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Concept 03 / 06Thermodynamics
Enthalpy, Heat Capacity and Calorimetry
Enthalpy as heat at constant pressure, the relation ΔH = ΔU + Δn(g)RT, exothermic and endothermic reactions, heat capacity, molar and specific heat, Cp - Cv = R for ideal gases, and measurement of ΔU in a bomb calorimeter and ΔH in a coffee-cup calorimeter.
Figure 1
Most chemistry happens in open vessels at constant pressure, where the system does expansion work. A new state function, enthalpy, is defined to account for this:
H=U+PV.
At constant pressure, ΔH=ΔU+PΔV, and from the first law ΔU=qP−PΔV, so
ΔH=qP: the heat absorbed at constant pressure.
Exothermic reactions release heat, ΔH<0; endothermic reactions absorb heat, ΔH>0 (see figure).
Relation between ΔH and ΔU. For reactions involving gases (treated as ideal), PΔV=ΔngRT, so
ΔH=ΔU+ΔngRT,
where Δng= moles of gaseous products − moles of gaseous reactants. Solids and liquids contribute negligibly. If Δng=0 (for example H2(g)+Cl2(g)→2HCl(g)), ΔH=ΔU.
Heat capacity. The heat needed to raise the temperature of a system by one kelvin: C=ΔTq (extensive). Molar heat capacity Cm is per mole; specific heat c is per gram (intensive). q=mcΔT=nCmΔT. Water has a high specific heat, 4.18Jg−1K−1.
CV=(∂T∂U)V and CP=(∂T∂H)P, so ΔU=nCVΔT and ΔH=nCPΔT for an ideal gas (any process).
For one mole of an ideal gas, CP−CV=R: at constant pressure extra heat is needed to do expansion work.
Monatomic gas: CV=23R, CP=25R, γ=1.67. Diatomic (near room temperature): CV=25R, CP=27R, γ=1.40.
Calorimetry.
Bomb calorimeter (constant volume): a sample burns in oxygen inside a sealed steel bomb immersed in water. No expansion work is done, so the heat measured is qV=ΔU. Heat released =CcalΔT, where Ccal is the heat capacity of the calorimeter, found by burning a standard (benzoic acid).
Coffee-cup calorimeter (constant pressure): an insulated cup open to the atmosphere measures qP=ΔH, typically for neutralisation or dissolution in solution.
Key Takeaways:
- ΔH=qP; ΔU=qV; ΔH=ΔU+ΔngRT.
- CP−CV=R (per mole, ideal gas); γ=CVCP.
- Bomb calorimeter gives ΔU; open calorimeter gives ΔH.
- Pitfall: count only gaseous species in Δng, and remember R=8.314×10−3kJmol−1K−1 when ΔH is in kilojoules.
Brain-Teaser 1: For the combustion of liquid benzene, C6H6(l)+215O2(g)→6CO2(g)+3H2O(l), find ΔH−ΔU at 298K.
Solution: Δng=6−7.5=−1.5. ΔH−ΔU=−1.5×8.314×298=−3716J=−3.72kJmol−1. The gas volume shrinks, the surroundings do work on the system, so slightly more heat is released at constant pressure than at constant volume.
Brain-Teaser 2: 1.0g of graphite is burnt in a bomb calorimeter with heat capacity 20.7kJK−1. The temperature rises from 298K to 299K. Find ΔU for the combustion of one mole of graphite.
Solution: Heat released =20.7×1=20.7kJ for 1g, that is 121mol. Per mole: ΔU=−20.7×12=−248kJmol−1. (Negative because the reaction releases heat.)
Brain-Teaser 3: How much heat is needed to raise the temperature of 2mol of N2 from 300K to 400K at (a) constant volume and (b) constant pressure?
Solution: (a) qV=nCVΔT=2×25×8.314×100=4157J. (b) qP=nCPΔT=2×27×8.314×100=5820J. The difference, nRΔT=1663J, is the expansion work at constant pressure.
