Chem 221 Notes 
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1 Gases
2 Microscopic    Energies
3 First Law
4 2nd & 3rd    Law
5 Phase    transitions
6 Mixtures
7 Phase    Diagrams
8 Equilibrium
9 Molecular    Interactions

Internal Energy | Enthalpy | Thermochemistry | Change Functions


If we are not restricting ourselves to constant volume conditions, we need to include the pV work term in the energy equation.  So, if we restrict ourselves to systems where only expansion work is done, we get

Energy in as heat but lost as work
Fig. 2.12  (from Atkins) When a system is subjected to constant pressure and is free to change its volume, some of the energy supplied as heat may escape back into the surroundings as work. In such a case, the change in internal energy is smaller than the energy supplied as heat.
dU = dq + dwexp


clearly, we see that if the system can expand, then the amount of heat gained or lost by the system is no longer equal to the change in internal energy.

We define this new heat as enthalpy.  The defining equation is

H = U + pV


This defining equation is not useful since we do not know the absolute value of internal energy but, as long as enthalpy is a state function, we can always determine changes in enthalpy dH by measuring changes in our other system measurables

H + dH = (U+dU) + (p+dp)(V+dV)


H + dH = U+dU + pV+ dpV+ pdV + dpdV H = U + pV so we can rewrite this as:
       dH = dU + dpV+ pdV + dpdV dpdV is infinitesimally small so drop it from the sum.
       dH = dU + dpV+ pdV


add in equation 3.19, where we recognize

dwexp = –pdV
pressure is constant

       dH = dq + dpV but we've already said pressure is constant so dp=0
       dH = dq


at constant pressure
with no additional we can finally write.
       ΔH = qp


So the heat exchanged between the system and the surroundings at constant pressure is the enthalpy change of the system.

There are several ways we can measure and/or calculate enthalpy changes. 

  • We could use a bomb calorimeter (isochoric) to measure a temperature change and thus calculate heat, qv, which gives us internal energy change. Then we can calculate the enthalpy change by figuring out volume/pressure changes.  [Assumed Cv is constant].

  • We could use a constant pressure calorimeter (isobaric) to measure a temperature change and hence calculate the heat, qp, which gives us enthalpy change.  [Assumed Cp is constant].

  • More sophisticated methods can also be used that don't assume constant heat capacity and even some that don't require calorimetric measurements.

No matter which method we calculate, at all but very high pressure changes, we generally can assume that the difference between the enthalpy change and internal energy change for condensed phases (solid, liquids) are negligibly small.

 CaCO3 exists in several different polymorphs: aragonite, vaterite and calcite.  At p = 1.0 bar, the density of calcite is 2.71 g cm-3 and the density of aragonite is 2.93 g cm-3.   What is the difference between enthalpy change and internal energy change for the process whereby 1.0 mol of CaCO3 is converted from calcite to aragonite.  Given that the change in internal energy for this process has been measured as ΔU = 0.21 kJ, what is the value of ΔH?

  ΔH = H(aragonite) – H(calcite)

        = {U(a) + pV(a)} – {U(b) + pV(b)}



  ΔH = ΔU + pΔV


so the difference between enthalpy and entropy is just p
ΔV.  From the densities, we can calculate the volume for 1.0 mol of CaCO3 as 34 cm3 and 37 cm3 for aragonite and calcite, respectively

so ΔU = 0.21 kJ, and if we add to that the pΔV value of –0.0003 kJ, we get ΔH = 0.21 kJ (after rounding).  So, to about 3 sig figs, we can generally ignore differences between enthalpy changes and entropy changes for condensed phase systems.

We already saw that we cannot ignore the pΔV term in cases where gas is a product.  To generalize, we cannot ignore in cases where either gas is used up or gas is produced.   As long as we assume ideal gas behavior, we can also eliminate consideration of systems that use up and produce the same amount (moles) of gas.

our equation for these special cases is derived from equation 3.22

ΔHm = ΔUm + ΔngRT


At this point, I need to digress for a bit.  There is much confusion about units with regards to enthalpy and internal energy, etc. in most text books.  On page 43 of Atkins, The units for  moles is used in one place (illustration 2.4) but not used in another place (Example 2.3).  So which is correct?  Does n have units of moles?  What are the units of ΔH and ΔU ?

I prefer to stay with the convention that n is unitless (it's really a ratio) and that ΔH and ΔU ultimately have units of kJ/mol

For further discussion on this, click here

where Δng is the change in the number of moles of gas molecules per mole of equation.  Thus, we MUST use molar values of ΔHm and ΔUm when we use equation 3.23.&n. 

Example:  one mole of water is vaporized at 100 ºC and 1 bar pressure.  What are the values of ΔHb>m and ΔUm for the process?

Since we have done this process at standard conditions, we can simply look up the tabulated enthalpy of vaporization for water at these conditions 
ΔHm = ΔHoV = 40.656 kJ/mol.  Since exactly one mole of water is involved in the reaction and all of it (1 mol) vaporized, Δng = 1.  So we can rearrange equation 3.23 to solve for ΔUm .

ΔUm = ΔHmΔngRT = 40.656 kJ/mol – 1 × 8.3145 J/molK × 373.15 K

ΔUm = 37.553 kJ/mol

Clearly, although the difference between the change in internal energy  ΔUm and the change in enthalpy ΔHm for a system that produces gas, it is not insignificant.

Equation 3.23 then, can be used to convert between  ΔHm and ΔUm for any process where gas is used up or is evolved.  Generally, we set up the experiment to measure one of these and then calculate the other.

Water, at 100ºC and 1.0 atm is heated with an electric heater current of 0.50 A at 12 V  for 300 s.  The mass of water that evaporates is 0.798 g.  What are the molar internal energy and enthalpy changes.

The reaction happening here is

Since this was done at constant pressure, we can calculate the enthalpy change by calculating the heat from the electric circuit.

ΔH = qp = IVt = 0.50 A × 12 V × 300 s = 1.8 kJ

Now, we need the molar value of the enthalpy change so

Finally, we can use equation 3.23 to calculate the change in internal energy for the equation as written. 

The parameter Δng is 1 so

ΔUm = ΔHm ΔngRT  =  41 kJ – 1 × 8.3145 J/molK × 373.15 K = +38 kJ/mol

Variation of Enthalpy with Temperature

Just as we were able to define heat capacity at constant volume, we can define the heat capacity at constant pressure to be


This property, Cp is an extensive property, just as is Cv.  We often tabulate intensive analogues of these, such as the molar heat capacity at constant pressure.

we can rearrange equation 3.24 to get

dH = Cp dT


and as before, if we are able to assume that the heat capacity is constant over the range of the temperature change then we can write

ΔH = Cp ΔT


which can also be written as

qp = Cp ΔT


since at constant pressure, we can equate the change in enthalpy to the energy transferred as heat into or out of the system.

a common empirical equation that allows us to adjust the heat capacity as the temperature changes is

Cp,m = a + bT + c/T2


where parameters a,b,c are independent of temperature but are unique for each substance.  Here is a table (from Atkins) containing A,B,C values for four pure materials.

Example:  What is the change in the molar enthalpy of N2 when it it heated from 25ºC to 100ºC?

Since N2 is not an atomic ideal gas, we cannot assume the heat capacity is constant with temperature. We must therefore incorporate equation 3.27 with equation 3.25 and integrate

Now we can solve this equation, if we remember three simple integral solutions from calculus.

The solution for the equation is

Now, if we substitute in all the values and do the math we find

H(373 K) – H(298 K) = 2.20 kJ mol–1

If we had assumed that the value of Cp (=29.14 J K-1), calculated at 25ºC from 3.27, did not change over the whole temperature range, we would have found.

H(373 K) – H(298 K) = 2.19 kJ mol–1

Recall recall the derivation of heat capacity at constant volume (eq. 3.16),  Cv = 3/2R (for one mole).  If we heat a material at constant pressure, it generally will expand, which means some of the energy that went into the system as heat is converted to work of expansion, hence, the temperature will rise less for work done at constant pressure than at constant volume.  For an ideal gas, we can derive the equation

Cp – Cv= nR .


 If our perfect gas is expanded adiabatically, then by the same logic, we would expect the temperature of the gas to decrease.  If we assume the heat capacity is independent of temperature,

ΔU = CvΔT.

Because the work was done adiabatically, q = 0 so the change in internal energy is simply the work done (eq. 3.12).  So we can rewrite the equation as

wad = CvΔT.


where subscript ad means adiabatic.  We know the expression for work so

–pΔV= CvΔT

In infinitesimal steps of a reversible process, we get  


Example:  How much work is done when 0.020 mol of Ar, initially at 25ºC expands adiabatically from 0.50 dm3 to 1.00 dm3?  The molar heat capacity of argon is CV = 12.48 J K-1 mol-1

From equation 3.31, we get c = 1.501


So ΔT = –110 K and so we can find the work from equation 3.29

w = 0.020 mol × 12.48 J K-1 mol-1 ×(–110 K) = –27 J

Notice that the temperature change depended only on the volume change were as the work was a function of the amount of gas.

In a similar derivation to what we did to get eq. 3.31, we can derive an expression for the pressure volume relationship under adiabatic conditions


Thus, for adiabatic conditions, our analogue to Boyle's Law (isothermal) is .

Example:  What is the final pressure when a sample of argon, initially at 100 kPa expands reversibly and adiabatically to twice its initial volume.

From equation 3.16 we know CV,m = 3/2 R and therefore, using eq. 3.29 we can get Cp,m = 5/2 R.  

Thus γ = 5/3.


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