Chem 221 Notes  
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 EnthalpyIf 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
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
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
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.
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. CaCO_{3} 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 CaCO_{3} 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?
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
where Δn_{g} is the change in the number of moles of gas molecules per mole of equation. Thus, we MUST use molar values of ΔH_{m} and ΔU_{m} 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 ΔH_{b}>m and ΔU_{m} 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
ΔU_{m} = ΔH_{m} – Δn_{g}RT = 40.656 kJ/mol – 1 × 8.3145 J/molK × 373.15 K ΔU_{m} = 37.553 kJ/mol Clearly, although the difference between the change in internal energy ΔU_{m} and the change in enthalpy ΔH_{m} for a system that produces gas, it is not insignificant. Equation 3.23 then, can be used to convert between ΔH_{m} and ΔU_{m} 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. Example: 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 = q_{p} = 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 Δn_{g } is 1 so ΔU_{m} = ΔH_{m} – Δn_{g}RT = 41 kJ – 1 × 8.3145 J/molK × 373.15 K = +38 kJ/mol Variation of Enthalpy with TemperatureJust as we were able to define heat capacity at constant volume, we can define the heat capacity at constant pressure to be This property, C_{p} is an extensive property, just as is C_{v}. We often tabulate intensive analogues of these, such as the molar heat capacity at constant pressure. we can rearrange equation 3.24 to get
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
which can also be written as
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
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 N_{2} when it it heated from 25ºC to 100ºC? Since N_{2} 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 C_{p} (=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), C_{v} = 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
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 = C_{v}ΔT. 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
where subscript ad means adiabatic. We know the expression for work so –pΔV= C_{v}ΔT. 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 dm^{3} to 1.00 dm^{3}? The molar heat capacity of argon is C_{V} = 12.48 J K^{1} mol^{1}. From equation 3.31, we get c = 1.501 and
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 C_{V,m} = 3/2 R and therefore, using eq. 3.29 we can get C_{p,m} = 5/2 R. Thus γ = 5/3.

Last updated:
07Apr2010