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Molarity equilibrium

Figure 2.10 provides a thermodynamic equilibrium molar fraction of the products of CPO of methane as a function of temperature. It is evident that at temperatures above 800°C, hydrogen and CO (in molar ratio of 2 1) are two major products of the reaction. The oxidant (oxygen or air) and the hydrocarbon feedstock (e.g., methane) are premixed in a mixer... [Pg.51]

Thermodynamic equilibrium molar fraction of hydrogen at different pressures as a function of temperature. [Pg.74]

The [HA] is the equilibrium molar concentration of the undissociated weak acid, not its initial concentration. The exact expression would then be [HA] = Minitia y — [H+], where Minitia y is the initial concentration of the weak acid. This is true because for every H+ that is formed an HA must have dissociated. However, many times if the Ka is small, you can approximate the equilibrium concentration of the weak acid by its initial concentration, [HA] Minitia y. [Pg.225]

Step 2 Write the equilibrium expression. Then substitute the equilibrium molar concentrations into the expression. [Pg.337]

Figures 1 to 3 present calculated equilibrium molar ratios of products to reactants as a function of temperature and total pressure of 1 and 100 atm. for the gas-carbon reactions (4), (7), and (5), (6), (4), (7), respectively. Up to 100 atm. over the temperature range involved, the fugacity coefficients of the gases are close to 1 therefore, pressures can be calculated directly from the equilibrium constant. From Fig. 1, it is seen that at temperatures above 1200°K. and at atmospheric pressure, the conversion of carbon dioxide to carbon monoxide by the reaction C - - COj 2CO essentially is unrestricted by equilibrium considerations. At elevated pressures, the possible conversion markedly decreases hence, high pressure has little utility for this reaction, since increased reaction rate can easily be obtained by increasing reaction temperature. On the other hand, for the reaction C -t- 2H2 CH4, the production of methane is seriously limited at one atmosphere pressure and practical operating temperatures, as seen in Fig. 2. Obviously, this reaction must be conducted at elevated pressures to realize a satisfactory yield of methane. For the carbon-steam reaction. Figures 1 to 3 present calculated equilibrium molar ratios of products to reactants as a function of temperature and total pressure of 1 and 100 atm. for the gas-carbon reactions (4), (7), and (5), (6), (4), (7), respectively. Up to 100 atm. over the temperature range involved, the fugacity coefficients of the gases are close to 1 therefore, pressures can be calculated directly from the equilibrium constant. From Fig. 1, it is seen that at temperatures above 1200°K. and at atmospheric pressure, the conversion of carbon dioxide to carbon monoxide by the reaction C - - COj 2CO essentially is unrestricted by equilibrium considerations. At elevated pressures, the possible conversion markedly decreases hence, high pressure has little utility for this reaction, since increased reaction rate can easily be obtained by increasing reaction temperature. On the other hand, for the reaction C -t- 2H2 CH4, the production of methane is seriously limited at one atmosphere pressure and practical operating temperatures, as seen in Fig. 2. Obviously, this reaction must be conducted at elevated pressures to realize a satisfactory yield of methane. For the carbon-steam reaction.
Connections to other types of phase diagrams can be obtained if order parameters are exchanged for intensive variables. Figure 17.2 is replotted with the order parameter V as the ordinate in Fig. 17.36. The diagram predicts the phases that would exist for a molar volume fixed by a rigid container at different temperatures. The tie-lines connect equilibrium molar volumes at the same temperature... [Pg.421]

The equilibrium state is entirely determined by the minima of G as a function of pressure and temperature. The equilibrium order parameter rf is determined by the minima of G and the equilibrium molar free energy can be calculated explicitly,... [Pg.422]

SOLUTION The chemical equation and the expression for the equilibrium constant Kc are given in reaction C and Eq. 12, respectively. We substitute the equilibrium molar concentrations (without their units) into the expression for Kc ... [Pg.560]

Step 3 Write the equilibrium molar concentrations or partial pressures by adding the change in molar concentration or partial pressure (from step 2) to the initial value for each substance (from step 1). [Pg.569]

Step 4 Substitute these equilibrium molarities into the expression for 4 the acidity constant ... [Pg.615]

Step 4 Substitution of the equilibrium molarities into the expression for the basicity constant (Kh = 3.6 X 10-4) yields... [Pg.617]

Step 3 Write the equilibrium molarities of the species in terms of x. [Pg.620]

Step 1 Initial molarity Step 2 Change in molarity Step 3 Equilibrium molarity... [Pg.622]

Step 4 Substitute the equilibrium molarities into the acidity constant expression ... [Pg.630]


See other pages where Molarity equilibrium is mentioned: [Pg.100]    [Pg.508]    [Pg.509]    [Pg.509]    [Pg.510]    [Pg.577]    [Pg.594]    [Pg.188]    [Pg.177]    [Pg.29]    [Pg.565]    [Pg.565]    [Pg.565]    [Pg.565]    [Pg.568]    [Pg.570]    [Pg.572]    [Pg.576]    [Pg.576]    [Pg.586]    [Pg.587]    [Pg.588]    [Pg.614]    [Pg.615]    [Pg.617]    [Pg.621]    [Pg.629]    [Pg.630]    [Pg.633]    [Pg.634]    [Pg.635]    [Pg.649]   
See also in sourсe #XX -- [ Pg.77 ]




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