Forward and Reverse Reactions

We call this a dynamic equilibrium because the evaporation and condensation do not stop at equilibrium. The rates of these competing processes become equal to one another, but they do not go to zero. So on the microscopic level, individual molecules continue to move from the liquid to the vapor phase and back again. But from a macroscopic perspective, the observable amounts of liquid and vapor no longer change. 

Turning to chemical reactions, the manner in which we have written chemical equations thus far has been to consider only the forward direction of the reaction. In the reverse reaction, the substances we ve been calling products react to form reactants. In some cases, the reverse reaction may occur to a large degree and in other cases to an infinitesimal and immeasurable degree, but in principle it can occur in every chemical reaction. This means that in a closed system, any chemical reaction will eventually reach a dynamic equilibrium analogous to that we just described for evaporation. 

This graph implies that the reaction does not proceed very far toward products. We ll see next how this type of observation 

The rate laws of the forward and reverse reactions suggest that a mathematical relationship can be written to describe equilibrium. For any reaction involving reactants R and products P, we write the chemical equation at equilibrium with two arrows, one pointing in each direction, to emphasize the dynamic character of the 

For simplicity, we will assume that this reaction proceeds in each direction through a single elementary step. (See Section 11.6. The result of our derivation here does not actually depend on this assumption, but it is easier to understand this way.) We can write the rate law for the forward direction of this reaction as  

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Complex chemical mechanisms are written as sequences of elementary steps satisfying detailed balance where tire forward and reverse reaction rates are equal at equilibrium. The laws of mass action kinetics are applied to each reaction step to write tire overall rate law for tire reaction. The fonn of chemical kinetic rate laws constmcted in tliis manner ensures tliat tire system will relax to a unique equilibrium state which can be characterized using tire laws of tliennodynamics. 

Although a system at equilibrium appears static on a macroscopic level, it is important to remember that the forward and reverse reactions still occur. A reaction at equilibrium exists in a steady state, in which the rate at which any species forms equals the rate at which it is consumed. 

First, it should be noted that the /V-methy1o1 group is activated by the carbonyl group. This reactive group is present in almost all /V-methy1o1 systems. Second, the reaction is an equiUbrium reaction so that both forward and reverse reactions can occur. Third, the agent is not simply a dimethylol agent, but is predominandy a mixture of mono- and di-substituted ureas. 

Activation Processes. To be useful ia battery appHcations reactions must occur at a reasonable rate. The rate or abiUty of battery electrodes to produce current is determiaed by the kinetic processes of electrode operations, not by thermodynamics, which describes the characteristics of reactions at equihbrium when the forward and reverse reaction rates are equal. Electrochemical reaction kinetics (31—35) foUow the same general considerations as those of bulk chemical reactions. Two differences are a potential drop that exists between the electrode and the solution because of the electrical double layer at the electrode iaterface and the reaction that occurs at iaterfaces that are two-dimensional rather than ia the three-dimensional bulk. 

At equihbrium there is no net current flow, the rate of the forward and reverse reactions are equal and E = E. The rate of the forward and reverse reactions at equihbrium... 

V. For reversible reactions the net rate of reaction can be expressed as the difference between forward and reverse reactions ... 

The first step occurs rapidly and reversibly, a dynamic equilibrium is set up in which the rates of forward and reverse reactions are equal. Because the second step is slow and ratedetermining, it follows that... 

The rate expression just written is unsatisfactory in that it cannot be checked against experiment. The species NOCl2 is a reactive intermediate whose concentration is too small to be measured accurately, if at all. To eliminate the [NOCl2] term from the rate expression, recall that the rates of forward and reverse reactions in step 1 are equal, which means that... 

Strategy Write the rate expression for the rate-determining second step. This will involve the unstable intermediate, O. To get rid of [O], use the feet that reactants and products are in equilibrium in step 1, so forward and reverse reactions occur at the same rate. 

Chemical reactions involving gases carried out in closed containers resemble in many ways the H20(/)-H20(g) system. The reactions are reversible reactants are not completely consumed. Instead, an equilibrium mixture containing both products and reactants is obtained. At equilibrium, forward and reverse reactions take place at the same rate. As a result, the amounts of all species at equilibrium remain constant with time. 

The characteristics just described are typical of all systems at equilibrium. First, the forward and reverse reactions are taking place at the same rate. This explains why die concentrations of species present remain constant with time. Moreover, these concentrations are independent of the direction from which equilibrium is approached. 

These pressures don t change, because forward and reverse reactions occur at the same rate. 

The equilibrium constants for forward and reverse reactions are rite reciprocals of each other. ... 

The species shown in red, HB and HA, act as Bronsted-Lowry acids in the forward and reverse reactions, respectively A and B (blue) act as Bronsted-Lowry bases. (We will use this color coding consistently throughout the chapter in writing Bronsted-Lowry acid-base equations.)... 

The relationship between activation energies for the forward and reverse reactions can be expressed mathematically. The activation energy is denoted by the symbol A// (read delta-//-cross ) and the heat of the reaction by AH. Hence we may write ... 

Catalysts increase the rate of reactions. It is found experimentally that addition of a catalyst to a system at equilibrium does not alter the equilibrium state. Hence it must be true that any catalyst has the same effect on the rates of the forward and reverse reactions. You will recall that the effect of a catalyst on reaction rates can be discussed in terms of lowering the activation energy. This lowering is effective in increasing the rate in both directions, forward and reverse. Thus, a catalyst produces no net change in the equilibrium concentrations even though the system may reach equilibrium much more rapidly than it did without the catalyst. 

Forward and reverse reaction rates. The rate of isotopic exchange between U4+(aq) and U02+ in dilute perchloric acid is given by... 

Like physical equilibria, all chemical equilibria are dynamic equilibria, with the forward and reverse reactions occurring at the same rate. In Chapter 8, we considered several physical processes, including vaporizing and dissolving, that reach dynamic equilibrium. This chapter shows how to apply the same ideas to chemical changes. It also shows how to use thermodynamics to describe equilibria quantitatively, which puts enormous power into our hands—the power to control the And, we might add, to change the direction of a reaction and the yield of products,... 

Like phase changes, chemical reactions tend toward a dynamic equilibrium in which, although there is no net change, the forward and reverse reactions are still taking place, but at matching rates. What actually happens when the formation of ammonia appears to stop is that the rate of the reverse reaction,... 

All chemical equilibria are dynamic equilibria. Although there is no further net change at equilibrium, the forward and reverse reactions are still taking place. 

This step also is bimolecular. The last two equations add up to give the reverse of the overall equation, 3 02(g) —> 2 03(g). The forward and reverse reactions jointly provide a mechanism for reaching dynamic equilibrium between the reactants and the products in the overall process. 

At equilibrium, the rates of the forward and reverse reactions are equal. Because the rates depend on rate constants and concentrations, we ought to be able to find a relation between rate constants for elementary reactions and the equilibrium constant for the overall reaction. 

FIGURE 13.21 The equilibrium constant for a reaction is equal to the ratio of the rate constants for the forward and reverse reactions, (a) A forward rate constant (A) that is relatively large compared with the reverse rate constant means that the forward rate matches the reverse rate when the reaction has neared completion and the concentration of reactants is low. (b) Conversely, if the reverse rate constant (A ) is larger than the forward rate constant, then the forward and reverse rates are equal when little reaction has taken place and the concentration of products is low. 

A catalyst speeds up a reaction by providing an alternative pathway—a different reaction mechanism—between reactants and products. This new pathway has a lower activation energy than the original pathway (Fig. 13.34). At the same temperature, a greater fraction of reactant molecules can cross the lower barrier of the catalyzed path and turn into products than when no catalyst is present. Although the reaction takes place more quickly, a catalyst has no effect on the equilibrium composition. Both forward and reverse reactions are accelerated on the catalyzed path, leaving the equilibrium constant unchanged. 

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) For a reaction with a very large equilibrium constant, the rate constant of the forward reaction is much larger than the rate constant of the reverse reaction, (b) At equilibrium, the rate constants of the forward and reverse reactions are equal. 

A slightly related reaction involves the amino group of naphthylamines can be replaced by a hydroxyl group by treatment with aqueous bisulfite. The scope is greatly limited the amino group (which may be NH2 or NHR) must be on a naphthalene ring, with very few exceptions. The reaction is reversible (see 13-6), and both the forward and reverse reactions are called the Bucherer reaction. 

We assume the forward and reverse reactions have Arrhenius temperature dependences with Ef < Ef. Setting dbomldT = 0 gives... 

Many reactions show appreciable reversibility. This section introduces thermodynamic methods for estimating equilibrium compositions from free energies of reaction, and relates these methods to the kinetic approach where the equilibrium composition is found by equating the forward and reverse reaction rates. 

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. 

Equilibrium Compositions for Single Reactions. We turn now to the problem of calculating the equilibrium composition for a single, homogeneous reaction. The most direct way of estimating equilibrium compositions is by simulating the reaction. Set the desired initial conditions and simulate an isothermal, constant-pressure, batch reaction. If the simulation is accurate, a real reaction could follow the same trajectory of composition versus time to approach equilibrium, but an accurate simulation is unnecessary. The solution can use the method of false transients. The rate equation must have a functional form consistent with the functional form of K,i,ermo> e.g., Equation (7.38). The time scale is unimportant and even the functional forms for the forward and reverse reactions have some latitude, as will be illustrated in the following example. 

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