Chemical equations constants

The logarithm of the equilibrium constant, K,. for the chemical equation shown in Figure 3-8a for a substituted benzoic acid can be related to the logarithm of the... 

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. 

Results may be reported for any component. The functional form of the rate law and the exponents x,j, w,... are not affected by such an arbitrary choice. The rate constants, however, may change in numerical value. Similarly, the stoichiometric chemical equation may be written in alternative but equivalent forms. This also affects, at most, the numerical value of rate constants. Consequentiy, one must know the chemical equation assumed before using any rate constant. 

In these circumstances a decision must be made which of two (or more) kinet-ically equivalent rate terms should be included in the rate equation and the kinetic scheme (It will seldom be justified to include both terms, certainly not on kinetic grounds.) A useful procedure is to evaluate the rate constant using both of the kinetically equivalent forms. Now if one of these constants (for a second-order reaction) is greater than about 10 ° M s-, the corresponding rate term can be rejected. This criterion is based on the theoretical estimate of a diffusion-controlled reaction rate (this is described in Chapter 4). It is not physically reasonable that a chemical rate constant can be larger than the diffusion rate limit. 

The equilibrium constant expression for the dissolving of SrCr04 can be written following the rules in Chapters 12 and 13. In particular, the solid does not appear in the expression the concentration of each ion is raised to a power equal to its coefficient in the chemical equation. 

Charles s and Gay-Lussac s law Relation stating that at constant P and n, the volume of a gas is directly proportional to its absolute temperature, 106-107, 111 Chelating agent Complexing ligand that forms more than one bond with a central metal atom the complex formed is called a chelate, 411-412 natural, 424-425 synthetic, 424-425 Chemical equation Expression that describes the nature and relative amounts of reactants and products in a reaction, 60-61. See also Equation, net ionic. 

Clausius-Clapeyron equation An equation expressing the temperature dependence of vapor pressure ln(P2/Pi) = AHvapCl/Tj - 1/T2)/R, 230,303-305 Claussen, Walter, 66 Cobalt, 410-411 Cobalt (II) chloride, 66 Coefficient A number preceding a formula in a chemical equation, 61 Coefficient rule Rule which states that when the coefficients of a chemical equation are multiplied by a number n, the equilibrium constant is raised to the nth power, 327... 

In this generalized equation, (75), we see that again the numerator is the product of the equilibrium concentrations of the substances formed, each raised to the power equal to the number of moles of that substance in the chemical equation. The denominator is again the product of the equilibrium concentrations of the reacting substances, each raised to a power equal to the number of moles of the substance in the chemical equation. The quotient of these two remains constant. The constant K is called the equilibrium constant. This generalization is one of the most useful in all of chemistry. From the equation for any chemical reaction one can immediately write an expression, in terms of the concentrations of reactants and products, that will be constant at any given temperature. If this constant is measured (by measuring all of the concentrations in a particular equilibrium solution), then it can be used in calculations for any other equilibrium solution at that same temperature. 

He also notes a situation where the rate constant for product buildup appears to be larger than that for reactant consumption. This signals intervention of an intermediate and is a special case of Eq. (4-25). The chemical equations are... 

The equilibrium composition of a reaction mixture is described by the equilibrium constant, which is equal to the activities of the products (raised to powers equal to their stoichiometric coefficients in the balanced chemical equation for the reaction) divided by the activities of the reactants (raised to powers equal to their stoichiometric coefficients). 

Because equilibrium constants must be raised to a power when a chemical equation is multiplied by a factor, and therefore change in magnitude, these rules are only general guidelines. 

SOLUTION First convert the units of partial pressures to bars using 1 bar = 105 Pa P i7 = 4.2 X 10 8 bar, Pcli = 8.3 X 10 8 bar. Write the expression for the equilibrium constant from the chemical equation given. 

The powers to which the activities are raised in the expression for an equilibrium constant must match the stoichiometric coefficients in the chemical equation, which is normally written with the smallest whole numbers for coefficients. Therefore, if we change the stoichiometric coefficients in a chemical equation (for instance, by... 

If we rewrite the chemical equation by multiplying through by 2, the equilibrium constant becomes... 

A note on good practice As these examples have shown, it is important to specify the chemical equation to which the equilibrium constant applies. 

If a chemical equation can be expressed as the sum of two or more chemical equations, the equilibrium constant for the overall reaction is the product of the equilibrium constants for the component reactions. For example, consider the three gas-phase reactions... 

Step 1 Write the balanced chemical equation for the equilibrium and the corresponding expression for the equilibrium constant. Then set up an equilibrium table as shown here, with columns labeled by the species taking part in the reaction. In the first row, show the initial composition (molar concentration or partial pressure) of each species... 

A catalyst speeds up both the forward and the reverse reactions by the same amount. Therefore, the dynamic equilibrium is unaffected. The thermodynamic justification of this observation is based on the fact that the equilibrium constant depends only on the temperature and the value of AGr°. A standard Gibbs free energy of reaction depends only on the identities of the reactants and products and is independent of the rate of the reaction or the presence of any substances that do not appear in the overall chemical equation for the reaction. 

The following plot shows how the partial pressures of reactant and products vary with time for the decomposition of compound A into compounds B and C. All three compounds are gases. Use this plot to do the following (a) Write a balanced chemical equation for the reaction, (h) Calculate the equilibrium constant for the reaction, (c) Calculate the value of Kc for the reaction at 25°C. 

STRATEGY First, we write the chemical equation for the equilibrium between the solid solute and the complex in solution as the sum of the equations for the solubility and complex formation equilibria. The equilibrium constant for the overall equilibrium is therefore the product of the equilibrium constants for the two processes. Then, we set up an equilibrium table and solve for the equilibrium concentrations of ions in solution. 

The interhalogen IFT can be made only by indirect routes. For example, xenon difluoride gas can react with iodine gas to produce 1FV and xenon gas. In one experiment xenon difluoride is introduced into a rigid container until a pressure of 3.6 atm is reached. Iodine vapor is then introduced until the total pressure is 7.2 atm. Reaction is then allowed to proceed at constant temperature until completion by solidifying the IF as it is produced. The final pressure in the flask due to the xenon and excess iodine vapor is 6.0 atm. (a) What is the formula of the mterhalogen (b) Write the chemical equation for its formation. 

In principle, Equation (7.28) is determined by equating the rates of the forward and reverse reactions. In practice, the usual method for determining Kkinetic is to run batch reactions to completion. If different starting concentrations give the same value for Kkinetic, the functional form for Equation (7.28) is justified. Values for chemical equilibrium constants are routinely reported in the literature for specific reactions but are seldom compiled because they are hard to generalize. 

These four equations are perfectly adequate for equilibrium calculations although they are nonsense with respect to mechanism. Table 7.2 has the data needed to calculate the four equilibrium constants at the standard state of 298.15 K and 1 bar. Table 7.1 has the necessary data to correct for temperature. The composition at equilibrium can be found using the reaction coordinate method or the method of false transients. The four chemical equations are not unique since various members of the set can be combined algebraically without reducing the dimensionality, M=4. Various equivalent sets can be derived, but none can even approximate a plausible mechanism since one of the starting materials, oxygen, has been assumed to be absent at equilibrium. Thermodynamics provides the destination but not the route. 

C06-0057. Acetylene (C2 H2) Is used In welding torches because it has a high heat of combustion. When 1.00 g of acetylene bums completely in excess O2 gas at constant volume, it releases 48.2 kJ of energy, (a) What Is the balanced chemical equation for this reaction (b) What is the molar energy of combustion of acetylene (c) How much energy is released per mole of O2 consumed ... 

The proportionality constant k, called the rate constant, has a constant value for a given reaction at a given temperature. The terms in square brackets are concentration terms (compare Chap. 14), and x and y are exponents which are often integral. The exponent x is called the order with respect to A, and y is called the order with respect to B. The sum x +y is called the overall order of the reaction. The values for x and y can be 0, 1, 2, 3 or 0.5, 1.5, or 2.5, but never more than 3. These values must be determined by experiment, and do not necessarily equal the values of a and b in the chemical equation. 

When the chemical equation is more complex, the equilibrium constant expression is also more complex and the deductions about the equilibrium concentrations of reactants and products are more involved too. 

The equilibrium constant expressions arc the same because the chemical equations arc the same. It is easy to see why the coefficients of the chemical equation are used as exponents in the equilibrium constant expression by writing out the equation and expression as in part (a). 

As for all chemical kinetic studies, to relate this measured correlation function to the diffusion coefficients and chemical rate constants that characterize the system, it is necessary to specify a specific chemical reaction mechanism. The rate of change of they th chemical reactant can be derived from an equation that couples diffusion and chemical reaction of the form (Elson and Magde, 1974) ... 

First we write the balanced chemical equation for the reaction. Then we write the equilibrium constant expressions, remembering that gases and solutes in aqueous solution appear in the Kc expression, but pure liquids and pure solids do not. 

Because the reaction quotient has a smaller value than the equilibrium constant, a net reaction will occur to the right. We now set up this solution as we have others, based on the balanced chemical equation. 

Then we calculate equilibrium partial pressures, organizing our calculation around the balanced chemical equation. We see that the equilibrium constant is not very large, meaning that we must solve the equation exactly (or by using successive approximations). 

B We need the balanced chemical equation in order to write the equilibrium constant expression. We start by translating names into formulas. 

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