Internal energy of the system is represented by the symbol U. It is a more general description of the energy of the system than enthalpy. We can never measure the absolute energy of a system but we can measure changes in the internal energy of the system using the state function
ΔH2U = Ufinal – Uinitial
Since we can never actually measure either Uinitial nor Ufinal we measure, instead, the energy transferred into or out of the system in the form of heat and work to obtain.
ΔU = q + w ( = heat transferred + work done)
This is, in fact, the mathematical expression of the First Law of Thermodynamicsand is completely general, not relying on any of the restrictions we placed on enthalpy measurements. We can measure work and heat under any conditions to get the change in internal energy because it turns out that while neither q nor w are state functions, their sum is.
If 15 kJ of work is done by the surroundings on the system and the system loses 10 kJ of energy as heat to the surroundings then
|ΔU||= -10 kJ + 15 kJ|
|= +5 kJ|
Chemical reactions often involve changes in Volume (exp. gases)
|C3H8(g) + 5 O2(g)||→ 3 CO2(g) + 4 H2O(g)|
|6 mol gas||7 mol gas||→ increase in volume at const. P|
If this reaction is to happen, it needs to push back the rest of universe (atmosphere) to make room for itself. Let’s put the reaction system into a cylindrical apparatus like that pictured at the beginning of this unit. As gas molecules are produced, the system expands to accommodate them and keep the pressure (inside and out) constantly the same. The amount of work done as the system expanded against constant external pressure can be calculated as:
|work| = force (a.k.a. pressure × area) × displacement (a.k.a. ΔX).
or |w| = P × A × ΔX or just P × ΔV
to fully define w, we need to recognize that if the system expands (does the work), the system lost energy so by definition, w is negative when volume expands (ΔV positive). So we can remove the magnitude formulation and put in a negative sign.
w = – PsurrΔV (gas expanding against an external pressure Psurr.)
ΔU = q + w = q – PsurrΔV
or qp = ΔU + Psurr ΔV
in this case, we can use the ideal gas law to get a more useful equation as long as the gas molecules all behave ideally.
qp = ΔU + ΔngRT.
which is also written as
ΔH = ΔU + ΔngRT.
We can use this formulation when the only volume increase comes from the fact that the number of moles of gas is different in products than in reactants. Note that the parameter ΔU is a value that is tied to the balanced chemical equation (just as we did for enthalpy changes). Thus, the value we used for ΔU and the value we used for Δng must be both for the same balanced chemical equation.
In a closed container (no volume change) there would be no work so
ΔV = 0 so: ΔU = qv.
subscript v refers to constant volume conditions.
Enthalpy is now better defined as.
H = U + PV
Changes in enthalpy are
ΔH = ΔU + ΔPV since we normally work in a fixed pressure (atmosphere) situation, we can rewrite this as
ΔH =ΔU + PΔV.
To do these measurements under general conditions, we need to expand a few previous definitions. First, let’s look at heat capacity.
- At constant pressure, we measure heat as ΔH = qp,and hence the heat capacity we need to use is Cp, where we define
- At constant volume, the heat change we measure is ΔU = qv and the heat capacity Cv is defined as
In many substances, the two heat capacities differ considerably. Consider one mole of ideal gas. We have
ΔH – ΔU = Δ(PV) = RΔT
Therefore: Cp – Cv = R for one mole of ideal gas.
Solids and liquids often have values of Cp and Cv that don’t differ much. On the other hand there is no consistent rule. Take for example water and benzene; for water,Cp – Cv = R = 0.075R, while for benzene we have Cp – Cv = R = 5.1R.
Let’s do an example where we look at the PV work done.
When 2.00 mol of SO2(g) react completely with 1.00 mol O2(g) to form 2.00 mol of SO3(g) at 25℃ and constant pressure of 1.00 atm, 198 kJ of energy is released as heat. Calculate ΔU and ΔH for this reaction.
2 SO2(g) + 1 O2(g) → 2 SO3(g)
The heat released is |qp| since it was measured under constant pressure conditions. so we can quickly say
ΔH‘ = -198 kJ (negative sign since the heat is released, i.e., exothermic)
ΔH = -198 kJ/mol (trivial in this case since the molar quantities given exactly match the numbers in the equation)
ΔH’ = qp + w = qp – PΔV (total PV work)
or (since we have the molar values)
ΔU = ΔH – PΔV. (remember, this is the PV work per mole which just happens to equal the total PV work in this case)
Since we have constant P and T conditions, the only change in PΔV will be due to change in numbers of moles of gas molecules. We can use the stoichiometry from the equation as written to determine PΔV:
|ΔU =ΔH – ΔngRT||or||ΔH =ΔU + ΔngRT|
|Note: Either of these two equations above are valid for any situation where the T and P are constant but where the number of moles of gas molecules changes because of stoichiometry. Pick one and learn it.|
Δng = (#moles product gas – #moles reactant gas) per mole of equation
Δng = 2 – 3 = -1.
Note that the coefficients used here to calculate Δn are really unitless as they represent ratios of moles, not actual measures of moles.
|ΔU||= -198 kJ/mol + -(-1 × 8.31451 J K-1mol-1 × 298 K)|
|= -198 kJ/mol + 2.48 kJ/mol|
|= -196 kJ/mol|