As we progress through Thermodynamics we will have more and more equations for you to learn and memorize. This page is a very condensed version of all our thermodynamic concepts and equations. It should be used in a review type manner and not used as your primary source of the material.
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:
And, thanks to the ideal gas law, that change in volume could easily be because of a change in moles of gas (Avogadro's Law) which means work can also be:
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 a few sections later (look ahead).
The internal energy (\(U\)) 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:
And now, applying the first law we get...
Internal Energy, \(U\), is also shown as \(E\) in many books and on Quest - this is "old school" and not IUPAC recommended but you should be aware that it is still in use.
It is very important to get the sign right on heat and work. Here is the breakdown:
\(+q\) heat flows INTO the system (adsorbed)
\(-q\) heat flows OUT OF the system (released)
\(+w\) work is done ON the system (volume compression)
\(-w\) work is done BY the system (volume expansion)
Enthalpy is defined: \(H = U + PV\) which means that (at constant pressure)
\(\Delta U = q + w \), and at constant pressure becomes,
This means that we now have the definition of enthalpy change:
And, conversely, if you hold volume constant, work is zero and we learn that:
For any substance not changing phase,
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:
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:
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:
Where \(P\) is power in watts (1 watt = 1 J/s) and \(t\) is time in seconds (s).
Remember there are two formulas relating \(\Delta U\) to \(\Delta H\):
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 \(\Delta n_{\rm gas}\).
DO check out the Calorimetry Help Sheet that is available in our eBook.
For constant pressure calorimetry (isobaric or "coffee-cup"):
For constant volume calorimetry (isochoric or "bomb"):
Note how you must also account for the hardware component of the calorimeter in bomb calorimetry.
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:
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.
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.
From Aktins/Jones, they use "mean bond enthalpies", \(\Delta H_{\rm B}\)
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.
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:
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:
Substituting in for \(q_{\rm rev}\) in the above equation and considering only infintesimal change gives the following equation:
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:
When heating or cooling the entropy change results only from temperature change.
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:
"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).
Entropy (\(S\)) is really a measure of the number of microstates (\(\Omega\)) in the system:
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, \(N_{\rm A}\). Note that this equation will give you ZERO entropy if you have only ONE microstate. That is the idea behind the "3rd Law" of thermodynamics.
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:
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:
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:
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:
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.
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.
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.