Reaction must be first order

The simplest way to demonstrate that a reaction is first-order in A, is to double the concentration of A and note the effect on the reaction s rate. If the observed rate doubles, then the reaction must be first-order in A. Alternatively, we can derive a relationship between the [A] and time by rearranging equation A5.6... 

B A plot of ln(CH3N2CH,) against time is linear, showing that the reaction must be first order k = 3.60 X 10 4 s l. 

Since the rate doubles as the concentration doubles, the reaction must be first order. 

It is important that we don t make the expression more complex than necessary. If we know the reaction must be first order in one of the components or zero th order in another component, this information should be fixed a priori in the kinetic expression we will use in the following. 

This problem is similar to Problem 20.5. Since doubling the time reduces the amount of sugar by half, it must be a first-order reaction. Alternatively, the reduction of the sugar concentration from 0.06 M to 0.03 M may be thought of as a new experiment with an initial concentration of 0.06 M. Since the same half-life (10 hours) was observed in both experiments, the reaction must be first-order because only in a first-order reaction is the half-life independent of the initial concentration. The rate constant may be evaluated from the half-life using equation (20-3). 

To determine m, the concentrations and reaction rates in Trials 1 and 2 are compared. As you can see from the data, while the concentration of B remains constant, the concentration of A in Trial 2 is twice that of Trial 1. Note that the initial rate in Trial 2 is twice that of Trial 1. Because doubling [A] doubles the rate, the reaction must be first order in A. That is, because 2 " = 2, m must equal 1. The same method is used to determine n, only this time Trials 2 and 3 are compared. Doubling the concentration of B causes the rate to increase by four times. Because 2" = 4, must equal 2. This information suggests that the reaction is second order in B, giving the following overall rate law. 

Qualitatively, doubling [A] doubles the rate therefore, the reaction must be first order with respect to A. Moreover, doubling [B] quadruples the rate therefore, the reaction must be second order with respect to B. These same results can be determined quantitatively by setting up and solving pairs of simultaneous equations, one pair of which is... 

Figure 2-7 Plots of Ca versus 1 for an irreversible reaction for = 0,1, and 1. The kinetics for all reactions must approach first order as the reactant concentration approaches ro to be consistent with equilibrium requirements.

The log-log plot shown in Fig. 3.5 is indeed linear with gradient = 1.1 0.1, which shows that the reaction is first order with respect to [Met] and, therefore (since it is second order overall), it must be first order with respect to [HOCI]. Thus, the complete rate equation is r = k0bs [HOCI] [Met], and 7c0bs = (40 7) dm3 mol-1 min-1 under the conditions of the reaction. 

In terms of chemical reactivity, we can now ask how rapid an intracellnlar reaction must be in order to prevent this escape . To a first approximation, the half-life of such a reaction should be in the range of this escape time, and so we calculate the rate for a reaction with this half-life ... 

We shall assume that the amount of SH+ present is throughout much smaller than [S] or [A,]. If the first step is rate determining, then the reaction will be first order with respect to both catalyst and substrate, with a catalytic constant depending on the nature of Aj. It is reasonable to suppose that, for a series of similar catalysts the catalytic constants will run parallel with their acid strengths, but since acids do not dissociate in the aprotic solvents being considered, some other reaction must be... 

A second requirement is that the rate law for the chemical reaction must be known for the period in which measurements are made. In addition, the rate law should allow the kinetic parameters of interest, such as rate constants and concentrations, to be easily estimated. For example, the rate law for a reaction that is first order in the concentration of the analyte. A, is expressed as... 

At the outset it should be clear that, in order to apply these methods, the order of reaction must be known. Most of the following applies to first-order reactions. 

A more serious problem is that we lose all kinetic information about the system until the data collection begins, and ultimately this limits the rates that can be studied. For first-order reactions we may be able to sacrifice the data contained in the first one, two, or three half-lives, provided the system amplitude is adequate that is, the remaining extent of reaction must be quantitatively detectable. However, this practice of basing kinetic analyses on the last few percentage of reaction is subject to error from unknown side reactions or analytical difficulties. 

How does one show that a reaction is truly first-order The plot of In [A], versus time must be linear. This is a necessary condition, but not a sufficient proof, unless quite... 

This expression suggests a rate-controlling step in which RM reacts with an intermediate. If so, [Int] °c [RM] /2. To be consistent with this, the initiation step should be first-order in [RM] and the termination step second-order in [Int]. Since O2 is not involved in the one propagation step deduced, it must appear in the other, because it is consumed in the overall stoichiometry. On the other hand, given that one RM is consumed by reaction with the intermediate, another cannot be introduced in the second propagation step, since the stoichiometry [Eq. (8-3)] would disallow that. Further, we know that the initiation and propagation steps are not the reverse of one another, since the system is not well-behaved. From this logic we write this skeleton ... 

Similar expressions can be written for third-order reactions. A reaction whose rate is proportional to [A] and to [B] is said to be first order in A and in B, second order overall. A reaction rate can be measured in terms of any reactant or product, but the rates so determined are not necessarily the same. For example, if the stoichiometry of a reaction is2A-)-B—>C- -D then, on a molar basis, A must disappear twice as fast as B, so that —d[A]/dt and -d[B]/dr are not equal but the former is twice as large as the latter. 

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