta H_{\rm transition}\over T_{\rm transition}}\]
"Transition" here means vaporization, or fusion, or sublimation, etc... and the temperature at transition is just the melting point (fusion) or the boiling point (vaporization).