Zero-order rate laws

Alternatively, if the reactions at the surface are slow in comparison with diffusion or other reaction steps, the dissolution processes are controlled by the processes at the surface. In this case the concentrations of solutes adjacent to the surface will be the same as in the bulk solution. The dissolution kinetics follows a zero-order rate law if the steady state conditions at the surface prevail ... 

The conventional mechanism and mathematical treatment for solvent extraction kinetics was proposed by Oele et al. in 1951 (11) and has since been accepted and used by others. Oele assumed that the coal will enter the liquid solvent in accordance with a zero-order rate law up to a certain time ... 

Some ketones such as /3-dicarbonyls contain substantial amounts of the enol at equilibrium. For example, acetylacetone in aqueous solutions contains 13% of 4-hydroxypent-3-en-2-one, which is stabilized both by an intramolecular hydrogen bond and the inductive effect of the remaining carbonyl group.17 When bromine is added to such a solution, a portion is initially consumed very rapidly by the enol that is already present at equilibrium. The ketone remaining after consumption of the enol reacts more slowly via rate-determining enolization. The slow consumption of bromine is readily measured by optical absorption. In acidic solutions containing a large excess of the ketone the slow reaction follows a zero-order rate law the rate is independent of bromine concentration, because any enol formed is rapidly trapped by bromine (Scheme 1). In this case, the amount of enol present at equilibrium may be determined as the difference between the amount of bromine added and that determined by extrapolation of the observed rate law to time zero, as is shown schematically in Fig. 2. 

A catalytic hydrogenation is performed at constant pressure in a semi-batch reactor. The reaction temperature is 80 °C. Under these conditions, the reaction rate is lOmmolT s-1 and the reaction may be considered to follow a zero-order rate law. The enthalpy of the reaction is 540 kj moT1. The charge volume is 5 m3 and the heat exchange area of the reactor 10 m2. The specific heat capacity of water is 4.2kJkg 1K 1. 

The solution of this differential equation, when it exists, describes the temperature profile in the solid. In order to solve this equation, we must assume that the exothermal reaction taking place in the solid follows a zero-order rate law, that is, the reaction rate is independent of the conversion. Then the variables can be changed to dimensionless coordinates. Thus, generalizing the soluhons ... 

For example, if the reaction A —> B obeys a zero-order rate law, the differential rate... 

Some chemical reactions obey what is called a zero-order rate law —the rate of the reaction is independent of concentration, for a limited time. A typical example might be an enzyme which has a limited number of active sites, but which binds the reactant so tightly that all the sites are filled if any significant concentration of the reactant is present in solution. Writing the concentration of the reactant as [A], this means that d A /dt = —k. 

Ivakin s kinetic investigation75 showed that the oxysulfates in reaction (b) decompose in accordance with a simple zero-order rate law, whereas reaction (a) is more complex kinetically. Thus a plot of the equation... 

Note that, if H2 followed a Langmuir isotherm, 1/(8) = Xg (Hj), we would obtain on substitution in Eq. (XVII.6.1) a zero-order rate law, which is in fact observed. The true isotherms, however, follow 1/(S) =ks(H2) , and it is only vrhen allowance is... 

Dissolution of silicate minerals is now believed to follow a zero-order rate law (Wood and Walther, 1983) ... 

Tlie integrated zero-order rate law is [A] = -kt + [A]q. Tlierefore, a plot of [A] versus time should be a... 

In the case where the gas is strongly adsorbed or the pressure is high, the process may follow a zero-order rate law. From Eq. (4.172), we see that the rate law can be written as... 

This zero-order rate law has been found to correctly model the reaction of certain gases on the surfaces of soUds. 

Because the forms of these expressions match, we can see that if we plot [A] (on the y axis) as a function of t (on the x axis), we will get a straight line. We can also see that the slope of that line (yn) must be equal to -k and the y intercept (b) must be equal to [A]q, the initial concentration of reactant A. Equation 11.4 provides us a model of the behavior expected for a system obeying a zero-order rate law. To test this model, we simply need to compare it with data for a particular reaction. So we could measure the concentration of reactant A as a function of time, and then plot [A] versus t. If the plot is linear, we could conclude that we were studying a zero-order reaction. The catalytic destruction of N2O in the presence of gold is an example of this type of kinetics. A graphical analysis of the reaction is shown in Figure 11.6. 

Figure 11.7 I This plot of [O3] vs. time shows that the dissociation of ozone does not follow a zero-order rate law, because the graph is clearly nonUnear.

As we did with the zero-order rate law, this differential equation can be integrated to obtain an equation that directly expresses how the concentration of the reactant varies as a function of time during the reaction process. For a general first-order reaction involving the consumption of a reacting species A, this integration yields. 

Types of Rate Laws Determining the Form of the Rate Law Method of Initial Rates Half-Life of a First-Order Reaction Second-Order Rate Laws Zero-Order Rate Laws Integrated Rate Laws for Reactions with More Than One Reactant 12.7 Catalysis Heterogeneous Catalysis Homogeneous Catalysis... 

---