3 Property relations and Thermochemistry

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1 3 Property relations and Thermochemistry 3.1 Intensive and extensive properties Extensive properties depend on the amount of mass or number of moles in the system. The extensive properties are usually in capital letters: U,, S etc. Intensive properties are expressed per unit mass or mole (kmole). The intensive properties are in lower-case letters. The mass intensive property is written u, h, s... and the molar intensive property as ū, h, s Ideal-gas mixtures The system contains a mixture of ideal gases. Each component or species X i has a molar mass M i. The number of moles of component i is N i. The total mass of species i is: The sum of partial pressures p i is the total pressure p: m i = N i M i (37) p = i p i (38) The mole fraction, mass fraction and mixture molar weight are defined in the following way: Mole fraction χ i : Mass fraction Y i : Mixture molar weight: χ i = Y i = M = i N i i N. (39) i m i i m. (40) i χ i M i. (41) The partial pressure p i may be expressed using molar fraction (39): p i = χ i p (42) Ideal gas mixtures properties could be calculated as sums of the individual mass or mole fraction i.e.: enthalpy/mass: h = i Y i h i (43) enthalpy/mole: h = i χ i hi (44) The same for internal energy u and s. Though entropy is: 18

2 s(t, p) = i Y i s i (T, p i ) (45) s(t, p) = i χ i s i (T, p i ) (46) (note partial pressure p i ) We have: ( pi ) s i (T, p i )=s i (T, p ref ) R ln p ref ( pi ) s i (T, p i ) = s i (T, p ref ) Rln p ref (47) (48) If the molar enthalpy h is given for a system the enthalpy/mass, h is readily given provided that the molar fraction of the different species is known: h = Σm i = h = ΣN i ΣM i N i and finally h = h ΣN i ΣM i N i h = h 1 ΣM i N i ΣN i h = h 1 (49) ΣM i χ i 3.3 Molar absolute (or standardized) enthalpy and molar enthalpy of formation The molar absolute enthalpy of a species i at temperature T is h i (T ). The superscript denotes that the pressure is at standard state pressure (p ), usually 1 atm. The molar absolute enthalpy at T of species i is related to the molar absolute enthalpy at an other temperature T through: h i (T )= h i (T )+ d h i (50) T Now we define the enthalpy difference (which is tabulated) h i (T ) as ence the integral T d h i T d h i = T ref d h i + h i (T ) in eq.(50) may be expressed: ref T ref d h i (51) T d h i = h i (T ) h i (T ) 19

3 and used in eq.(50): h i (T )= h i (T )+ h i (T ) h i (T ). (52) The term h i (T ) comes in part from the forming of the species and is best explained in the context of an example. Let us look at the simple situation where water is formed from the elements hydrogen and oxygen O 2 2 O and write the absolute enthalpy for water (eq.(52)) O(T )= O(T )+ O(T ) O(T ). (53) The absolute enthalpy O (T ) is related to the enthalpy of the elements of water in the following way O(T )= h f, 2 O(T )+ (T )+ 1 2 h O 2 (T ). (54) h f, 2 O (T ) is called the enthalpy of formation of water at temperature T and is the difference between the enthalpy of the product ( 2 O) and the enthalpy of the elements ( 2, O 2 ) forming the product at T and pressure p. The term arises from the shift in the enthalpy scales between product and reactants and is due to the difference in the chemical energy stored in the constituents. There is no natural zero level for the enthalpy scale (as it is for entropy (perfect crystals of a single pure isotope of a single element at T = 0K)) and an absolute scale of the enthalpy has to be decided. The convention is that the absolute enthalpy is zero for an element in its most stable state at T ref (= K) and at p ref (= 1atm) (for hydrogen and oxygen it is 2 and O 2 in gas phase). For such an element i the absolute enthalpy is: h i (T )= h i (T ) (55) and hence zero for T = T ref. Now we can write the absolute enthalpy for water at temperature T. Use eq.(55) in eq.(54) and put the result into equation (53) and we have: O(T )= h f, 2 O(T )+ (T )+ 1 2 h O 2 (T )+ O(T ) O(T ) (56) If we put T = T ref ( h i (T ref ) = 0) eq.(56) becomes: O(T )= h f, 2 O(T ref )+ O(T ) (57) which gives us the common form of the absolute or standardized enthalpy: h i (T )= h f,i(t ref )+ h i (T ) (58) As the enthalpy is a state function the absolute enthalpy of a species is only dependent on its temperature, pressure and phase and route to the final temperature is of no importance. Figure(7) shows a simple example: hydrogen dissociating into atoms. Independent of at what temperature T the enthalpy 20

4 of formation of the monatomic hydrogen is taken we get the same absolute h (T ) as long as we take into account also the change of the sensible enthalpy of both and 2 when the temperature T varies. Thus eq.(56) rewritten for this case: h (T )= h f,(t )+ 1 2 (T )+ h (T ) h (T ) (59) h o (T') 0 T ref =298 T' T'' T Figure 7: Figure showing the absolute enthalpy change due to dissociation of a hydrogen molecule into atoms (Exercise: Evaluate the reaction at a few different T and T ) When we have the absolute enthalpy of the reactants and products of a given reaction at a given temperature T and pressure we may determine the enthalpy change the mixture will encounter due to the reaction. The enthalpy of the products minus the enthalpy of the reactants is called the heath of reaction (or enthalpy of reaction, enthalpy of combustion) R (or per mass h R = R ) at constant pressure and temperature T for the reaction in prod νimi question. (Unfortunately there is also a heath of combustion (= h R )) An example: Determine R for the reaction: at T =298K and T =4000K. Thus C O 2 CO O (T )= h CO 2 (T )+2 O(T ) 1 h C 4 (T ) 2 h O 2 (T ) With equations (58) and (55) we get: (T )= h f,co 2 (T ref )+ h CO 2 (T ) +2( h f, 2 O(T ref )+ O(T )) ( h f,c 4 (T ref )+ h C 4 (T ) + 2 h O 2 (T )) 21

5 At T = 298K and T = 4000K we get = kcal and = kcal respectively. The mass involved is prod ν im i = 80g 3.4 Adiabatic flame temperature The adiabatic flame temperature is the temperature the products in the reaction raise to if we technically have a complete reaction to the right. From the first law of thermodynamics we have for an adiabatic system at constant pressure p: react (T, p) = prod (T ad,p) (60) The temperature T ad thus obtained is the adiabatic flame temperature at constant pressure. Continuing with the previous example of burning methane in oxygen we start with reactants at T = T ref and look for the T ad which satisfies the relation = 0: = 0 = h f,co 2 (T ref )+ h CO 2 (T ad ) +2( h f, 2 O(T ref )+ O(T ad )) ( h f,c 4 (T ref )+ h C 4 (T ref ) + 2 h O 2 (T ref )) The matching to get a T ad that sets to 0 is left as an exercise. The enthalpy tables can also be used to determine the adiabatic flame temperature at constant volume. Again from the first law of thermodynamics we have for an adiabatic system at constant volume: From the definition of enthalpy we have: U react (T, V )=U prod (T ad,v). (61) U = pv. We use the relation in eq.(61) and set T = T ref : or react (T ref ) (pv ) reac = prod (T ad ) (pv ) prod prod (T ad ) react (T ref ) = (pv ) prod (pv ) reac The ideal gas law: pv = NRT and the restriction V = constant simplifies the equation: prod (T ad ) react (T ref ) = (N prod T ad N reac T ref )R (62) where N prod and N reac are number of moles of products and reactants respectively. If we use the relation in the previous example we see that the number of moles does not change during the reaction N prod = N reac = N and the adiabatic flame temperature at constant volume can be determined from the following relation: prod (T ad ) react (T ref )=N(T ad T ref )R. Compared with the adiabatic temperature at constant pressure the adiabatic temperature at constant volume will be higher. 22

Reacting Gas Mixtures

Reacting Gas Mixtures Reading Problems 15-1 15-7 15-21, 15-32, 15-51, 15-61, 15-74 15-83, 15-91, 15-93, 15-98 Introduction thermodynamic analysis of reactive mixtures is primarily an extension of the principles