Phase Diagrams in Chemistry: Clapeyron and Clausius-Clapeyron Equations - Prof. Stefan Fra, Study notes of Physical Chemistry

Download Phase Diagrams in Chemistry: Clapeyron and Clausius-Clapeyron Equations - Prof. Stefan Fra and more Study notes Physical Chemistry in PDF only on Docsity! Chemistry 331 Lecture 20 Phase Diagrams NC State University Definition of a phase diagram A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function of temperature and pressure. In the Figure shown below the regions of space indicate the three phases of carbon dioxide. The curved lines indicate the coexistence curves. Note there is a unique triple point. P re ss ur e (a tm ) Degrees of freedom How many degrees of freedom are there in the liquid phase? A. 0 B. 1 C. 2 D. 3 Free energy dependence along the coexistence curve In a system where two phases (e.g. liquid and gas) are in equilibrium the Gibbs energy is G = Gl + Gg, where Gl and Gg are the Gibbs energies of the liquid phase and the gas phase, respectively. If dn modes (a differential amount of n the number of moles) are transferred from one phase to another at constant temperature and pressure, the differential Gibbs energy for the process is: The rate of change of free energy with number of moles is called the chemical potential. dG = ∂G g ∂ng P,T dng + ∂G l ∂nl P,T dnl The significance of chemical potential of coexisting phases We can write the Gibbs free energy change using the following notation: Note that if the system is entirely composed of gas molecules the chemical potential µg will be large and µl will be zero. Under these conditions the number of moles of gas will decrease dng < 0 and the number of moles of liquid will increase dnl > 0. Since every mole of gas molecules converted results in a mole of liquid molecules we have that: dng = -dnl dG = µgdng + µ ldnl Degrees of freedom What are the degrees of freedom in the liquid phase? A. Temperature and Volume B. Temperature and Pressure C. Pressure and Volume D. All of the above Degrees of freedom What are the degrees of freedom in the liquid phase? A. Temperature and Volume B. Temperature and Pressure C. Pressure and Volume D. All of the above The Clapeyron equation Substituting these factors into the total derivative above we have Solving for dP/dT gives This equation is known as the Clapeyron equation. It gives the two-phase boundary curve in a phase diagram with ∆trsH and ∆trsV between them. The Clapeyron equation can be used to determine the solid-liquid curve by integration. Starting with a known point along the curve (e.g. the triple point or the melting temperature at one bar) we can calculate the rest of the curve referenced to this point. Vm αdP – Sm αdT = Vm βdP – Sm βdT dP dT = Sm β – Sm α Vm β – Vm α = ∆ trsSm ∆ trsVm = ∆ trsHmT∆ trsVm dP P1 P2 = ∆ trsHm∆ trsVm dT TT1 T2 The liquid-vapor and solid- vapor coexistence curves The Clapeyron equation cannot be applied to a phase transition to the gas phase since the molar volume of a gas is a function of the pressure. Making the assumption that Vmg >> Vml we can use the ideal gas law to obtain a new expression for dP/dT. The integrated form of this equation yields the Clausius-Clapeyron equation. dP dT = ∆ trsHm TVm g = P∆ trsHm RT2 dP PP1 P2 = ∆ trsHm RT2 dT T1 T2 ln P2P1 = ∆ trsHmR 1 T1 – 1T2 = ∆ trsHmR T2 – T1 T1T2 Applying the Clausius- Clapeyron equation If we use ∆H of evaporation the C-C equation can be used to describe the liquid-vapor coexistence curve and if we use ∆H of sublimation this equation can be used to describe the solid-vapor curve. The pressure derived from the C-C equation is the vapor pressure at the given temperature. Applications also include determining the pressure in a high temperature vessel containing a liquid (e.g. a pressure cooker). If you are given an initial set of parameters such as the normal boiling point, for example you may use these as T1 and P1. Then if you are given a new temperature T2 you can use the C-C to calculate P2. Conceptual Question The key assumptions for the Clausius-Clapeyron equation that defines the liquid-vapor and solid-vapor coexistence curves are: A. Vmg >> Vml ,Vms >> Vmg B. Vml >> Vmg ,Vmg >> Vms C. Vms >> Vmg ,Vml >> Vmg D. Vmg >> Vml ,Vmg >> Vms Conceptual Question Which expression can be used to determine the coexistence boundary of two phases: A. dPdT = ∆ trsSm ∆ trsVm B. dPdT = ∆ trsHm ∆ trsVm C. dGdT = ∆ trsSm ∆ trsVm D. dGdT = ∆ trsHm ∆ trsVm Constructing the phase diagram for CO2 We can use the Clapeyron and Clausius-Clapeyron equations to calculate a phase diagram. For example, we can begin with the CO2 diagram shown above. The triple point for CO2 is 5.11 atm and 216.15 K. The critical point for for CO2 is 72.85 atm and 304.2 K. We also have the following data Note that we can calculate the enthalpy of sublimation from ∆vapHo = ∆subHo - ∆fusHo = 16.9 kJ/mol. ρsolid = 1.53 g/cm3 and ρliquid = 0.78 g/cm3, respectively. The density ρ = m/V = nM/V so the molar volume is Vm = V/n = M/ρ where M is the molar mass. In units of L/mole we have Vsm = 44 g/mole/[1530 g/L] = 0.0287 Vlm = 44 g/mole/[780 g/L] = 0.0564 ∆fusV = Vlm - Vsm = 0.0564 - 0.0287 = 0.0277 L/mole Transition ∆trsHo (kJ/mol) Ttrs (K) Fusion 8.33 217.0 Sublimation 25.23 194.6 Constructing the phase diagram for CO2 Starting with the triple point we use the Clausius-Clapeyron equation to calculate the liquid-vapor coexistence curve. P = 5.11exp{∆vapH/R[T – 216.15]/216.15T} P = 5.11exp{2,032[T – 216.15]/216.15T} Notice that if we were to calculate the critical pressure using this formula we would obtain 77.3 atm which is about 5 atm larger than the experimental number. There are several sources of inaccuracy including mainly our neglect of the temperature dependence of the enthalpy. We can also begin a the critical point P = 72.8 exp{∆vapH/R[T – 304.2]/304.2T} P = 72.8 exp{2,032[T – 304.2]/304.2T} Constructing the liquid-vapor curve Liquid-vapor P (atm) T (K) 5.11 216.15 6.03 220 9.0 230 13.0 240 P re ss ur e (a tm ) P = 5.11exp{2032[T – 216.15]/216.15T} Constructing the liquid-vapor curve Liquid-vapor P (atm) T (K) 5.11 216.15 6.03 220 9.0 230 13.0 240 24.9 260 P re ss ur e (a tm ) P = 5.11exp{2032[T – 216.15]/216.15T} Constructing the liquid-vapor curve Liquid-vapor P (atm) T (K) 5.11 216.15 6.03 220 9.0 230 13.0 240 24.9 260 41.0 280 P re ss ur e (a tm ) P = 5.11exp{2032[T – 216.15]/216.15T} Constructing the solid-vapor curve Starting again at the triple point P = 5.11exp{∆vapH/R[T – 216.15]/216.15T} P = 5.11exp{3034[T – 216.15]/216.15T} Solid-vapor P (atm) T (K) 5.11 216.15 3.38 210 1.64 200 P re ss ur e (a tm ) Constructing the solid-vapor curve Starting again at the triple point P = 5.11exp{∆vapH/R[T – 216.15]/216.15T} P = 5.11exp{3034[T – 216.15]/216.15T} Solid-vapor P (atm) T (K) 5.11 216.15 3.38 210 1.64 200 0.725 190 0.298 180 P re ss ur e (a tm ) Constructing the solid-vapor curve Starting again at the triple point P = 5.11exp{∆vapH/R[T – 216.15]/216.15T} P = 5.11exp{3034[T – 216.15]/216.15T} Solid-vapor P (atm) T (K) 5.11 216.15 3.38 210 1.64 200 0.725 190 0.298 180 0.111 170 P re ss ur e (a tm ) Constructing the solid-liquid curve Using the Clapeyron equation we calculate: P = 5.11 + [∆fusH/∆fusV] ln{T/216.15} P = 5.11 + 2,967 ln{T/216.15} Solid-liquid P (atm) T (K) 5.11 216.15 58.0 220 124.2 230 P re ss ur e (a tm ) Constructing the solid-liquid curve Using the Clapeyron equation we calculate: P = 5.11 + [∆fusH/∆fusV] ln{T/216.15} P = 5.11 + 2,967 ln{T/216.15} Solid-liquid P (atm) T (K) 5.11 216.15 58.0 220 124.2 230 319 240 559 260 780 280 987 300 1180 320