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Chapter 11: Problem 58

For a certain reaction involving a single reactant, it is found that \(C_{0} \sqrt{T}\) is constant, where \(C_{0}\) is the initial concentration of the reactant and \(T\) is the half-life. What is the order of the reaction? (a) 1 (b) \(1.5\) (c) 2 (d) 3

Short Answer

Expert verified
The order of the reaction is 2.

Step by step solution

01

Identify the Given Information

The problem states that the product of the initial concentration of the reactant (\(C_{0}\)) and the square root of the half-life (\(T\)) is constant. Mathematically, this can be expressed as \(C_{0} \sqrt{T} = k\), where \(k\) is a constant.
02

Recall the Half-Life Formulas for Different Order Reactions

For zero-order reactions, the half-life is given by \( t_{1/2} = \frac{C_{0}}{2k} \). For first-order reactions, \(t_{1/2} = \frac{0.693}{k}\), which does not depend on \(C_{0}\). For second-order reactions, \(t_{1/2} = \frac{1}{kC_{0}}\), where \(k\) is the rate constant.
03

Compare Given Information to Known Half-Life Formulas

As the relationship given is \(C_{0} \sqrt{T} = k\), let's find the order for which the half-life \(T\) has a dependency on \(C_{0}^{-1/2}\) as this will be consistent with \(\sqrt{T}\) being a constant multiplied by \(C_{0}\). Examining the half-life formulas, only the second-order reaction has a half-life inversely proportional to the initial concentration, which is \(t_{1/2} = \frac{1}{kC_{0}}\), as this can be rearranged into the form \(C_{0}\cdot t_{1/2}^{1/2} = constant\), which matches the given relationship.
04

Determine the Order of the Reaction

Based on the comparison, the reaction that fits the given relationship is the second-order reaction where the half-life depends inversely on the initial concentration. Therefore, the order of the reaction is 2.

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Key Concepts

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

Reaction Order
Understanding the reaction order is crucial for predicting how the concentration of reactants affects the speed of chemical reactions. It reflects the dependency of the rate of reaction on the concentration of reactants. In chemical kinetics, reaction orders are typically whole numbers, although they can also be fractions or zero, and they determine the mathematical relationship between the concentration of reactants and the rate of reaction.

For example, a first-order reaction implies that the rate is directly proportional to the concentration of one reactant. So, if you double the concentration, the rate also doubles. In contrast, a second-order reaction indicates that the rate is proportional to the square of the concentration of one reactant or to the product of two reactant concentrations. This means that doubling one reactant's concentration quadruples the reaction rate.

Identifying the reaction order is not just a piece of theoretical knowledge; it can tremendously impact industrial processes where time and efficiency are of utmost importance. To determine the reaction order from an experimental standpoint, one would typically measure how the rate of the reaction changes as concentrations are varied over time.
Half-life of Reaction
The half-life of a reaction, symbolized by the term length is one of those intriguing concepts that intersect chemistry and physics. In the realm of chemical kinetics, the half-life of a reaction refers to the time required for the concentration of a reactant to decrease by half. This is not only essential when dealing with radioactive decay but also when analyzing the degradation of pharmaceuticals, pollutants, and other chemicals.

What's interesting is that the half-life can vary greatly depending on the order of the reaction. For instance, a zero-order reaction has a half-life that is dependent on the initial concentration of the reactant. However, for a first-order reaction, the half-life remains constant no matter the concentration. This is valuable in predicting how long it takes for a substance to reduce to a specific level within a reaction, regardless of how much was initially present.

In our step-by-step solution, we navigated through different half-life equations to deduce the correct reaction order by matching the format of the provided relationship, showcasing how integral the concept of half-life is in chemical kinetics.
Rate Laws
Rate laws are mathematical equations that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. These laws are pivotal to chemical kinetics, providing critical insights into the mechanisms underlying chemical reactions, and are derived from experimental observations.

The general form of a rate law is rate = k[Reactant]n, where 'k' is the rate constant, '[Reactant]' is the concentration of the reactant, and 'n' represents the reaction order. Keep in mind though that the rate constant 'k' is not truly constant; it can vary with temperature and must be determined experimentally for each reaction at a given temperature.

An example is the rate law for a simple one-step reaction with one reactant, A, yielding product B: rate = k[A]. This showcases a first-order reaction, where the rate is directly proportional to the concentration of A. However, as we have seen in our exercise, if the rate doubles as the concentration of reactant doubles, we are dealing with a second-order reaction. Consequently, understanding rate laws not only helps in calculating how fast a reaction will proceed under different conditions but also aids in determining the reaction order by examining concentration and rate changes.

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Most popular questions from this chapter

For a reaction \(2 \mathrm{~A}+\mathrm{B}+3 \mathrm{C} \rightarrow \mathrm{D}+3 \mathrm{E}\), the following date is obtained: $$ \begin{array}{ccccc} \hline \text { Reaction } & \multicolumn{2}{c} {\text { Concentration in }} & \text { Initial rate of } \\ & \multicolumn{2}{c} {\text { mole per litre }} & \text { formation of } \\ & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D}\left(\text { torr } \mathbf{s}^{-1}\right) \\ \hline 1 & 0.01 & 0.01 & 0.01 & 2.5 \times 10^{-4} \\ 2 & 0.02 & 0.01 & 0.01 & 1.0 \times 10^{-3} \\ 3 & 0.01 & 0.02 & 0.01 & 2.5 \times 10^{-4} \\ 4 & 0.01 & 0.02 & 0.02 & 5.0 \times 10^{-4} \\ \hline \end{array} $$ The order with respect to \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are, respectively, (a) \(0,1,2\) (b) \(2,0,1\) (c) \(1,0,2\) (d) \(2,1,1\)

Rate of a reaction: \(\mathrm{A}+2 \mathrm{~B} \rightarrow \mathrm{P}\) is \(2 \times 10^{-2} \mathrm{M} / \mathrm{min}\), when concentrations of each \(A\) and \(B\) are \(1.0 \mathrm{M}\). If the rate of reaction, \(r=K[\mathrm{~A}]^{2}[\mathrm{~B}]\), the rate of reaction when half of the \(\mathrm{B}\) has reacted should be (a) \(5.625 \times 10^{-3} \mathrm{M} / \mathrm{min}\) (b) \(3.75 \times 10^{-3} \mathrm{M} / \mathrm{min}\) (c) \(9.375 \mathrm{M} / \mathrm{min}\) (d) \(2.5 \times 10^{-3} \mathrm{M} / \mathrm{min}\)

Iodide ion is oxidized to hypoiodite ion, \(\mathrm{IO}^{-}\), by hypochlorite ion, \(\mathrm{ClO}^{-}\), in basic solution as: $$ \begin{array}{ccccc} & \mathbf{I}^{-} & \mathbf{C l O}^{-} & \mathbf{O H}^{-} & \left(\mathrm{mol} \mathbf{L}^{-1} \mathbf{s}^{-1}\right) \\ \hline 1 & 0.010 & 0.020 & 0.010 & 12.2 \times 10^{-2} \\ 2 & 0.020 & 0.010 & 0.010 & 12.2 \times 10^{-2} \\ 3 & 0.010 & 0.010 & 0.010 & 6.1 \times 10^{-2} \\ 4 & 0.010 & 0.010 & 0.020 & 3.0 \times 10^{-2} \\ \hline \end{array} $$ The correct rate law for the reaction is (a) \(r=K\left[\mathrm{I}^{-}\right]\left[\mathrm{ClO}^{-}\right]\left[\mathrm{OH}^{-}\right]^{0}\) (b) \(r=K\left[\mathrm{I}^{-}\right]^{2}\left[\mathrm{ClO}^{-}\right]^{2}\left[\mathrm{OH}^{-}\right]^{0}\) (c) \(r=K\left[\mathrm{I}^{-}\right]\left[\mathrm{ClO}^{-}\right]\left[\mathrm{OH}^{-}\right]\) (d) \(r=K\left[\mathrm{I}^{-}\right]\left[\mathrm{ClO}^{-}\right]\left[\mathrm{OH}^{-}\right]^{-1}\)

Two substances, 'A' and 'B' are initially present as \(\left[A_{0}\right]=8\left[B_{0}\right]\) and \(t_{1 / 2}\) for the firstorder decomposition of 'A' and 'B' are 10 and \(20 \mathrm{~min}\), respectively. If they start decomposing at the same time, after how much time, the concentration of both of them would be same? (a) \(20 \mathrm{~min}\) (b) \(40 \mathrm{~min}\) (c) \(60 \mathrm{~min}\) (d) \(200 \mathrm{~min}\)

For the consecutive unimolecular-type first-order reaction: \(\mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{R} \stackrel{k_{2}}{\longrightarrow} \mathrm{S}\), the concentration of component ' \(\mathrm{R}\) ', \(C_{\mathrm{R}}\), at any time, ' \(t\) ' is given by \(C_{\mathrm{R}}=C_{\mathrm{AO}} \cdot K_{1}\left[\frac{e^{-k_{1} t}}{\left(k_{2}-k_{1}\right)}+\frac{e^{-k_{2} t}}{\left(k_{1}-k_{2}\right)}\right]\) If \(C_{\mathrm{A}}=C_{\mathrm{AO}}, C_{\mathrm{R}}=C_{\mathrm{s}}=0\) at \(t=0\), the time at which the maximum concentration of 'R' occurs is (a) \(t_{\max }=\frac{k_{2}-k_{1}}{\ln \left(k_{2} / k_{1}\right)}\) (b) \(t_{\max }=\frac{\ln \left(k_{2} / k_{1}\right)}{k_{2}-k_{1}}\) (c) \(t_{\max }=\frac{e^{k_{2} / k_{1}}}{k_{2}-k_{1}}\) (d) \(t_{\max }=\frac{e^{k_{2}-k_{1}}}{k_{2}-k_{1}}\)
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