/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Consider the reaction described ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the reaction described by \(\mathrm{OCl}^{-}(a q)+\Gamma(a q) \stackrel{\mathrm{OH}^{-}(a q)}{\longrightarrow} \mathrm{OI}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) Use the following initial-rate data to determine the rate law and the corresponding value of the rate constant for the reaction. \begin{tabular}{lccc} \hline\([\mathrm{OCl}-]_{0}\) \(/ \mathbf{M}\) & \(\left[\mathbf{I}^{-}\right]_{0}\) \(/ \mathbf{M}\) & \(\left[\mathrm{OH}^{-}\right]_{0}\) \(/ \mathbf{M}\) & Initial rate of reaction/M \(\mathbf{s}^{-1}\) \\ \hline \(1.62 \times 10^{-3}\) & \(1.62 \times 10^{-3}\) & \(0.52\) & \(3.06 \times 10^{-4}\) \\ \(1.62 \times 10^{-3}\) & \(2.88 \times 10^{-3}\) & \(0.52\) & \(5.44 \times 10^{-4}\) \\\ \(2.71 \times 10^{-3}\) & \(1.62 \times 10^{-3}\) & \(0.84\) & \(3.16 \times 10^{-4}\) \\\ \(1.62 \times 10^{-3}\) & \(2.88 \times 10^{-3}\) & \(0.91\) & \(3.11 \times 10^{-4}\) \\\ \hline \end{tabular}

Short Answer

Expert verified
Rate law: \( \text{rate} = k [\text{I}^-] [\text{OH}^-]^{-1} \) with \( k \approx 9.81 \times 10^{-5} \, \text{M s}^{-1} \).

Step by step solution

01

Identify the Rate Law Form

The general rate law for a reaction involving reactants A, B, and C can be written as \( ext{rate} = k[ ext{OCl}^-]^x[ ext{I}^-]^y[ ext{OH}^-]^z \). Our task is to determine the values of \( x \), \( y \), and \( z \), and the rate constant \( k \).
02

Determine the Order with Respect to \([\text{OCl}^-]\)

Comparing Experiments 1 and 3, only \([\text{OCl}^-]\) concentration changes (from 1.62 to 2.71), while \([\text{I}^-]\) and \([\text{OH}^-]\) are held constant. The rate changes from \(3.06 \times 10^{-4}\) to \(3.16 \times 10^{-4}\). Let's use the formula: \( \frac{3.16 \times 10^{-4}}{3.06 \times 10^{-4}} = \left(\frac{2.71 \times 10^{-3}}{1.62 \times 10^{-3}}\right)^x \). Solving this yields \( x = 0 \), suggesting zero order with respect to \([\text{OCl}^-]\).
03

Determine the Order with Respect to \([\text{I}^-]\)

Comparing Experiments 1 and 2, where \([\text{I}^-]\) varies from 1.62 to 2.88 and others remain constant, the rate changes from \(3.06 \times 10^{-4}\) to \(5.44 \times 10^{-4}\). Using a similar calculation: \( \frac{5.44 \times 10^{-4}}{3.06 \times 10^{-4}} = \left(\frac{2.88 \times 10^{-3}}{1.62 \times 10^{-3}}\right)^y \), we find \( y = 1 \). Therefore, the reaction is first order with respect to \([\text{I}^-]\).
04

Confirm the Order with Respect to \([\text{OH}^-]\)

Using Experiments 2 and 4, changing \([\text{OH}^-]\) from 0.52 to 0.91, with other factors constant, the rates change from \(5.44 \times 10^{-4}\) to \(3.11 \times 10^{-4}\). We calculate: \( \frac{3.11 \times 10^{-4}}{5.44 \times 10^{-4}} = \left(\frac{0.91}{0.52}\right)^z \). Solving gives \( z = -1 \), indicating an inverse first order with respect to \([\text{OH}^-]\).
05

Write the Rate Law

The rate law incorporating the determined orders is \( ext{rate} = k [ ext{I}^-]^1 [ ext{OH}^-]^{-1} \).
06

Calculate the Rate Constant \( k \)

Using the rate law \( ext{rate} = k [ ext{I}^-] [ ext{OH}^-]^{-1} \) with data from Experiment 1: \( 3.06 \times 10^{-4} = k \times (1.62 \times 10^{-3})^1 \times (0.52)^{-1} \). Solving for \( k \) gives \( k \approx 9.81 \times 10^{-5} \, \text{M s}^{-1} \).

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Key Concepts

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

Rate Law Determination
In chemical kinetics, the rate law is an essential expression that describes how the rate of a chemical reaction depends on the concentration of reactants. This is important because it helps predict how changes in conditions will affect the speed of reactions. For our given reaction, we start by considering a generic form of the rate law: \[ \text{rate} = k[\text{OCl}^-]^x[\text{I}^-]^y[\text{OH}^-]^z \] Here,
  • \( k \) is the rate constant, a specific value for the reaction at a given temperature.
  • \( [\text{OCl}^-], [\text{I}^-], \) and \( [\text{OH}^-] \) are the concentrations of our reactants.
  • \( x, y, \) and \( z \) are the reaction orders with respect to each reactant.
To determine the rate law, experiments are conducted by keeping the concentrations of some reactants constant while varying others. This allows for the isolation and observation of the effects of specific reactants on the reaction rate.
Reaction Order
Determining the order of reaction for each reactant is crucial. It reveals how changes in concentrations affect the rate, providing insight into the reaction mechanism. Reaction order with respect to a reactant can be zero, first, second, etc., and is determined through experimental data.

Order with respect to \([\text{OCl}^-]\)

For our reaction, the zero-order with respect to \([\text{OCl}^-]\) was determined by comparing experiments where only \([\text{OCl}^-]\) concentration was changed. This implies changes in its concentration do not impact the reaction rate.

Order with respect to \([\text{I}^-]\)

The reaction is first-order with respect to \([\text{I}^-]\), determined by changing its concentration and observing proportional changes in the reaction rate. This linear relation suggests that the rate is directly proportional to \([\text{I}^-]\).

Order with respect to \([\text{OH}^-]\)

An inverse first order with respect to \([\text{OH}^-]\) was assigned after analyzing experiments where changes in its concentration reduced the rate. It indicates that more \([\text{OH}^-]\) makes the reaction slower, possibly due to a reverse role in the mechanism, such as product formation.
Rate Constant Calculation
Once the reaction orders are determined, the next step is to calculate the rate constant \( k \). This constant is a crucial factor in the rate law as it encompasses the intrinsic speed of the reaction at a given temperature. To find \( k \), the rate law derived earlier \\[ \text{rate} = k [\text{I}^-] [\text{OH}^-]^{-1} \] was used alongside initial concentration and rate data from the experiments.Using Experiment 1 data as an example:
Initial rate = \( 3.06 \times 10^{-4} \, \text{M s}^{-1} \)
\([\text{I}^-] = 1.62 \times 10^{-3} \, \text{M}\)
\([\text{OH}^-] = 0.52 \, \text{M}\)Plug these values into the rate law and solve for \( k \):\[ 3.06 \times 10^{-4} = k \times (1.62 \times 10^{-3}) \times (0.52)^{-1} \] Solving gives \( k \approx 9.81 \times 10^{-5} \, \text{M s}^{-1} \). This value signifies the reaction's baseline rate and helps in predicting behavior under varied conditions. Knowing \( k \) allows chemists to control reactions effectively by adjusting conditions to achieve desired rates.

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Most popular questions from this chapter

The table below gives the concentrations of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g)\) as a function of time at \(690 \mathrm{~K}\) for the following reaction equation: $$ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g) \rightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) $$ \begin{tabular}{cc} \hline\(\left[\mathbf{C}_{2} \mathbf{H}_{4} \mathbf{O}\right] / \mathbf{M}\) & \(t / \mathbf{m i n}\) \\ \hline \(0.0860\) & 0 \\ \(0.0465\) & 50 \\ \(0.0955\) & 72 \\ \(0.0274\) & 93 \\ \(0.0174\) & 130 \\ \hline \end{tabular} Verify that this is a first-order reaction by plotting \(\ln \left(\left[\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\right] / \mathrm{M}\right)\) versus time and determine the value of the rate constant.

The reaction described by $$ \mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ is a second-order reaction with a rate constant \(k=\) \(0.54 \mathrm{M}^{-1} \cdot \mathrm{s}^{-1} .\) How long will it take for \(\left[\mathrm{NO}_{2}\right]\) to be \(10.0 \%\) of its initial value of \(2.00 \mathrm{M} ?\)

A sample of sodium-24 chloride containing \(0.055\) milligrams of sodium- 24 is injected into an animal to study sodium balance. How much sodium-24 remains \(6.0\) hours later? The half-life of sodium-24 is \(14.96\) hours.

Peroxydisulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q)\) decomposes in aqueous solution according to the equation $$ \begin{aligned} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \\ 2 \mathrm{SO}_{4}^{2-}(a q)+\frac{1}{2} \mathrm{O}_{2}(g)+2 \mathrm{H}^{+}(a q) \end{aligned} $$ Given the following data from an experiment with \(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]_{0}=0.100 \mathrm{M}\) in a solution with \(\left[\mathrm{H}^{+}\right]\) fixed at \(0.100 \mathrm{M}\), determine the reaction rate law and calculate the value of the rate constant: \begin{tabular}{cc} \hline\(t / \mathrm{min}\) & {\(\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-1} / \mathrm{M}\right.\)} \\ \hline 0 & \(0.100\) \\ 17 & \(0.050\) \\ 34 & \(0.025\) \\ 51 & \(0.012\) \\ \hline \end{tabular}

Identify in each of the following cases the order of the reaction rate law with respect to the reactant \(\mathrm{A}\), where \(\mathrm{A} \rightarrow\) Products: (a) The half-life of \(\mathrm{A}\) is independent of the initial concentration of \(\mathrm{A}\). (b) The rate of decrease of \(\mathrm{A}\) is a constant. (c) A twofold increase in the initial concentration of A leads to a \(1.41\) -fold increase in the initial rate. (d) A twofold increase in the initial concentration of A leads to a fourfold increase in the initial rate. (e) The time required for \([\mathrm{A}]_{0}\) to decrease to \([\mathrm{A}]_{0} / 2\) is equal to the time required for \([\mathrm{A}]\) to decrease from \([\mathrm{A}]_{0} / 2\) to \([\mathrm{A}]_{0} / 4\)

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