Quantities 1 Using Chemical Equations

It is important to be able to calculate how much of a particular product will be produced when certain quantities of the reactants are consumed. In the next section, we will see how to use chemical equations to set up conversion factors that we can use for these and related calculations. 

The rate of a process is expressed by the derivative of a concentration (square brackets) with respect to time, d[ ]/dt. If the concentration of a reaction product is used, this quantity is positive if a reactant is used, it is negative and a minus sign must be included. Also, each derivative d[ ]/dt should be divided by the coefficient of that component in the chemical equation which describes the reaction so that a single rate is described, whichever component in the reaction is used to monitor it. A rate law describes the rate of a reaction as the product of a constant k, called the rate constant, and various concentrations, each raised to specific powers. The power of an individual concentration term in a rate law is called the order with respect to that component, and the sum of the exponents of all concentration terms gives the overall order of the reaction. Thus in the rate law Rate = k[X] [Y], the reaction is first order in X, second order in Y, and third order overall. 

SOLUTION Use Eq. 1 to determine a reaction Gibbs free energy—a thermodynamic quantity—from a cell emf—an electrical quantity. From the chemical equation for the cell reaction (reaction A), we see that n = 2 mol. 

Given this context, the use of chemical symbols, formulae and equations can be readily misinterpreted in the classroom, because often the same representations can stand for both the macroscopic and sub-microscopic levels. So H could stand for an atom, or the element hydrogen in an abstract sense H2 could mean a molecule or the substance. One common convention is that a chemical equation represents molar quantities, so in Example 9 in Table 4.1,... 

When two substances react, they react in exact amounts. You can determine what amounts of the two reactants are needed to react completely with each other by means of mole ratios based on the balanced chemical equation for the reaction. In the laboratory, precise amounts of the reactants are rarely used in a reaction. Usually, there is an excess of one of the reactants. As soon as the other reactant is used up, the reaction stops. The reactant that is used up is called the limiting reactant. Based on the quantities of each reactant and the balanced chemical equation, you can predict which substance in a reaction is the limiting reactant. 

The balanced equation expresses quantities in moles, but it is seldom possible to measure out quantities in moles directly. If the quantities given or required are expressed in other units, it is necessary to convert them to moles before using the factors of the balanced chemical equation. Conversion of mass to moles and vice versa was considered in Sec. 4.5. Here we will use that knowledge first to calculate the number of moles of reactant or product, and then use that value to calculate the number of moles of other reactant or product. 

If the quantities of both reactants are in exactly the correct ratio for the balanced chemical equation, then either reactant may be used to calculate the quantity of product produced. (If on a quiz or examination it is obvious that they are in the correct ratio, you should state that they are so that your instructor will understand that you recognize the problem to be a limiting quantities problem.)... 

Sulfurous acid reacts with sodium hydroxide to produce sodium sulfite and water, (a) Write a balanced chemical equation for the reaction. (b) Determine the number of moles of sulfurous acid in 50.0g sulfurous acid, (c) How many moles of sodium sulfite will be produced by the reaction of this number of moles of sulfurous acid (d) How many grams of sodium sulfite will be produced (e) How many moles of sodium hydroxide will it take to react with this quantity of sulfurous acid (/) How many grams of sodium hydroxide will be used up ... 

The total quantity of reactant is limited to 5.000 g. If either reactant is in excess, the amount in excess will be wasted, because it cannot be used to form product. Thus, we obtain the maximum amount of product when neither reactant is in excess (i.e., when there is a stoichiometric amount of each present). The balanced chemical equation for this reaction, 2 KI + Pb (N03 )2 -> 2 KN03 + Pbl2, shows that... 

The relationship above gives a way of converting from grams to moles to particles, and vice versa. If you have any one of the three quantities, you can calculate the other two. This becomes extremely useful in working with chemical equations, as we will see later, because the coefficients in the balanced chemical equation are not only the number of individual atoms or molecules at the microscopic level, but also the number of moles at the macroscopic level. 

Of these, the first reaction is the one usually recorded, and in the majority of cases it is the one that occurs to the largest extent. The presence of alkali pushes the equilibrium well to the left, so that in alkaline solution the thiosulphate is stable. This explains the formation of thiosulphates on boiling alkaline sulphite solutions with sulphur. The laws of chemical equilibria, however, demand the presence of perfectly definite although perhaps very small quantities of sulphite and free sulphur in solution, and if alkaline thiosulphate solutions containing alkali sulphide are boiled in the absence of air, they become deep yellow, owing to polysulphide formation, the extra sulphur for which is obtained from the thiosulphate. Assuming the sodium derivatives to be used, the equation may be written... 

Quantities of chemicals, expressed in moles, can be used to find the formula of a compound, to establish an equation and to determine reacting masses, a A compound contains 72% magnesium and 28% nitrogen. What is its empirical formula [2]... 

The quantity gk T) in Equation (7.67) is again a molar quantity, characteristic of the individual gas, and a function of the temperature. It can be related to the molar Gibbs energy of the fcth substance by the use of Equation (7.67). The first two terms on the right-hand side of this equation are zero when the gas is pure and ideal and the pressure is 1 bar. Then gk(T) is the chemical potential or molar Gibbs energy for the pure fcth substance in the ideal gas state at 1 bar pressure. We define this state to be the standard state of the fcth substance and use the symbol 1 bar, yk = 1] for the... 

It is apparent from this equation that we can determine only the difference between the two excess chemical potentials at the experimental conditions. If one is known from other studies, then the other can be determined. It must be pointed out that the difference is not isothermal and, if isothermal quantities are desired, the same corrections as discussed in Section 10.12 must be made by the use of Equation (10.87). 

The quantity AH° is the change of enthalpy for the change of state represented by a balanced chemical equation under the condition that all substances are in their standard states at the temperature and pressure in question. The overbar is used here because some of the individual enthalpies, Hf, may be partial molar quantities. 

An equilibrium constant must always be accompanied by a chemical equation. This equation is often used without the subscript eq that reminds us that the concentrations are equilibrium values. Strictly speaking, this equation should be written as K = TT(ci/c0), but the standard concentration c° = 1 M will be omitted, as mentioned before equation 3.1-8. Thus the equilibrium constant will be treated as a dimensionless quantity, as, of course, it must be if we are going to take its logarithm. 

Constituents can either be elements or compounds. While writing the chemical equations, we can use zero as well as negative quantities of the constituents, besides the positive ones (as is the convention). 

Errors and confusion in modelling arise because the complex set of coupled, nonlinear, partial differential equations are not usually an exact representation of the physical system. As examples, first consider the input parameters, such as chemical rate constants or diffusion coefficients. These input quantities, used as submodels in the detailed model, must be derived from more fundamental theories, models or experiments. They are usually not known to any appreciable accuracy and often their values are simply guesses. Or consider the geometry used in a calculation. It is often one or two dimensions less than needed to completely describe the real system. Multidimensional effects which may be important are either crudely approximated or ignored. This lack of exact correspondence between the model adopted and the actual physical system constitutes the basic problem of detailed modelling. This problem, which must be overcome in order to accurately model transient combustion systems, can be analyzed in terms of the multiple time scales, multiple space scales, geometric complexity, and physical complexity of the systems to be modelled. 

The balanced chemical equation for this reaction shows that 1 mol of nitric acid reacts with 1 mol of sodium hydroxide. If equal molar quantities of nitric acid and sodium hydroxide are used, the result is a neutral (pH 7) aqueous solution of sodium nitrate. In fact, when any strong acid reacts with any strong base in the mole ratio from the balanced chemical equation, a neutral aqueous solution of a salt is formed. Reactions between acids and bases of different strengths usually do not result in neutral solutions. 

Balancing a chemical equation requires an understanding of the Law of Conservation of Mass, which says that mass cannot be created or destroyed. The amount of mass in the reactants will be the amount of mass in the products. The credit for this discovery is given to Antoine Lavoisier, who took very careful measurements of the quantities of chemicals and equipment that he used. Conservation of mass also holds true when balancing equations. The number of atoms of each element in the reactants will be equal to the number of atoms of each element in the products. A useful mnemonic device for conservation of mass is What goes in, must come out. ... 

Stoichiometric quantities of reactants are almost always used, and only in the most unusual cases is it necessary to use more or less of a reactant than the quantity demanded by the chemical equation. On the other hand, diluents must often be used in order to have the correct mixture during the reaction. It should also be emphasized that intensive mixing (stirring) is usually necessary for the satisfactory progress of the reaction. Also, the reaction temperature often plays an important role, for example, in preventing the formation of undesirable isomers. 

Lesson 4 described "How much is there and Lesson 5 expands to elucidate "How much is supposed to be there Reaction stoichiometry establishes the quantities ot reactants (used) and products (obtained) based on a balanced chemical equation. 

Stoichiometry establishes the quantities of reactants (used) and products (obtained) based on a balanced chemical equation. With a balanced equation, you can compare reactants and products, and determine the amount of products that might be formed or the amount or reactants needed to produce a certain amount of a product. However, when comparing different compounds in a reaction, you must always compare in moles (i.e., the coefficients). The different types of stoichiometric calculations are summarized in Figure 5.1. 

Stoichiometry The quantities of reactants (used) and products (obtained) based on a balanced chemical equation. 

Step 4 Complete the change row by writing the number of moles of each substance that would react with or be produced from the quantity in step 3. Use a minus sign with each quantity of reactant. The magnitudes in the change row are in the same ratio as the coefficients in the balanced chemical equation. 

Using this method, we calculate the quantity of each product produced and the quantity of any excess reactant all in one calculation. Note that the quantities in the initial and final rows are not in the ratio of the balanced chemical equation only the magnitudes of the ratios in the change row are in the same ratio as those in the balanced equation. 

Balanced chemical equation, using limiting quantity... 

Step 2 Start the second row by entering the moles of limiting quantity—the same as the number of moles of that substance in the first row, since all of the limiting quantity is used up. The other entries in the second row are calculated using the technique of Sec. 10.1. The entries in row 2 will always be in the same mole ratio as the coefficients in the balanced chemical equation. 

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