To picture the setup of a constant-pressure calorimeter, just think of the coffee-cup calorimeter: an insulated container covered with a lid and filled with a solution in which a reaction or some physical process, such as dissolution, is occurring. The pressure, which is just the atmospheric pressure, remains constant throughout the process. The setup of a constant-volume calorimeter is perhaps a little less familiar to you or harder to picture. Furthermore, the term bomb calorimeter
sounds rather ominous. Well, don’t worry—this isn’t some relic from the Cold War. Perhaps “bomb” is a bit misleading: A more accurately descriptive term is “decomposition vessel.” This better reflects what is actually taking place. A sample of matter, typically a hydrocarbon, is placed in the steel decomposition vessel, which is then filled with almost pure O2 gas. The decomposition vessel is then placed in an insulated container holding a known mass of water. The contents of the decomposition vessel are ignited by an electric ignition mechanism. The material combusts (burns) in the presence of the oxygen, and the heat that is evolved in the combustion is the heat of the reaction. Because W = , no work is done in an isovolumic (V = 0) process, so Wcalorimeter = 0. Furthermore, because of the insulation, the whole calorimeter can be considered isolated from the rest of the universe, so we can identify the “system” as the sample plus oxygen and steel and the surroundings as the water. Because no heat is exchanged between the calorimeter and the rest of the universe, Qcalorimeter is 0. So Usystem + Usurroundings = Ucalorimeter = Qcalorimeter-Wcalorimeter = 0. Therefore, Usystem = -Usurroundings. Because no work is done , qsystem = -qsurroundings, and msteelcsteelT + moxygencoxygenT = -mwatercwaterT.MCAT Expertise
We need a different formula to calculate q
during phase changes, when T = 0. If we used q = mcT, we’d erroneously think q= 0.Note that by using the layer of insulation to isolate the entire calorimeter from the rest of the universe, we’ve created an adiabatic process. This means that no heat is exchanged between the calorimeter and the rest of the universe, but it is
exchanged between the steel decomposition vessel and the surrounding water. As the previous derivation shows, heat exchange between the system and its surroundings makes it possible for us to calculate the heat of the combustion.
Enthalpy
Most reactions in the lab occur under constant pressure (at 1 atm) in closed thermodynamic systems. To express heat changes at constant pressure, chemists use the term enthalpy (
H). Enthalpy is a state function, so we can calculate the change in enthalpy (H) for a system that has undergone a process—for example, a chemical reaction—by comparing the enthalpy of the final state to the enthalpy of the initial state, irrespective of the path taken. The change in enthalpy is equal to the heat transferred into or out of the system at constant pressure. To find the enthalpy change of a reaction, Hrxn, you must subtract the enthalpy of the reactants from the enthalpy of the products:H
rxn = Hproducts-HreactantsA positive H
rxn corresponds to an endothermic process, and a negative Hrxn corresponds to an exothermic process. It is not possible to measure enthalpy directly; only H can be measured, and only for certain fast and spontaneous processes. Several standard methods have been developed to calculate H for any process.Standard Heat of Formation
The standard enthalpy of formation
of a compound, H°f, is the enthalpy change that would occur if one mole of a compound in its standard state were formed directly from its elements in their respective standard states. Remember that standard state is the most stable physical state of an element or compound at 298 K and 1 atm. Note that H°f of an element in its standard state, by definition, is zero. The H°f ’s of most known substances are tabulated. You do not need to memorize these values, as they will be provided for you as necessary.
Standard Heat of Reaction
The standard heat of a reaction
, H°rxn, is the hypothetical enthalpy change that would occur if the reaction were carried out under standard conditions. What this means is that all reactants must be in their standard states and all products must be in their standard states. This can be calculated by taking the difference between the sum of the standard heats of formation for the products and the sum of the standard heats of formation of the reactants:H
°rxn = Σ(H°f of products)-Σ(H°f of reactants)Key Concept
State functions are path-independent. Always.
Hess’s Law