Partial equilibrium

In order to define completely the solubility of a gas in a liquid, it generally is necessary to state the temperature, the equilibrium partial pressure of the solute gas in the gas phase, and the concentration of the solute gas in the liquid phase. Stric tly speaking, the total pressure on the system also should be stated, but for low total pressures, less than about 507 kPa (5 atm), the solubihty for a particular partial pressure of solute gas normally will be relatively independent of the total pressure of the system. 

Reaction between an absorbed solute and a reagent reduces the equilibrium partial pressure of the solute, thus increasing the rate of mass transfer. The mass-transfer coefficient hkewise is enhanced, which contributes further to increased absorption rates. Extensive theoretical analyses of these effects have been made, but rather less experimental work and design guidehnes. 

In the first step, in which the molecules of the fluid come in contact with the adsorbent, an equihbrium is established between the adsorbed fluid and the fluid remaining in the fluid phase. Figures 25-7 through 25-9 show several experimental equihbrium adsorption isotherms for a number of components adsorbed on various adsorbents. Consider Fig. 25-7, in which the concentration of adsorbed gas on the solid is plotted against the equilibrium partial pressure p of the vapor or gas at constant temperature. At 40° C, for example, pure propane vapor at a pressure of 550 mm Hg is in equilibrium with an adsorbate concentration at point P of 0.04 lb adsorbed propane per pound of silica gel. Increasing the pressure of the propane will cause... 

This is a 4 2 reaction, and is thus pressure dependent. However, it is necessary to compute the equilibrium partial pressure of some alternative gaseous species, such as SiCls, and other hydrocarbons such as C2H2 and for this a Gibbs energy minimization calculation should be made. 

Table 6-5 shows the equilibrium eonstant with the equilibrium partial pressure of NH3 starting with a stoiehiometrie mixture of Hj and Nj at pressures of 1, 10, and 100 atm. Figure 6-12 shows the relationship between equilibrium eonversion X versus temperature and pressure for stoiehiometrie feed. 

The oxidation rates in carbon monoxide (Fig. 5.12) are less than those for carbon dioxide. They increase steadily with temperature up to 800°C but then decrease markedly by a factor of 100 up to 1 000°C. The decrease in rate can be attributed to the beneficial factors operating at the higher temperatures with carbon dioxide and to the unfavourable thermodynamics for reaction 5.2 resulting in low equilibrium partial pressures of carbon dioxide. 

Looking at the data in Table 12.2, you might wonder whether these three experiments have anything in common. Specifically, is there any relationship between the equilibrium partial pressures of N02 and N204 that is valid for all the experiments It turns out that there is, although it is not an obvious one. The value of the quotient (F>No2)2/f,N2o4 is the same, about 11, in each case ... 

Numerical values of equilibrium constants can be calculated if the partial pressures of products and reactants at equilibrium are known. Sometimes you will be given equilibrium partial pressures directly (Example 12.3). At other times you will be given the original partial pressures and the equilibrium partial pressure of one species (Example 12.4). In that case, the calculation of K is a bit more difficult, because you have to calculate the equilibrium partial pressures of all the species. 

Strategy First write the expression for K. Then substitute equilibrium partial pressures into that expression and solve. 

Note that H2 and I2 are being made (pressure increases), and HI is being used up. Furthermore, AP for H2(g) must be 0.10 atm to get from an original pressure of zero to an equilibrium partial pressure of 0.10 atm. Your table should then look like ... 

In air, the partial pressures of N2 and 02 are about 0.78 and 0.21 atm, respectively. The expression for K can be used to calculate the equilibrium partial pressure of NO under these conditions ... 

More commonly K is used to determine the equilibrium partial pressures of all species, reactants, and products, knowing their original pressures. To do this, you follow what amounts to a four-step path. 

Express the equilibrium partial pressures of all species in terms of a single unknown, x. To do this, apply the principle mentioned earlier The changes in partial pressures of reactants and products are related through tite coefficients of the balanced equation. To keep track of these values, make an equilibrium table, like the one illustrated in Example 12.4. 

Having found x, refer back to the table and calculate the equilibrium partial pressures of all species. 

You must express all equilibrium partial pressures in terms of a single unknown, x. A reasonable choice for x is... 

Reality Check Note that the equilibrium partial pressure of HI is intermediate between its value before equilibrium was established (0.80 atm) and that immediately afterward (1.00 atm). This is exactly what Le Chatelier s principle predicts part of the added HI is consumed to re-establish equilibrium. 

Calculate the equilibrium partial pressure of hydrogen if the equilibrium partial pressures of ammonia and nitrogen are 0.015 atm and 1.2 atm, respectively. 

At a certain temperature, K is 0.040 for the decomposition of two moles of bromine chloride gas (BrCl) to its elements. An equilibrium mixture at this temperature contains bromine and chlorine gases at equal partial pressures of0.0493 atm. What is the equilibrium partial pressure of bromine chloride ... 

Reaction quotient (Q) An expression with the same form as Kbut involving arbitrary rather than equilibrium partial pressures, 333-334 Reaction rate The ratio of the change in concentration of a species divided by the time interval over which the change occurs, 285 catalysis for, 305-307 collision model, 298-300 concentration and, 287-292,314q constant, 288 enzymes, 306-307 egression, 288... 

Chapter 4 presents the Third Law, demonstrates its usefulness in generating absolute entropies, and describes its implications and limitations in real systems. Chapter 5 develops the concept of the chemical potential and its importance as a criterion for equilibrium. Partial molar properties are defined and described, and their relationship through the Gibbs-Duhem equation is presented. 

Self-Test 9.9A The initial partial pressures of nitrogen and hydrogen in a rigid, sealed vessel are 0.010 and 0.020 bar, respectively. The mixture is heated to a temperature at which K = 0.11 for N2(g) + 3 H2(g) 2 NH3(g). What are the equilibrium partial pressures of each substance in the reaction mixture ... 

Sllf-Tfst 9.9B Hydrogen chloride gas is added to a reaction vessel containing solid iodine until its partial pressure reaches 0.012 bar. At the temperature of the experiment, K = 3.5 X 10-32 for 2 HCl(g) + I2(s) 2 HI(g) + Cl2(g). Assume that some 12 remains at equilibrium. What are the equilibrium partial pressures of each gaseous substance in the reaction mixture ... 

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