Exponent equilibrium-constant expression

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). 

The rules for writing the equilibrium constant expression apply to reactions in solution as well as to gas-phase reactions. Put the concentrations of the products in the numerator and the concentrations of the reactants in the denominator. No exponents are needed because all the coefficients in the balanced chemical equation equal unity. 

The exponents in the equilibrium constant expression for Ksp are the coefficients in the balanced equation. Therefore,... 

In Eq. (16-4), k is called the rate constant, the exponents a, b, c, and x are called orders. The orders are usually integers, but may be fractions such as 1/2 or 2/3. A positive order a means that the rate increases with [A], a negative order c means that the rate decreases with increasing [C], If the rate depends on [X] and x > 0, we say that X is a catalyst which increases the rate if x < 0, we say that X is an inhibitor which decreases the rate. The sum of all the individual orders is called the overall order of the reaction. Despite the apparent similarity of an empirical rate law to an equilibrium constant expression, the orders are not necessarily equal to stoichiometric coefficients. 

In this reaction Equation 13.1, the coefficients from the balanced equation become the exponents in the equilibrium constant expression. The concentrations of the products go in the numerator, and the concentrations of the reactants go in the denominator. 

I In equilibrium constant expressions, the concentration terms are multiplied and divided, not added or subtracted. The concentrations in the numerator are those of the substances on the right side of the chemical equation, and those in the denominator are for the substances on the left side. The exponents in the equilibrium constant expression... 

Note that the two chemical equations state the same facts, as do the two equilibrium constant expressions. This pair of equations shows why die coefficient in the balanced chemical equation becomes an exponent in the equilibrium constant expression. 

An equilibrium-constant expression can be written only after a correct, balanced chemical equation that describes the equilibrium system has been developed. A balanced equation is essential because the coefficients in the equation become the exponents in the equilibrium-constant expression. 

The correct equilibrium constant expression is b. The products are written in the numerator and the reactants are written in the denominator. Because H2 has a coefficient of 3 in the balanced equation, an exponent of 3 is used with the concentration... 

Thermodynamics defines the relationship governing the concentrations at equilibrium. The exponents of the equilibrium concentrations in the expression for the equilibrium constant are the same as the stoichiometric coefficients in the net equation, as in... 

Another problem which can appear in the search for the minimum is intercorrelation of some model parameters. For example, such a correlation usually exists between the frequency factor (pre-exponential factor) and the activation energy (argument in the exponent) in the Arrhenius equation or between rate constant (appears in the numerator) and adsorption equilibrium constants (appear in the denominator) in Langmuir-Hinshelwood kinetic expressions. 

A modest value of E° produces a large equilibrium constant. The value of K is correctly expressed with one significant figure, because E° has three digits. Two are used for the exponent (14), and only one is left for the multiplier (4). 

Thus the ratio of the forward and back rate constants. (K2IK ) is not the equilibrium constant for the overall reaction, but K, the exponent being the reciprocal of the stoicheiometric number of the limiting reaction. This significance and its exploitation have been examined by Horiuti [10. Frank-Kamenetsky [11] has developed some safe conditions for the applicability of certain forms of the hypothetical method, in the form of rather strong inequalities on the partial derivatives of the kinetic expressions. 

When we consider a system at equilibrium, we find that if, at a specific temperature, we arrange the reactant and product concentrations expressed in moles per liter in a particular way, no matter what these concentrations are, a constant value results. This constant value is called the equilibrium constant and, for solubility, is given the symbol Ksp. The general formula to find the is the product (multiplication) of the concentrations of the products expressed in moles per liter over the product of the concentrations of the reactants expressed in moles per liter. Also, to find the constant, coefficients in the dissolving equation become exponents. Once we know the Ksp for the solubility of a substance, we can know precisely how soluble that substance is. When we are looking for a fixing agent for film development, we need a substance that will combine with silver ions and form a very insoluble product. The Ksp for a substance is determined from an equation as follows For... 

The correct answer is (B). This is a matter of remembering the proper procedures for writing an equilibrium constant. In the expression, the product(s) is the numerator and the reactant(s) is the denominator. The coefficients for each should become exponents. Therefore, in the reaction shown, the equilibrium expression should be ... 

C is correct. The equilibrium constant is products over reactants with the coefficients as exponents. However, reactants and products in pure liquid and solid phases generally have an exponent of zero, so they are not included in the equilibrium expression. 

The expression obtained here is the correct expression for the equilibrium constant, but the method of derivation has no general validity. This is because reaction rates actually depend on the mechanism of the reaction, determined by the number of colliding species, whereas the equihbrium constant expression depends only on the stoichiometry of the chemical reaction. The sum of the exponents in the rate constant gives the order of the reaction, and this may be entirely different from the stoichiometry of the reaction (see Chapter 22). An example is the rate of reduction of S20g with I ... 

Fits of two principal reaction mechanisms, both of which have the above general form, were made, after initial trials of rate expressions corresponding to mechanisms with other forms of rate expression had resulted in the rejection of these forms. In the above equation the Molecular Adsorption Model (MAM) predicts n=2, m=l while the Dissociative Adsorption Model (DAM) leads to n=2, m=l/2. The two mechanisms differ in that MAM assumes that adsorbed molecular oxygen reacts with adsorbed carbon monoxide molecules, both of which reside on identical sites. Alternatively, the DAM assumes that the adsorbed oxygen molecules dissociate into atoms before reaction with the adsorbed carbon monoxide molecules, once more both residing on identical sites. The two concentration exponents, referred to as orders of reaction, are temperature independent and integral. All the other constants are temperature dependent and follow the Arrhenius relationship. These comprise lq, a catalytic rate constant, and two adsorption equilibrium constants K all subject to the constraints described in Chapter 9. Notice that a mechanistic rate expression always presumes that the rate is measured at constant volume. 

This expression is sometimes called the Law of Mass Action. Note that the equilibrium constant, Kc, always places products in the numerator of the expression and reactants in the denominator this criterion alone eliminates choice (D). The second feature to note is that the coefficients in the chemical equation become exponents in the equilibrium expression. The only coefficient here is the 2 in front of the reactant F2 molecule, and that becomes the exponent shown in choice (C), the correct choice. Choice (A) does not include the coefficient, and choice (B) uses the coefficient as a multiplier rather than as an exponent. 

As you can see, K2 is the square root of Ki. Thus, dividing by two in the stoichiometry leads to a division by two in the exponents of the equilibrium expression. It s clear that we must look carefully at the way a chemical equation is written when we consider the value of the equilibrium constant. 

Notice that the coefficients in the chemical equation become the exponents in the expression of the equilibrium constant. 

The exponent Mk depends on the mean square displacement of the atom from its equilibrium position and hence upon temperature. It is linear with (kT/m Xsin / where k is the Boltzmann constant, T the absolute temperature, the scattering angle, the wavelength and m the atomic mass (for a monatomic material). In addition there are complicated expressions dependent upon the crystal symmetry. As an example, for silicon at room temperature the /, are reduced by approximately 6%. With this correction all the equations of dynamical theory still apply. 

The variable k represents the rate constant. Note the order of each reactant is 1. The reaction order, which describes the order of the entire reaction, can be determined by adding the order of each reactant. For instance, in this example each reactant is first order (meaning each has an understood exponent of 1). The reaction order is the sum of the exponents, or 1 + 1=2. This is a second-order reaction. Most reactions have an order of 0, 1, or 2, but some have fractional orders or larger numbers (though these are quite rare). The order of the reaction must be determined experimentally. Unlike equilibrium expressions, the exponents have nothing to do with the coefficients in the balanced equations. 

Rate is the disappearance of reactants with time expressed as the derivative —d[A ld(t). [A] and [5] are the concentrations of reactants and [C] and [D] are the concentrations of the products. The exponents a, b, c, and d are the reaction orders for each compound. The rate constants are kf for the forward and kREV for the reverse reaction. For those cases where the reaction is far away from equilibrium the reverse rate is negligible and this term is dropped from the rate expression. 

According to these expressions, the intensity of the OH emission will decay as a biexponent, the rapid initial phase 72 represents the reaction as it proceeds until the velocity of dissociation and recombination become equal. The slower phase 71 represents the decay when the two populations (< >OH and 0 ) are in equilibrium with each other. The relative amplitudes of the two phases Ar = (a2i — 7i)/(72 7i) and the macroscopic rate constants (71,72) allow one to calculate the rate of all partial reactions. The agreement between rate constants calculated by time-resolved measurements and steady-state kinetics is usually good. In a limiting case, where the rate of recombination is much slower than dissociation pKo > pH >> pK, the amplitude of the slow phase representing recombination will diminish to zero and the emission of the < >OH state will decay in a single exponent curve with a macroscopic rate constant 72 = k + %,nr) k. ... 

If valence is thought of as a measure of a bond s attractive force, it should, at equilibrium, be proportional to the repulsive overlap force between the atom cores. Not surprisingly, expressions that have been proposed for the repulsive component of interatomic forces are similar in form to Equations 10.1 and 10.2. Biirgi and Dunitz [16] have shown that Equation 10.1 can be derived from the Morse potential, and Brown and Shannon [8] have shown that the constant, W, in Equation 10.2 is related to the Born exponent [17]. 

Square brackets indicate concentrations (as they do in an equilibrium expression - see Chapter 7), the exponents a and b are the individual orders of the reaction with respect to the reactants A and B, and k is the proportionality constant, known as the rate constant. The values ofa,b and k have to be determined experimentally. The sum of the individual orders, a + b, is known as the overall order of the reaction. The order of reaction with respect to a particular reactant indicates precisely what happens to the rate of reaction if that concentration is changed. 

The denominator, moreover, contains the reactant concentration raised to a power equal to its stoichiometric coefficient in the balanced chemical equation. (In this equation the coefficient of N2O4 is 1, which generally is not written either as a coefficient or as an exponent.) Table 15.1 lists the starting and equilibrium concentrations of N2O4 and NO2 in a series of experiments carried out at 25°C. Using the equilibrium concentrations from each of the experiments in the table, the value of the expression [N02]V[N204] is indeed constant (the average is 4.63 x 10 ) within the limits of experimental error. 

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