The reaction follows a first-order rate law. Calculate the rate constant. Assume that after 50,000 s all the N203(g) had decomposed. [Pg.646]

It was pointed out that a bimolecular reaction can be accelerated by a catalyst just from a concentration effect. As an illustrative calculation, assume that A and B react in the gas phase with 1 1 stoichiometry and according to a bimolecular rate law, with the second-order rate constant k equal to 10 1 mol" see" at 0°C. Now, assuming that an equimolar mixture of the gases is condensed to a liquid film on a catalyst surface and the rate constant in the condensed liquid solution is taken to be the same as for the gas phase reaction, calculate the ratio of half times for reaction in the gas phase and on the catalyst surface at 0°C. Assume further that the density of the liquid phase is 1000 times that of the gas phase. [Pg.740]

The resulting rate law agrees with the fonn found experimentally. Of course the postidated mechanism can only be proven by measuring the rate constants of the individual elementary steps separately and comparing calculated rates of equation (A3.4.148) with observed rates of HBr fomiation. [Pg.792]

Another method for determining rate law parameters is to employ a search for those parameter values that minimize the sum of the squared difference of measured reaction rate and the calculated reaction rate. In performing N experiments, the parameter values can be determined (e.g., E, Cg, Cj, and C2) that minimize the quantity ... [Pg.173]

Let us now assume that these matters have been attended to properly. At this stage we can but assume that the reaction orders were correctly identified and correct mathematical procedures followed. During the course of the work, the investigator should make the occasional quick calculation to show the values are roughly correct. (Does the rate constant yield the correct half-time ) Also, one should examine the experimental data fits to see that the data really do conform to the selected rate equation. Deviations signal an incorrect rate law or complications, such as secondary reactions. [Pg.11]

Second-order kinetics. Prove that a reaction following the rate law v = fc[A]2 is characterized by a linear plot of [P] 1 versus t 1, where P is the product of the stoichiometric reaction A = P. Show how k is calculated by this method. [Pg.41]

Second-order kinetics, (a) Derive expressions for the half-time and lifetime of A if the rate law for its disappearance is v = fc[A]2 (b) calculate t]/i and t for the data presented in Section 2.2 (c) calculate the second half-life, t /i(2), i.e., the time elapsed between 50 percent and 75 percent completion, for the same reaction (d) compare fj/2(l) and fi/>(2), and contrast this result with that from first-order kinetics. [Pg.41]

The reaction follows a mixed second-order rate law. The progress was monitored spec-trophotometrically at 723 nm, where Np4+ has a maximum absorption. The following data refer to an experiment with [Np3+]o = 1.53 x 10-4 M, [Fe3+]o = 2.24 x 10-4 M (taken at 298.0 K, [H+] = 0.400 M, and ionic strength = 2.00 M). Calculate the rate constant either taking the end point value as 0.351 or, if a suitable program is available, allowing it to be found in the calculation. [Pg.41]

The reaction follows first-order kinetics when studied at constant [Melm] and [CO]. Formulate the rate law and calculate any constants from the following data obtained at 23 °C in benzene with [CO] = 3.17 X 10 3 M ... [Pg.42]

Calculate t p, Re, and k, assuming a second-order rate law for exchange. Also calculate the expected value of tip in a similar experiment but with 5.0 x 10-3 M Fe2+. [Pg.67]

Derive the steady-state rate law for each, and the expression for the product ratio, [RN3]/[ROH], in terms of the rate constants and [Nj ]. Use the following data (in 25 percent aqueous dioxane at 36.2 °C) to decide between them by means of graphs or calculations ... [Pg.119]

Bond energies. The net reaction CD + RH = RC1 + HC1 proceeds by a chain mechanism in which the propagators are Cl and R (but not H ), and chain-breaking occurs by dimerization of Cl. Write a scheme consistent with this and derive its rate law. Show how one can use E and AH for the bond dissociation of CP to calculate an activation energy for an elementary reaction. [Pg.194]

STRATEGY The level of mercury(II) in the urine can be predicted by using the integrated first-order rate law, Eq. 5b. To use this equation, we need the rate constant. Therefore, start by calculating the rate constant from the half-life (Eq. 7) and substitute the result into Eq. 5b. [Pg.664]

The assumption that the rate of consumption of the intermediate in the slow step is insignificant relative to its rates of formation and decomposition in the first step is called a pre-equilibrium condition. A pre-equilibrium arises when an intermediate is formed and sustained in a rapid formation reaction and its reverse. The calculation of the rate law is then much simpler. For instance, if we propose that the first step in the NO oxidation gives rise to a pre-equilibrium, we would write... [Pg.671]

Calculate a concentration, time, or rate constant by using an integrated rate law (Examples 13.3, 13.4, and 13.5). [Pg.690]

The rate law of a reaction is an experimentally determined fact. From this fact we attempt to learn the molecularity, which may be defined as the number of molecules that come together to form the activated complex. It is obvious that if we know how many (and which) molecules take part in the activated complex, we know a good deal about the mechanism. The experimentally determined rate order is not necessarily the same as the molecularity. Any reaction, no matter how many steps are involved, has only one rate law, but each step of the mechanism has its own molecularity. For reactions that take place in one step (reactions without an intermediate) the order is the same as the molecularity. A first-order, one-step reaction is always unimolecular a one-step reaction that is second order in A always involves two molecules of A if it is first order in A and in B, then a molecule of A reacts with one of B, and so on. For reactions that take place in more than one step, the order/or each step is the same as the molecularity for that step. This fact enables us to predict the rate law for any proposed mechanism, though the calculations may get lengthy at times." If any one step of a mechanism is considerably slower than all the others (this is usually the case), the rate of the overall reaction is essentially the same as that of the slow step, which is consequently called the rate-determining step. ... [Pg.291]

We need to determine whether the rate law predicted by each mechanism matches the experimental rate law by calculating the rate law predicted by each. [Pg.1089]

Then the chemist performs another experiment with I B] = 1.50 M and [A] = 0.050 M. Again, it takes 3.0x10 seconds for the concentration of A to fall to 0.025 M. (a) What is the rate law of the reaction Show your reasoning, (b) Calculate the rate constant. [Pg.1131]

If mechanism (A) applied the Cr(VI)+V(IV) system would be anomalous when compared with the Cr(VI) + Fe(II) and Ce(IV) + Cr(III) reactions which have similar rate laws and Cr(V) -> Cr(IV) transformations as rate-controlling steps. Apart from this there are other good reasons for rejecting mechanism (+). At 25 °C, K is 10 ° and k is 0.56 l.mole sec , allowing At2 to be calculated as... [Pg.163]

data analysis for reactions in solids are somewhat different from those used in other types of kinetic studies. Therefore, the analysis of data for an Avrami type rate law will be illustrated by an numerical example. The data to be used are shown in Table 8.1, and they consist of (a,t) pairs that were calculated assuming the A3 rate law and k = 0.025 min-1. [Pg.262]

Method 2 A mathematical solution is obtained by substituting the experimental values of Experiments 1 and 3 into rate-law expressions and dividing the latter by the former. Note the calculations are easier when the experiment with the larger rate is in the numerator. [Pg.260]

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