The Balance Equation

In the formulation of the microscopic balance equations, the molecular nature of matter is ignored and the medium is viewed as a continuum. Specifically, the assumption is made that the mathematical points over which the balance field-equations hold are big enough to be characterized by property values that have been averaged over a large number of molecules, so that from point to point there are no discontinuities. Furthermore, local equilibrium is assumed. That is, although transport processes may be fast and irreversible (dissipative), from the thermodynamics point of view, the assumption is made that, locally, the molecules establish equilibrium very quickly. 

Real flows are not frictionless. As a result of friction, a flow field is formed with local and time related changes in the velocity. The temperature and concentration fields will not only be determined by conduction and diffusion but also by the flow itself. The form and profiles of flow, temperature and concentration fields are found by solving the mass, momentum and energy balances, which are the subject of the next section. 

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The differential material balances contain a large number of physical parameters describing the structure of the porous medium, the physical properties of the gaseous mixture diffusing through it, the kinetics of the chemical reaction and the composition and pressure of the reactant mixture outside the pellet. In such circumstances it Is always valuable to assemble the physical parameters into a smaller number of Independent dimensionless groups, and this Is best done by writing the balance equations themselves in dimensionless form. The relevant equations are (11.20), (11.21), (11.22), (11.23), (11.16) and the expression (11.27) for the effectiveness factor. 

NOTE Alternately, you may want to try using a Tin/HC system which will give an equivalent yield in a much shorter time with the disadvantage that Tin is a much more expensive metal. The balanced equation for the reduction follows ... 

The balanced equation for turbulent kinetic energy in a reacting turbulent flow contains the terms that represent production as a result of mean flow shear, which can be influenced by combustion, and the terms that represent mean flow dilations, which can remove turbulent energy as a result of combustion. Some of the discrepancies between turbulent flame propagation speeds might be explained in terms of the balance between these competing effects. 

In Chapter 2 we developed models based on analyses of systems that had simple inputs. The right-hand side was either a constant or it was simple function of time. In those systems we did not consider the cause of the mass flow—that was literally external to both the control volume and the problem. The case of the flow was left implicit. The pump or driving device was upstream from the control volume, and all we needed to know were the magnitude of the flow the device caused and its time dependence. Given that information we could replace the right-hand side of the balance equation and integrate to the functional description of the system. 

The calculation of the two-zone model is based on the balance equations for air mass flow, contaminant mass flow, water vapor mass flow, and heat flow of both zones. 

The model equations are determined by writing the balance equations based on the conservation of mass and energy. Tlie balance equations have the following basic form ... 

The balance equations for water vapor flows are similar to balance equations for contaminant flows, but in addition possible condensation and evaporation must be calculated. Also they must be considered in heat flow equations. 

With this as introduction, let us now look at the individual steps in more detail for the case of hydroboration-oxidation of 1-decene. A boron hydride that is often used is dibomne (B2H6). Diborane adds to 1-decene to give tridecylborane according to the balanced equation ... 

Consider the balanced equation for glnconeogenesis in Section 23.1. Account for each of the components of this equation and the indicated stoichiometry. 

Every chemical reaction can go in either forward or reverse direction. Reactants can go forward to products, and products can revert to reactants. As you may remember from your general chemistry course, the position of the resulting chemical equilibrium is expressed by an equation in which /Cec], the equilibrium constant, is equal to the product concentrations multiplied together, divided by the reactant concentrations multiplied together, with each concentration raised to the power of its coefficient in the balanced equation. Eor the generalized reaction... 

Chemical reactions are represented by chemical equations, which identify reactants and products. Formulas of reactants appear on the left side of the equation those of products are written on the right In a balanced chemical equation, there are the same number of atoms of a given element on both sides. The same situation holds for a chemical reaction that you carry out in the laboratory atoms are conserved. For that reason, any calculation involving a reaction must be based on the balanced equation for that reaction. 

As you found in Example 4.10, the balanced equation for the reaction between Mn04 and Fe2+ in acidic solution is... 

Strategy (1) Start by calculating the number of moles of Fe2+. Then (2) use the coefficients of the balanced equation to find the number of moles of Mn04. Finally, (3), use molarity as a conversion factor to find the volume of KMn04 solution. 

The balanced equation for the reaction that takes place is... 

Laundry bleach is a solution of sodium hypochlorite (NaCIO). To determine the hypochlorite (CIO-) content of bleach (which is responsible for its bleaching action), sulfide ion is added in basic solution. The balanced equation for the reaction is... 

The law of combining volumes, like so many relationships involving gases, is readily explained by the ideal gas law. At constant temperature and pressure, volume is directly proportional to number of moles (V = kin). It follows that for gaseous species involved in reactions, the volume ratio must be the same as the mole ratio given by the coefficients of the balanced equation. 

As you can see from Figure 11.1 (p. 286), the concentration of N2Os decreases with time the concentrations of N02 and 02 increase. Because these species have different coefficients in the balanced equation, their concentrations do not change at the same rate. When one mole of N2Os decomposes, two moles of N02 and one-half mole of 02 are formed. This means that... 

The order of a reaction must be determined experimentally it cannot be deduced from die coefficients in the balanced equation. This must be true because there is only one reaction order, but there are many different ways in which the equation for the reaction can be balanced. For example, although we wrote... 

Sometimes the rate expression obtained by the process just described involves a reactive intermediate, that is, a species produced in one step of the mechanism and consumed in a later step. Ordinarily, concentrations of such species are too small to be determined experimentally. Hence they must be eliminated from the rate expression if it is to be compared with experiment. The final rate expression usually includes only those species that appear in the balanced equation for the reaction. Sometimes, the concentration of a catalyst is included, but never that of a reactive intermediate. 

Reality Check Notice that the concentration of 02, a product in the reaction, appears in the denominator. The rate is inversely proportional to the concentration of molecular oxygen, a feature that you would never have predicted from the balanced equation for the reaction. As the concentration of 02 builds up, the rate slows down. 

Note from the balanced equation that one mole of I2 is formed for every mole of H2. Putting it another way, as the reaction proceeds from left to right, the increase in the number of moles of I2 is the same as that for H2 ... 

Note that the numbers for AP are related through the coefficients of the balanced equation (2HI, I2, H2)... 

Example 12.4 illustrates a principle that you will find very useful in solving equilibrium problems throughout this (and later) chapters. As a system approaches equilibrium, changes in partial pressures of reactants and products—like changes in molar amounts—are related to one another through the coefficients of the balanced equation. 

Using the balanced equation for the reaction, write the expression for K. 

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. 

All the coefficients in the balanced equation are the same, 1. It follows that all the partial pressures change by the same amount, x. 

The quantities E°, E x, and Eed are independent of how die equation for the cell reaction is written. You never multiply the voltage by the coefficients of the balanced equation. 

Strategy First (1), write a balanced equation for the reaction, which is very similar to that for ZnS, except that Zn2+ is replaced by Bi3+. (2) Using the balanced equation, calculate the number of moles of S02. Finally (3), use the ideal gas law to calculate the volume of S02. 

Avogadro s Hypothesis provides a method for identifying the molecules present in a gas. Also, it explains why the volumes of gases that react with each other are in the same simple ratio as are the moles in the balanced equation. The importance of these results makes the explana-... 

If we start with hydrogen and oxygen, equilibrium is attained after most of the hydrogen and oxygen have united to form water. More important, though, the partial pressures at equilibrium are the same as those obtained beginning with pure H20. The equilibrium pressures are fixed by the temperature, the composition, and the total pressure they do not depend upon the direction from which equilibrium is approached. The balanced equation does not indicate the concentrations (or partial pressures) at equilibrium. 

The solubility product is learned from measurements of the solubility. In turn, it can be used as a basis for calculations of solubility. Suppose we wish to know how much cuprous chloride, CuCl, will dissolve in one liter of water. We begin by writing the balanced equation for the reaction ... 

Will a precipitate form in either case In both The first step is to write the balanced equation for the reaction of calcium sulfate dissolving in water and then use the Equilibrium Law ... 

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