From WolfWikis

LECTURE 6: Chemical thermodynamics

Enthalpy

Section 19-7

An important point is that volume work -PdV is only one kind of work. It is the important one for gases but for most other systems we are interested in other kinds of work (e.g. electrical work in a battery).

A good way to measure ΔU's is to make sure there are no work terms at all. If so:

ΔU_{no work} = q +w = q+0 = q.

However, this means that the -PdV volume work term should also be zero and this implies we must keep volumes the same. That can actually be hard. Therefore we define at new state function ENTHALPY

H ≡ U + PV; ( The ≡ symbol is used to show that this equality is actually a definition.)

If we differentiate we get:

dH = dU + d(PV) = dU + PdV + VdP

We know that under reversible conditions we have

dU = δw +δq = -PdV + δq (+ other work terms that we assume zero)

Thus,

dH = -PdV +δq + PdV + VdP

dH = δq + VdP

That means that as long as there is no other work and we keep the pressure constant:

ΔH = q_P, instead of ΔU= q_V

Working at constant P is a lot easier to do than at constant V. This means that the enthalpy is a much easier state function to deal with than the energy U.

For example when we melt ice volumes change whether we like or not, but at long as the weather does not change too much pressure is constant. So if we measure how much heat we need to add to melt a mole of ice we get the molar heat of fusion:

Such enthalpies are measured and tabulated.

In this case the volume change is actually quite small, as it usually is for condensed matter. Only if we are dealing with gases is the difference between enthalpy and energy really important

So:

U ~ H for condensed matter

U and H are different for gases.

A good example of this is the difference between the heat capacity at constant V and at constant P. For most materials there is not much of a difference, but for an ideal gas (see 19.42) we have

C_p = C_V + nR

Needless to say that the heat capacity is a path function: it depends on what you keep constant.

Determining enthalpies from heat capacities.

The functions H and C_p are related by differentiation:

This means that we can:

1. measure C_p as a function of temperature
2. integrate this function and find H(T)

However, there are problems with this approach:

1. Reference point: we have to deal with the lower limit of integration.

   Ideally we start at zero Kelvin (but we cannot get there), but how do we compare one compound to the other?

2. At temperatures where there is a phase transition there is a sudden jump in enthalpy. E.g. when ice melts we have to first add the heat of fusion until all ice is gone before the temperature can go up again (assuming all is done under reversible well-equilibrated conditions).
3. At the jumps in H, the C_p is infinite.

It should be stressed that there are no absolute enthalpies. All that is properly defined are differences in enthalpy ΔH and these are only defined for processes

For example for the process of:

Process 1: heating ice from -20^oC to -0^oC

We could write ΔH₁ = ∫ C_P,ice. dT from T=253K to T=273K = H(273) -H(253).

But before moving beyond the melting point first a different process needs to take place, that of

Process 2melting

This gives us Δ_fusH = H_liquid - H_solid (both at 273K!).

Whe we heat the liquid water further to say +20^oC we would have to integrate over the heat capacity of the liquid.

Process 3: heating water from 0^oC to +20^oC

We could write ΔH₃ = ∫ C_P,water. dT from T=273K to T=293K = H(293)-H(273).

The total change in enthalpy between -20 and +20 would be the sum of the three enthalpy changes.

ΔH_{total process} = ΔH₁ + Δ_fusH + ΔH₃

Of course we could consider doing the same calculation for any temperature between -20 and +20 and summarize all our results in a graph. The three processes can thus schematically be shown in the following graph.

Schematic enthalpy function showing the jump at the melting point

Notice that the slopes (i.e. the heat capacities!) before and after the melting point differ. The slope for the liquid is a little steeper because the liquid has more degrees of freedom and therefore the heat capacity of the liquid tends to be higher than of the solid. In the figure the enthlapy curves are shown as straight lines. This would be the case if the heat capacities are constant over the temperature interval. Althoug C_p is typically a 'slow' or 'weak' function of temperature it usually does change a bit, which means that the straight lines for H become curves.

Notice that for process two, the temperature is constant, that means dat ΔT or dT is zero, but ΔH is finite, consequently ΔH/ΔT is infinitly large. Taking the limit for ΔT going to zero (eh, we are actually already there..) we get a derivative:

This derivative, the heat capacity must undergo a singularity: the slope is infinitely large (The H curve goes straight up).

When there are more phase transtitions, more discontinuties in H and singularities in C^p result. See figure 19.7 for an example. (Note that H(T)-H(0), not H(T) is plotted to avoid the question what the absolute enthalpy is)

Scanning calorimetry

There is technique that allows us to measure the heat capacity as a function of temperature fairly directly. It is called Differential Scanning Calorimetry (DSC). You put a sample in a little pan and put the pan plus an empty reference pan in the calorimeter. The instrument heats up both pans with a constant heating rate. Both pans get hotter by conduction, but the heat capacity of the filled pan is obviously bigger. This means that the heat flow into the sample pan must be a bit bigger than into the empty one. This differential heat flow induces a tiny temperature difference ΔT between the two pans that can be measured. This temperature difference is proportional to the heat flow difference which is proportional to the heat capacity difference.

ΔT ~ ΔΦ ~ Δ_{between pans}C_p = C_p^sample (if the pans cancel)

However, there are number of serious broadening issues with the technique. If you melt something you will never get to see the infinite singularity of the heat capacity. Instead it broadens out into a peak. If you integrate the peak you get the Δ_fusionH and the onset is calibrated to give you the melting point.

The ideal heat capacity signal and its broadened DSC signal

It is even possible to heat the sample with a rate that fluctuates with a little sine wave. This version (Modulated DSC) can even give you the (small) difference in C_p before and after the melting event.

Heat of chemical reactions

Of course we can consider the ΔH for a whole bunch of different processes, like changing temperature, changing pressure, changing composition, changing phase (melting, evaporating, subliming etc.). (See the table 19.1 for notations). A particulary important application -particularly for us chemists- of thermodynamics is the one that describes the heat and work of chemical reactions.

Chemical reactions can both give off ('exothermic) and absorb heat (endothermic). We usually study them at constant pressure so it is easier to consider reaction enthalpies than energies. If during the reaction the number of moles of gas is subject to change there may be a considerable difference (otherwise not).

If heat is produced we assign a negative value to the reaction enthalpy

Δ_rH= H_prod - H_reactants

Formation enthalpy

A special case of a reaction enthalpy is the enthalpy of formation. Δ_fH This is the reaction enthalpy of a (hypothetical) reaction where one mole of a substance is formed out of its constituent elements. As a matter of convention the enthalpy of formation of the elements are arbitrarily assumed to be zero, at least in their most stable form (for carbon that is graphite not diamond).

Heat of combustion

Another common reaction enthalpy that has its own symbol Δ_cH is the combustion enthalpy. This is a complete reaction with oxygen that produces nothing but water, carbon dioxide (no soot or monoxide!) and other oxides.

Hess's law

As enthalpy and energy are state functions we should expect additivity of U and H when we study chemical reactions. This additivity is expressed in Hess's Law. The additivity has important consequences and the law finds wide spread application in the prediction of heats of reaction.

1. The reverse reaction has the negative enthalpy of the forward one.
2. If we can do a reaction in two steps we can calculate the enthalpy of the combined reaction by adding up:

By this mechanism it is often possbile to calculate the heat of a reaction even if this reaction is hard to carry out. E.g. we could burn both graphite and diamond and measure the heats of combustion for both. The difference would give us the heat of the transformation reaction from graphite to diamond.

Reaction-as-written convention (caution!)

The enthalpy is for the reaction-as-written. That means that if we write:

Reverse reactions

Because H is a state function the reverse reaction has the same enthalpy but with opposite sign

Combining values

It is quite possible that you cannot really do a certain reaction in practice. For many reactions we can arrive at enthalpy values by doing some bookkeeping

A good example is found in the book on 789ff. We can calculate the enthalpy for the reaction of PCl₃ with chlorine if we know the two reactions that the elements phosphorous and chlorine can undergo

You do have to make sure you balance your equations properly! (But you can do that by now, you're are chemist).

Tabulation and standard states

The Plimsoll waterline symbol

Heats of reaction are an important tool in the prediction of what will happen when you do chemistry. There are extensive tables available to do such predictions. They give standard reaction enthalpies (Δ_rH^o) defined for standard state conditions

Standard states

Standard state values are marked by a superscript plimsoll ^o

Plimsoll was a British MP in the 19th century who put forward legislation to have marks on the sides of all ships to indicate how heavily they could be loaded. The symbol was adopted in thermodynamic notation to indicate an arbitrarily chosen non-zero standard state.

The reason for taking a non-zero standard state is that we want to make quantities dimensionless by dividing them by the standard value, e.g. P/P^o [bar/bar] is dimensionless. Obviously the standard value should not be zero to make this work.

The fact that a non-zero state is taken makes it unfortunate that in recent texts on thermodynamics the plimsoll is often replaced by a superscript ^o erroneously suggesting a natural zero point.

Because the chosen standard is non-zero there is an infinite number of possible standard states. They are all equally legitimate, although some of them are more usual than others. This is connected with the choice of units.

• For pressure a non-zero value of 1 bar is usually the standard,
• For temperature the value is the (non-zero) temperature of interest, this is not necessarily 25^oC. Often values for a range of temperatures of interest are tabulated.
• What exactly the standard state of composition (e.g. concentration) is needs to be specified upon tabulation. For example it is quite possible in a chemical reaction to express one component, say A, in molality, the other, say B, in mole fraction. The standard state is where these measures are unity (non-zero!), even if this leads to a rather non-existant:

Of course this is rather bizarre, as putting A into B means B is no longer pure! Nevertheless this choice of standard state leads to a workable set of values as long as it is applied consistently. When switching to a different definition of the standard state, thermodynamic values cannot always be compared so easily.

Tabulation

Tabulation needs to follow certain conventions:

1. The reaction happens at standard pressure (usually one bar)
2. All concentrations -if applicable- are at unity in the units specified (but these units can vary: molarity, molality, mole fraction etc.)
3. The temperature is the one of interest (i.e. the values are typically tabulated as function of T, not just 25^oC).
4. The first reactant on the left has a stoichiometric coefficient unity (e.g. C_(s))

The last rule imples that a reaction like:

2C_(s) + 2O_2(g) -> 2CO_2{g)

is not a standard reaction because of the 2 in front of C. Its reaction enthalpy Δ_rH = 2Δ_rH^o

Notice also that it is customary (and imperative) to indicate the phases as solid, liquid or gaseous. Suppose you do a reaction between water and sodiumhydroxide at 0^oC

H₂O(s) (ice!) + Na -> NaOH + ½ H₂

H₂O(l) (water!) + Na -> NaOH + ½ H₂

The enthalpies of these two reactions will differ by the enthalpy of fusion of ice!

Another reason for indicating the phases is that it is sometimes desirable to convert ΔH into ΔU or vice versa. The difference lies in the volume work term PV that can be considerable for gases but it typically negligible for condensed phases. To a good approximation we can write:

Δ_rH = Δ_rU + RTΔn_gas

Δn_gas stands for the change in the total number of moles of gas during the reaction.

In the above case of the reaction of sodium with water Δn_gas is +½

Changing temperature

The definition of the standard state does not fix the temperature. Usually the temperature of tabulation is given in the table (usually 25^oC) and if we need to use the values at different temperatures we need to convert. The way to do that is to use:

H(T) = H(T1) + C_p*(T-T1) + ….

After all (∂H/∂T)_P = C_p!!

If the temperature change is small C_p can often be taken as a constant. If not we need it as function of the temperature in the range of interest and we must integrate.

Because the reaction enthalpy involves the enthalpies of all reactants and products, we must take into account all their changes, i.e. all the heat capacities.

If we can neglect temperature dependence of C_p we get

Δ_rH(T) = Δ_rH(To) + Δ_rC_p.(T-T_o)

Of course to calculate Δ_rC_p we need to use the appropriate stoiciometric coefficients e.g.:

H₂O_(l) + Na => NaOH + 0.5 H₂ (water)

Δ_rC_p= 0.5C_p(H₂) + C_p(NaOH) - C_pNa + C_pH₂OH_(l)

Changing pressure

The standard pressure in one bar. Yes Δ_rH also depends on pressure but typically not very strongly so. We will come back to that.