A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

Short Answer

Expert verified
The given reaction is a first-order reaction.

Step by step solution

01

Understanding Rate Laws and Reaction Order

Rate laws express the rate of a reaction in terms of the concentration of reactants. The rate of a reaction is generally represented as -d[A]/dt = k[A]^n, where 'A' is the reactant, 'k' is the rate constant, 'n' is the reaction order and 't' is time.
02

Observing The Given Data

The kinetic data provided for the reaction shows that when the time triples (from 500s to 1500s, and from 1500s to 3500s), the concentration of 'A' halves each time (from 2.00M to 1.00M, then to 0.50M, and finally to 0.25M).
03

Identifying The Reaction Order

The constant halving of concentration 'A' each time the time interval triples is characteristic of a first-order reaction. This is because, for first order reactions, the rate of the reaction is proportional to the first power of reactant concentration. The data fits this pattern, indicating that the reaction is first order with respect to 'A'.

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Key Concepts

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

Rate Laws
The study of how the speed of chemical reactions changes with varying conditions is a core principle of chemical kinetics. At the heart of this analysis lies the rate law, a mathematical equation that conveys the relationship between the concentration of reactants and the rate at which they transform into products.

Rate laws can be represented by the equation \( -d[A]/dt = k[A]^n \), where \( [A] \) designates the molar concentration of the reactant \( A \), \( k \) is the rate constant that provides the speed of the reaction at a particular temperature, and \( n \) stands for the reaction order. This order is an exponent that shows the dependence of the rate on the concentration of \( A \) and is determined empirically. Different reactions may exhibit zero, first, second, or mixed-order behaviors depending on how their rates relate to the concentration of reactants.

If the concentration of the reactant does not affect the rate, the reaction is zero-order (\( n = 0 \)). For a first-order reaction (\( n = 1 \)), the rate doubles if the concentration of the reactant doubles. A second-order reaction (\( n = 2 \)) would see the rate quadruple if the reactant concentration doubled.
Kinetic Study
A kinetic study involves analyzing how a chemical reaction proceeds over time to determine the speed and the mechanism of the reaction. By measuring changes in reactant or product concentrations at various times, chemists can unravel the complexity of the reaction pathway and deduce the reaction order from the rate law.

In practice, this involves conducting experiments and gathering data like concentration changes over time, as seen in the exercise problem. When interpreting this data, patterns emerge that indicate how the reaction rate is affected by the concentration of reactants. It's methodical detective work—plotting data on graphs, calculating rates, and comparing these rates under different conditions all contribute to building an understanding of a reaction's kinetics.

Critical to any kinetic study is maintaining controlled conditions. For instance, temperature is kept constant because it significantly affects the rate constant. Once data is collected, graphs such as concentration versus time for a first-order reaction or 1/concentration versus time for a second-order reaction, can reveal the reaction kinetics clearly and facilitate the determination of the rate law.
First-Order Reaction
For a first-order reaction, the rate is directly proportional to the concentration of one reactant. This direct proportionality implies that as the concentration of the reactant decreases by a certain factor, the time required for that decrease is constant.

Looking at the kinetic data provided in the exercise, the concentration of reactant \( A \) is halved repetitively over regular time intervals (500s to 1500s to 3500s). Such regular halving is indicative of a first-order reaction because it suggests a constant rate of reaction per unit concentration of \( A \). It aligns with the mathematical characteristic of a first-order rate law, where a plot of the natural logarithm of the concentration of \( A \) versus time yields a straight line, a hallmark of first-order kinetics.

Additionally, the integrated rate law for a first-order reaction, \( \ln[A] = -kt+ \ln[A]_0 \), where \( [A]_0 \) is the initial concentration of the reactant, shows that the time required to reach half the initial concentration (also known as the half-life) is constant regardless of the starting concentration. This property is unique to first-order reactions and becomes a powerful tool for identifying this type of reaction order from experimental data.

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Most popular questions from this chapter

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

The following first-order reaction occurs in \(\mathrm{CCl}_{4}(1)\) at \(45^{\circ} \mathrm{C}: \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) The rate constant is \(k=6.2 \times 10^{-4} \mathrm{s}^{-1} .\) An \(80.0 \mathrm{g}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}(\mathrm{l})\) is allowed to decompose at \(45^{\circ} \mathrm{C}.\) (a) How long does it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 \mathrm{g} ?\) (b) How many liters of \(\mathrm{O}_{2},\) measured at \(745 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C},\) are produced up to this point?

For the reaction \(\mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C},\) which proceeds by a single-step bimolecular elementary process, (a) \(t_{1 / 2}=0.693 / k ;\) (b) rate of appearance of C= - rate of disappearance of \(\mathrm{A} ;\) (c) rate of reaction = \(k[\mathrm{A}][\mathrm{B}] ;\) (d) \(\ln [A]_{t}=-k t+\ln [A]_{0}.\)

One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.

According to collision theory, chemical reactions occur through molecular collisions. A unimolecular elementary process in a reaction mechanism involves dissociation of a single molecule. How can these two ideas be compatible? Explain.

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