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For the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C},\) explain at least two ways in which the rate law could be zero order in chemical A.

Short Answer

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There are at least two ways in which the rate law could be zero order in chemical A for the reaction A + B → C. One way is the presence of a catalyst or enzyme that becomes saturated with reactant A, making the reaction rate independent of A's concentration. Another possibility is a fast equilibration of a pre-reaction step that determines the overall reaction rate, causing the concentration of an intermediate product to remain constant despite changes in A's concentration. Additionally, a competing reaction pathway with a different rate law may also lead to the overall rate law becoming zero order for A if the other pathway's rate is much greater and independent of A's concentration.

Step by step solution

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1. Presence of a catalyst or enzyme

A catalyst or an enzyme can make the reaction zero order with respect to A. When the catalyst or enzyme is present in the reaction mixture, it can become saturated with reactant A, causing it to bind with A at its maximum capacity. In this situation, the reaction rate will not increase with an increase in A's concentration since the catalyst or enzyme sites are fully occupied and functioning at their maximum rate.
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2. Fast equilibration of a pre-reaction step

One possibility is that the reaction involves a fast pre-equilibrium step that determines the overall reaction rate. Suppose there is a fast equilibrium reaction occurring before the slow, rate-determining step: \(A + B \rightleftharpoons D\) \(D \rightarrow C\) The overall reaction will still be A + B → C. However, as D forms quickly from A + B and is also consumed quickly in the slow step (D → C), the concentration of D will remain constant even if A's concentration changes. This scenario will lead to the rate law being zero order with respect to A, as the rate will be determined by the slow step and will not depend on A's concentration.
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3. A competing reaction pathway

Another possibility is that there exists a competing reaction pathway with a different rate law, also producing C. For example, consider the following two pathways: \(A + B \rightarrow C\) (pathway 1) \(A + X \rightarrow C\) (pathway 2) If the rate of pathway 2 is much greater than that of pathway 1 and pathway 2's rate doesn't depend on the concentration of A, the overall rate law may become zero order for A. For example, if pathway 2 is a first-order reaction concerning X and the concentration of X is much higher than B, the overall rate of the reaction will depend mostly on pathway 2, making it independent of reactant A's concentration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Reaction Rates
The reaction rate is a crucial concept in chemical kinetics. It measures how fast or slow a reaction occurs. This rate is defined by changes in the concentration of reactants or products over time. Reaction rates can be expressed as: \[\text{Rate} = -\frac{d[A]}{dt} = \frac{d[C]}{dt}\]A negative sign indicates the consumption of reactants like A and B, while the product formation like C is positive. A zero-order reaction means that the concentration of a specific reactant doesn't affect the rate at which the reaction proceeds. In these reactions, the rate remains constant, despite changes in concentration. They're often observed when surfaces or catalysts are saturated and cannot process more reactants at a faster rate.
Catalyst Mechanism and Its Impact
Catalysts are substances that speed up chemical reactions without being consumed in the process. They work by providing an alternative pathway with lower activation energy. Enzymes, a type of biological catalyst, play a similar role. • Catalysts can modify the reaction mechanism, making it possible for a reaction to be zero order with respect to a reactant. This is because the presence of a catalyst can cause saturation. • When the catalyst is fully saturated with a reactant, the reaction rate will not increase even if more reactant is added. The catalyst's efficiency depends on its active sites, and once these are fully occupied, the reaction rate plateaus.
Pre-Equilibrium Concepts
Pre-equilibrium involves a scenario in which a fast reaction step precedes the slower rate-determining step. This concept is often seen in complex reaction mechanisms. Here’s how it works:• The fast initial step reaches equilibrium quickly, allowing a constant intermediate concentration. • This intermediate typically doesn't change appreciably even with varying amounts of reactant A because the equilibrium is rapidly re-established. Thus, the rate law depends on the concentration of this intermediate, not directly on A.For example, if a fast equilibrium \[A + B \rightleftharpoons D\]is followed by a slow step \[D \rightarrow C\]then the reaction rate depends mainly on the second slow step, lessening A’s impact on the rate law.
Principles of Chemical Kinetics
Chemical kinetics is the branch of chemistry that studies the speed and pathways of chemical reactions. It involves looking at factors that influence reaction rates, such as temperature, concentration, and the presence of catalysts. • Kinetics helps understand reaction mechanisms by revealing which steps are fast or slow. • Zero-order kinetics is one of the simplest forms, where the rate is constant and independent of reactant concentration. • Such scenarios are common when reacting substances include catalysts, as well as where saturation occurs. Having a solid grasp of kinetic principles allows chemists to manipulate reaction conditions to maximize yield and efficiency, crucial across various manufacturing and biological processes.

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Most popular questions from this chapter

The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.

Consider a reaction of the type aA \(\longrightarrow\) products, in which the rate law is found to be rate \(=k[\mathrm{A}]^{3}\) (termolecular reactions are improbable but possible). If the first half-life of the reaction is found to be \(40 .\) s, what is the time for the second half-life? Hint: Using your calculus knowledge, derive the integrated rate law from the differential rate law for a termolecular reaction: $$ \text {Rate} =\frac{-d[\mathrm{A}]}{d t}=k[\mathrm{A}]^{3} $$

Consider the reaction $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ If, in a certain experiment, over a specific time period, 0.0048 mole of \(\mathrm{PH}_{3}\) is consumed in a \(2.0-\mathrm{L}\) container each second of reaction, what are the rates of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) in this experiment?

Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average interval between flashes of individual insects was 16.3 \(\mathrm{s}\) at \(21.0^{\circ} \mathrm{C}\) and 13.0 \(\mathrm{s}\) at \(27.8^{\circ} \mathrm{C}\) a. What is the apparent activation energy of the reaction that controls the flashing? b. What would be the average interval between flashes of an individual firefly at \(30.0^{\circ} \mathrm{C} ?\) c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.

The reaction $$ 0^{\circ} \mathrm{C}, $$ These relationships hold only if there is a very small amount of \(\mathrm{I}_{3}^{-}\) present. What is the rate law and the value of the rate constant? (Assume that rate \(=-\frac{\Delta\left[\mathrm{H}_{2} \mathrm{SeO}_{3}\right]}{\Delta t} )\)

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