- 247 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Useful Equations - The Clapeyron Equation' - adamdaniel

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Useful Equations - The Clapeyron Equation

- Gives the rate of change of the vapor pressure with temperature, dp/dT, in terms of the enthalpy of vaporisation, ∆Hvap, volume of the liquid, Vl, and volume of the vapor, Vv, at temperature T and at a pressure equal to the vapor pressure.

The Clapeyron Equation

- What is the change in boiling point of water at 100oC per mm change in atmospheric pressure. The heat of vaporization is 539.7 cal g-1, the molar volume of liquid water is 18.78 mL, and the molar volume of steam is 30.199 liters, all at 100oC and 1 atm.

The Clapeyron Equation

- Calculate the change in pressure required to change the freezing point of water 1oC. At 0o the heat of fusion of ice is 79.7 cal g-1, the density of water is 0.9998 g mL-1, and the density of ice is 0.9168. The reciprocals of the densities, 1.0002 and 1.0908 are the volumes in mL of 1 g. The volume change upon freezing (Vl – Vs) is therfore -9.056 x 10-5 L g-1. For small changes of ∆Hfus, T, and (Vl – Vs) are virtually constant, so that

The Clapeyron Equation

- A pressure of 133 atm lowers the freezing point of water 1oC.
- The reciprocal ∆T/∆V = -1/133 = - 0.0075 deg atm-1, shows that an increase in 1 atm lowers the freezing point 0.0075oC

Useful Equations -The Clausius-Clapeyron Equation

Clausius showed the Clapeyron equation could be simplified by assuming the vapor obeys the ideal gas law and by neglecting the volume of a mole of liquid in comparison with a mole of vapor. For example with water at 100oC the volume of vapor is 30.2 liters and the volume of liquid is 0.0188 liter

C is the integration constant

The Clausius-Clapeyron Equation

- Frequently a more convenient form of the equation to use is
- Obtained by integrating between the limits, P2 at T2 and P1 at T1.

The Clausius-Clapeyron Equation

Using this equation, it is possible to calculate the heat of vaporization or the heat of sublimation from the vapor pressure at two different temperatures.

The approximations involve the assumptions that the vapor is an ideal gas and the heat of vaporization is independent of temperature.

Over wide temperature ranges, plots of log P versus 1/T are somewhat curved because ∆Hvap varies with temperature. It is possible to calculate the heat of vaporization at any temperature from the slope of the curve by drawing a tangent to the curve at the required temperature.

Gibbs Free Energy

- For an irreversible or spontaneous process the entropy must increase.
- The quantity E – TS is referred to as the Helmholtz free energy, “A”. If the volume and temperature is constant:
- The Helmholtz free energy “A” decrease for a irreversible spontaneous process.

Gibbs Free Energy

- Physical and chemical process are usually carried out at constant pressure and temperature (P and T).
- d(E + PV – TS)T,P < 0
- The quantity E + PV – TS is referred to as the Gibbs free energy and is represented by G
- G = E + PV – TS = H – TS (H = enthalpy)
- (dG)TP = < 0
- Thus for an irreversible process at constant T and P in which only pressure volume work is done, the Gibbs free energy decreases.

Criteria for irreversibility and reversibility for processes involving no work or only pressure-volume work

These relationships may be applied to finite changes by replacing the d’s by ∆’s

Although the criteria show whether a change is spontaneous it does not provide any information on the speed (kinetics) of the process.

Gibbs Free Energy

- At constant temperature and pressure, the Gibbs free energy G = H – TS becomes:

ΔG = ΔH - T ΔS

- Whether a process is spontaneous at constant T & P depends on two terms, ΔH and TΔS.
- A change is favored if ΔH is negative and ΔS is positive. These changes correspond to a decrease in energy and an increase in disorder respectively.
- Note that at very high temperatures T ΔS can dominate

Criteria of Chemical Equilibrium

- Berthelot 1879 falsely concluded that reactions which evolve heat are spontaneous
- Spontaneous reactions can also absorb heat
- The spontaneous process leads to the minimum possible value of Gibbs free energy for the system.
- At equilibrium ΔG = 0
- If a system is in equilibrium it cannot undergo a spontaneous change under the given conditions

Gibbs Free Energy

- If ∆Go is negative, the equilibrium constant has a value greater than 1 and the reactants in their standard state will react spontaneously to give the products in their standard states.
- If ∆Go is 0, the reactants in their standard state will be in equilibrium with products in their standard states and K = 1.
- If ∆Go has a positive value, the reactants in their standard state will not react to give products in their standard states ( K < 1).
- NB: It is not necessary for ∆Go to be –ve for a reaction to be useful.

Limitations of Thermodynamics

- The Gibbs free energy can only predict if a chemical reaction is possible.
- It cannot predict how fast a reaction will occur.
- The Gibbs free energy for 1 mole of carbon and 1 mole of oxygen at one atmosphere, 25oC, is greater than the Gibbs free energy for carbon dioxide under the same conditions. However the reaction is so slow it virtually does not occur.
- C + O2 → CO2 (possible)
- The reverse reaction of CO2 forming C and O2 will not occur spontaneously.
- CO2 → C + O2 (not possible)
- Note at equilibrium all species would be present.

Chemical Equilibria

- (X) indicates the concentration of the reactants and products, but to be strictly correct it is the activity of reactants and products that should be used.

Since the chemical potentials of the reactants and products in their standard states are equal to their standard molar Gibbs fee energies, then

This equation is very important because it links the standard Gibbs free energy for a reaction with the value of the equilibrium constant.

Chemical EquilibriaGibbs Free Energy

- The standard Gibbs free energy change, ∆Go, for a reaction is the change in G when the indicated number of moles of reactants, each at unit activity, is converted into the indicated number of moles of products, each at unit activity.
- The standard Gibbs free energy can be calculated from calorimetric and thermodynamic data.
- Thus the equilibrium constant can be obtained without experimental information

Gibbs Free Energy

- Methods of determining ∆Go
- From the equilibrium constant ∆Go = -RTlnK
- ∆Ho obtained calorimetrically and ∆So obtained from third law entropies using ∆Go = ∆Ho - T∆So
- Using statistical mechanics and spectroscopic data

Gibbs Free Energy of Formation

- The Gibbs free energy of formation of a substance is the Gibbs free energy change for the reaction in which the substance in its standard state at 25oC is formed from its elements in their standard states at 25oC.
- As was the case for ∆Ho, ∆Go the Gibbs free energy for a reaction can be determined by adding and subtracting the reactions for which ∆Go is known.

Calc. of ∆G0 from

- Calculate ∆Go and Kp at 25oC for
- CO(g) + H2O(g) = CO2(g) + H2(g)
- ∆Go= (-94.2598 + 0) – (-32.8079 – 54.6357)

= -6.8162 kcal

- ∆Go= -(1.987)(298.1)(2.303)log Kp

Calculation of Gibbs Free Energy and Equilibrium Constant from Enthalpy of Formation and Entropy Values

- Calculate Kp for the following reaction at 25oC
- C (graphite) + 2H2 (g) = CH4 (g)
- ∆Ho = -17,889 cal
- ∆So = 44.5 – 2(31.211) – 1.3069

= -19.28 cal deg-1

- ∆Go = ∆Ho - T ∆So = -17,889 – (298.1)(-19.28)

= -12,140 cal

Calculation of Gibbs Free Energy and Equilibrium Constant

- ∆Go = -RTlnK
- -12,140 = -(1.987)(298.1)(2.303)logKP
- logKP = 8.91
- KP = 8.1 x 108
- Although the equilibrium constant for this reaction is large at ambient temperature it is not possible to carry out this reaction as no catalyst is known.
- At high temperature the reaction is unfavourable

Influence of Temperature on ∆G

- The values for ĈP for reactants and products are known over a wide range of temperature.
- With this information it is possible to calculate ∆Go at any temperature in the range that ĈP is known

Influence of Temperature on ∆G

- The enthalpy of reaction at temperature T
- The Gibbs free energy change for may be calculated from this equation if the heat capacity for each reactant and product is known as a function of temperature from 25oC to the desired temperature, ie. the value of the constants a, b and c are known

Influence of Temperature on ∆G

- Calculate ∆Go and KP for the reaction:
- C(graphite) + H2O(g) = CO(g) + H2(g)
- ∆Go298K = -32.8079 – (54.6357)

= 21.8278 kcal

- ∆Ho298K = -26.4157 – (-57.7979)

= 31.3822 kcal

- ĈP graphite = 3.81 + 1.56 x 10-3T and the values of the gases are obtained from the Table

Influence of Temperature on ∆G

- This general equation can now be used to calculate the Gibbs free energy change for the reaction at any temperature in the range for which the heat capacity equations are valid (300-1,500oK)
- At 1000oK, ∆Go1000 = -1,330
- NB: that the reaction occurs at 1000oK and not at 298oK

Download Presentation

Connecting to Server..