{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture 15 - Thermochemistry continued. Example 2. ...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Thermochemistry continued. Example 2. Calculate ΔH and ΔS for the following process applied to one mole of water at P = 1 atm: H2O(l,  ­10C) → H2O(s,  ­10C) This process describes freezing of 1 mole of supercooled water at –10C. Solution: Think of H2O(l,  ­10C) as “reactants” and H2O(l,  ­10C) as “products”. Then ΔH = H (H2O(s,  ­10C)) – H (H2O(l,  ­10C)) = [ H (H2O(s, 0C)) + C p (H2O(s))( ­10C – 0C)]  ­ [ H (H2O(l, 0C)) + C p (H2O(l))( ­10C – 0C)] =  ­ΔHmelting + { C p (H2O(s))  ­ C p (H2O(l))} ( ­10C ­0C) =  ­5615 J To calculate the entropy change note that our process is irreversible (freezing of supercooled water occurs spontaneously). In order to use the equations we know, we have to come up with a path that takes us from the initial state to the final state in a reversible way. Here is one: H2O(l,  ­10C) → H2O(l, 0C) → H2O(s, 0C) → H2O(s,  ­10C) so ΔS = C p (H2O(l)) ln(273K/263K)  ­ ΔHmelting/273K + C p (H2O(s)) ln(263K/273K) = =  ­20.5413 J/K The change of the entropy for the surroundings is equal to: ΔSsurr =  ­ΔH/T = 5615J/263K = 21.3498 J/K Note that ΔSsurr + ΔS > 0, which is consistent with the 2nd law. Phase transitions: melting, freezing, condensation, sublimation, etc. Liquid ­gas transition at constant P, T, or V Constant P experiment: Compressing a gas at a constant T (start at low pressure where the material is in the gaseous form) PV = nRT for an ideal gas. Constant V experiment: Equilibrium is established when the number of molecules leaving water is equal to the number of molecules entering water from vapor (unless all water is evaporated before then). Our goal now is to find out under which conditions (i.e., what P and T) two phases can be in equilibrium with one another. We know the general result: at given P and T equilibrium corresponds to the minimum of G. But now we have to write G for a system that generally consists of more than one phase (e.g., gas and liquid). Consider n moles of a chemical kept at a constant pressure. Assume that it can be in the form of vapor (g) or liquid (l). If ng is the number of moles of vapor and nl =n ­ng is the amount of liquid then the Gibbs energy can be written as: G(P,T) = nl µl(P,T) + ng µg(P,T), where µ = G/n is the molar Gibbs energy of liquid or gas called the chemical potential. (According to our convention, the chemical potential should be denoted G ; however because it is so important in chemistry, different notation is used). If µl(P,T) < µg(P,T) then the system can minimize its Gibbs energy by becoming liquid: nl = n If µl(P,T) > µg(P,T) then the liquid will evaporate. Finally, if µl(P,T) = µg(P,T) then both phases will coexist. The equation µl(P,T) = µg(P,T) determines the coexistence curve between liquid and its vapor. Some properties of the chemical potential. S = S/n, V = V/n. A different symbol is used for G/n: µ = G/n According to the definition of the Gibbs free energy, µ = H − TS and dµ = VdP − SdT We can use this expression to calculate the change of the chemical potential as a function of temperature and/or pressure. ...
View Full Document

{[ snackBarMessage ]}