Review Sheet - Exam 4 - McCord CH301

Hydrocarbons and their Combustion Reactions

Make sure you know how to completely write out the chemical equation (balanced) for a combustion reaction. Remember that all hydrocarbons (alkanes) react with oxygen gas to give carbon dioxide and water. When done at 25°C, the water will be liquid, at or above 100°C the water will be gas phase. All the C's become CO₂'s and all the H's become H₂O's. Any oxygen in the formulas just become one of the many oxygens in the CO₂ and H₂O.

Also know the "heat of combustion" is the amount of heat released from a combustion reaction. Heats of combustion are always positive because we define them as the "amount of heat" released. The actual \(\Delta H\) for a combustion reaction will be negative because they are all exothermic. Do not confuse the two - the amounts are identical but the signs are different.

State functions

State functions are qualities or states of a system that are independent of the path by which the system arrived. Anything you can measure as matter “just sits there” is a state function. The main state functions that we constantly concern ourselves with are composition (formula), mass, volume, pressure, and temperature. We have very specific ways to measure each of these. Some ways are easier than others – temperature is easier than composition. In our study of thermodynamics we introduce four more state functions, all of which are pieces of the energy puzzle of matter. The four are internal energy (\(E\)), enthalpy (\(H\)), entropy (\(S\)), and free energy (\(G\)). Each of these energy based state functions will have a unique value for a given set of all the other state functions previously mentioned. What I mean is that if I know that I’ve got say exactly 30 grams of CaCO₃ at 25°C under 1 atm pressure then I also am aware of the fact that it also has an exact amount of internal energy, enthalpy, entropy, and free energy also. Do I know those actual values? Maybe, but usually not. For example, I do know a specific formula for the internal energy of an ideal monatomic gas, it is simply

\[E = {3\over 2}RT \]

This is a rare case where we can get the absolute value for \(E\). Typically, real substances are not always gases and are certainly not “ideal”. Real substances have real intermolecular attractions/repulsions that push these energetic state functions to all sorts of possibilities. So how do we get these thermodynamic state functions to reveal themselves? We initiate a change and in doing so we measure \(\Delta E\), \(\Delta H\), \(\Delta S\), and \(\Delta G\). All these \(\Delta\)-values reveal themselves when matter undergoes change. Any physical or chemical change within matter will yield probable changes in these states functions (though not always).

Heat and Work

Heat (\(q\)) and work (\(w\)) are not state functions. They are very much path dependent in their values. We will only consider these two forms of energy in our studies. We will also only consider expansion work. Realize there are other forms of energy and different types of work. However, we need not concern ourselves with those until the need arises. For typical endo- and exothermic chemical reactions and physical changes, heat and expansion work are the only energies that we need to track and/or measure.

Heat Capacity

Heat capacity (\(C\)) is the ratio of how energy changes (heat flow) as temperature changes. There are 2 types of heat capacities that we use, one for constant volume processes (which tracks \(q_{\rm v}\)) and one for constant pressure processes (which tracks \(q_{\rm p}\)).

\[\Delta E = q_{\rm v} = n C_{\rm v} \Delta T\] \[\Delta H = q_{\rm p} = n C_{\rm p} \Delta T\]

It is these equations that are used for calorimetry (see Calorimetry help sheet). It is also these equations (the constant pressure one in particular) that we use to track heat in ANY substance (solid, liquid, or gas). Each phase will have its own unique heat capacity. Tables will list heat capacities in 2 different ways or amounts. Many are listed on a per mole basis – these are molar heat capacities. Many are also listed on a per gram basis – these are specific heat capacities. Always look at the units on a measurement to know whether it is per mole or per gram or even per device (like an entire calorimeter). Units will tell you what to do. Once again, knowing the difference in extensive properties (heat capacity, \(C_{\rm p}\) in J/K) and intensive properties (molar heat capacity, \(C_{\rm p}\) in J/mol K) will always help you out.

Thermochemistry

Do know how to calculate amount specific \(\Delta H\)’s when given the general \(\Delta H\) of a reaction. Enthalpies of combustion are like this. If fuel A burns to yield 900 kJ of heat per mole of A, then burning 2.5 moles of A would yield 2250 kJ of heat (\(900\times 2.5\)). This is simply reaction stoichiometry with energy listed as another reactant (endothermic) or product (exothermic). This also applies to physical changes (melting and boiling, Chapter 16 section 10 , “Changes of State”).

Hess' Law (3 versions)

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.

IMPORTANT: Just remember you must ADD energy to the system (+, endothermic) to break bonds (reactants) and then the system releases energy (-, exothermic) when new bonds are formed (products). It is this reason that the reactants are listed first in the equation above and the products last. This is unlike all the other equations we use.

Boltzmann’s Formula (statistical definition of entropy)

You want a straight up definition of entropy? Then here you go...

\[S = k {\rm ln}W\]

Where \(W\) is the number of microstates for a system and \(k\) is the Boltzmann constant – really just the same as \(R\) except for single molecules instead of moles of molecules. So all you need is the number of microstates and you can get the absolute entropy for any system. How do you count microstates you ask? Well, that is not really that easy. If you pick a real simple system and confine it greatly (say a temperature of absolute zero so that all thermal energies states are zeroed out) then you might have just one possible microstate and then have zero entropy (see “The 3rd Law” on the next page).

However, even at ZERO kelvin there can be more than one microstate based on different positional locations and orientations of molecules. This non-zero entropy at zero kelvin is called residual entropy. We can easily get bogged down here in the statistical complications of probabilities - so what to do? KNOW that yes, positive entropy changes (2nd Law) always favor the most probable outcome. Read through the derivation of all this in section 10.3 and get to the end of it all. Luckily, we (as chemists) can measure changes in entropy via reversible heat flow and temperature.

Isothermal change and Entropy

There are 2 basic equations for \(\Delta S\) for isothermal changes. First, if the change is isothermal expansion of a gas you get

\[\Delta S = nR \ln \left({V_2\over V_1}\right)\]

Note that \(\left({P_1\over P_2}\right)\) can be substituted in for \(\left({V_2\over V_1}\right)\) thanks to Boyle’s Law. Now, if the isothermal change is due to a phase change (a transition), then \(q_{\rm rev}\) is the same as \(\Delta H_{\rm trans}\) and we simply get

\[\Delta S_{\rm trans} = {\Delta H_{\rm trans}\over T}\]

Entropy via Look-up Tables

Also, like with Hess’s Law and \(\Delta H_{\rm f}^\circ\)’s, we can use table values for standard molar entropy (\(S^\circ\)) and get the change of entropy for a reaction:

\[\Delta S_{\rm rxn}^\circ =\sum{n S^\circ (\rm products)} - \sum{n S^\circ (\rm reactants)}\]

Note how those are absolute entropies (\(S\)’s not \(\Delta S\)’s). Absolute entropies are possible due to the 3rd law which establishes conditions for true ZERO entropy.

Equilibrium

Chemical equilbrium is an example of a dynamic equilibrium and not static equilibrium. Know the difference in the two. Static equilibrium is fixed and non-changing – like balancing weights on a balance beam. Dynamic equilibrium has no NET overall change but does have some given processes still proceeding. The process itself proceeds both forwards and backwards at exactly the same rate. Any thing that you are constantly depleting via one process is simultaneously being replenished by another process. Stated chemically, equilibrium is achieved when the forward rate of reaction equals the reverse rate of reaction. That is a purely kinetic argument for equilibrium and we will study reaction kinetics in Chapter 15 in our book (that’s CH302). A complete understanding of equilibrium requires knowledge of both arguments (definitions) for the equilibrium state. The end of Chapter 10 focuses on the other definition of equilibrium which is based purely on thermodynamic state functions. Lets get the thermodynamic argument for equilibrium established though.

The bottom line for the thermodynamic argument lies in the spontaneity of a reaction. 2nd Law dictates what direction of change is the spontaneous direction. We know that indicator to be universal entropy. If \(\Delta S_{\rm univ}\) is positive, you have found a spontaneous process as written. Whatever is positive one direction must be negative going the other direction. One way, spontaneous (downhill) and the other way, non-spontaneous (uphill). There IS a case right in between – neither up or downhill, flat so to speak. Both directions are equally likely to proceed forward. When conditions are met like this then you have a stalemate on universal entropy. It’s that special condition where \(\Delta S_{\rm univ}=0\) and it is the result of achieving equilibrium. Unfortunately tracking both the system and the surroundings to get the universal entropy is a bit tedious. Let’s get ourselves a new state function for the system that allows us to track in a relative way the universal entropy. That new state function is the Gibb's free energy, \(G\), defined as:

\[G = H - T\;S\]

First notice that \(G\) is made up from 3 other state functions. Also check out how \(G\) will change with \(T\). LOOK at the equation, as \(T\) increases, \(G\) must decrease. That was \(G\)... now, with a little reasoning, we find out that we can now switch to \(\Delta G_{\rm sys}\) to track spontaneity. If we hold temperature (\(T\)) and pressure (\(P\)) constant, we get the following for the change in free energy:

\[\Delta G = \Delta H - T\Delta S\]

This allows us to track spontaneity with a purely system state function, \(\Delta G\). It tracks via sign the opposite of the way that \(\Delta S_{\rm univ}\) does because \(\Delta G = -T\Delta S_{\rm univ}\) (see section 10.7 in your text for proof). Now consider the 3 possible outcomes for \(\Delta G\):

We now have a new standard to judge spontaneity and equilibrium. ALL equilibrium processes must have a free energy change equal to zero. This is the same as saying that all the free energies (that’s plain ol’ \(G\) here) of all the reactants must equal the free energies of all the products – our “stalemate” condition for equilibrium.

Also note how \(\Delta G\)’s sign is controlled by the signs on \(\Delta H\) and \(\Delta S\). There are 4 cases here which are analagous (but opposite sign!) to those for \(\Delta S_{\rm univ}\) which are given in Table 10.6 in section 10.6 of your book. Here is a plot showing the 4 possible cases:

Free energy is a state function and therefore can be calculated via the free energies of formation of the reactants and products just like the enthalpy of reaction was:

\[\Delta G_{\rm rxn}^\circ =\sum{n\Delta G^\circ_{\rm f} (\rm products)} - \sum{n\Delta G^\circ_{\rm f} (\rm reactants)}\]

Most thermodynamic tables include \(\Delta H^\circ_{\rm f}\), \(\Delta G^\circ_{\rm f}\), and \(\Delta S^\circ\), but sometimes you might NOT have \(\Delta G^\circ_{\rm f}\) (like on an exam) and you should know how to calculate \(\Delta G\) from \(\Delta H\) and \(\Delta S\) using the familiar equation:

\[\Delta G = \Delta H - T\Delta S\]

The standard version looks like this:

\[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]

And since \(\Delta H\) and \(\Delta S\) don’t change much with temperature, you can use any temperature (within reason) and calculate non-standard \(\Delta G\). That is:

\[\Delta G \approx \Delta H^\circ - T\Delta S^\circ \]

Remember if you have an equilibrium process occurring, then \(\Delta G = 0\) 
and therefore

\(\displaystyle{\Delta H = T\Delta S}\)        \(\displaystyle{\Delta S = {\Delta H \over T}}\)        \(\displaystyle{T = {\Delta H \over \Delta S }}\)