Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .\) This rapid reaction gives the following rate data: $$ \begin{array}{ccc} \hline\left[\mathrm{OCI}^{-}\right](M) & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\)

Short Answer

Expert verified
(a) The rate law for the reaction is: Rate = k[OCl⁻][I⁻] (b) The rate constant (k) is approximately 6.08 x 10² M⁻¹s⁻¹. (c) The reaction rate when [OCl⁻] = 2.0 x 10⁻³ M and [I⁻] = 5.0 x 10⁻⁴ M is approximately 6.08 x 10⁻⁵ M/s.

Step by step solution

01

Write the rate law for the reaction

To determine the rate law for the reaction, we want to find the orders of the reactants (OCl⁻ and I⁻). We can do this by examining how the initial concentrations affect the initial rate in the provided data. The rate law has the general form: Rate = k[OCl⁻]^m[I⁻]^n Where k is the rate constant, m and n are the orders of OCl⁻ and I⁻ respectively. Consider the two sets of initial concentrations: 1. [OCl⁻] = 1.5 x 10⁻³ M, [I⁻] = 1.5 x 10⁻³ M, Initial rate = 1.36 x 10⁻⁴ M/s 2. [OCl⁻] = 3.0 x 10⁻³ M, [I⁻] = 1.5 x 10⁻³ M, Initial rate = 2.72 x 10⁻⁴ M/s From these two sets of data, we can find the order m as follows: (2.72 x 10⁻⁴) / (1.36 x 10⁻⁴) = ([3.0 x 10⁻³]^m[I⁻]^n) / ([1.5 x 10⁻³]^m[I⁻]^n) 2 = (3.0 / 1.5)^m |_divide both sides by n as it cancels out_| 2 = 2^m Therefore, m = 1 Now, we look for the order n. Compare two other sets of initial concentrations: 1. [OCl⁻] = 1.5 x 10⁻³ M, [I⁻] = 1.5 x 10⁻³ M, Initial rate = 1.36 x 10⁻⁴ M/s 3. [OCl⁻] = 1.5 x 10⁻³ M, [I⁻] = 3.0 x 10⁻³ M, Initial rate = 2.72 x 10⁻⁴ M/s We find the order n as follows: (2.72 x 10⁻⁴) / (1.36 x 10⁻⁴) = ([OCl⁻]^m[3.0 x 10⁻³]^n) / ([OCl⁻]^m[1.5 x 10⁻³]^n) 2 = (3.0 / 1.5)^n |_divide both sides by m as it cancels out_| 2 = 2^n Therefore, n = 1 Now that we have determined the orders for both reactants, the rate law is: Rate = k[OCl⁻][I⁻]
02

Calculate the rate constant with proper units

To find the rate constant (k), we can use any set of initial concentrations and their corresponding initial rate from the given data. Let's use the first set: 1.36 x 10⁻⁴ M/s = k(1.5 x 10⁻³ M)(1.5 x 10⁻³ M) k = ((1.36 x 10⁻⁴) M/s) / ((1.5 x 10⁻³)^2 M²) k ≈ 6.08 x 10² M⁻¹s⁻¹ Therefore, the rate constant (k) is approximately 6.08 x 10² M⁻¹s⁻¹.
03

Calculate the reaction rate with given concentrations

Now that we have the rate law and the rate constant, we can calculate the reaction rate for the given concentrations: [OCl⁻] = 2.0 x 10⁻³ M [I⁻] = 5.0 x 10⁻⁴ M Rate = k[OCl⁻][I⁻] Rate = (6.08 x 10² M⁻¹s⁻¹)(2.0 x 10⁻³ M)(5.0 x 10⁻⁴ M) Rate ≈ 6.08 x 10⁻⁵ M/s Therefore, the reaction rate when [OCl⁻] = 2.0 x 10⁻³ M and [I⁻] = 5.0 x 10⁻⁴ M is approximately 6.08 x 10⁻⁵ M/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Rate Law
The rate law is a mathematical equation that describes how the concentration of reactants affects the speed or rate of a chemical reaction. The rate law is expressed with the equation: \[ \text{Rate} = k [\text{Reactant}_1]^m [\text{Reactant}_2]^n \] where:
  • Rate is the speed of the reaction.
  • k is the rate constant, specific to every reaction.
  • m and n are the orders of the reaction with respect to each reactant.
In our reaction between iodide ions and hypochlorite ions, the rate law is found by observing the change in initial rates with different reactant concentrations. By examining the given data, we find that the reaction is first order in both OCl⁻ and I⁻, indicated by: \[ \text{Rate} = k [\text{OCl}^-][\text{I}^-] \] This means doubling the concentration of either reactant will double the reaction rate.
Reaction Rate
Reaction rate refers to the speed at which reactants are converted to products in a chemical reaction. It is typically measured as the change in concentration of a reactant or product per unit time. Units are often in \( \text{moles per liter per second (M/s)} \). In the iodide and hypochlorite reaction we consider, the reaction rate can be explicitly calculated if the rate constant and correct concentrations are provided. For example, if given concentrations are \([\text{OCl}^-] = 2.0 \times 10^{-3} \text{ M}\) and \([\text{I}^-] = 5.0 \times 10^{-4} \text{ M}\), and a known rate constant \(k = 6.08 \times 10^2 \text{ M}^{-1}\text{s}^{-1}\), you would substitute into the rate law formula to find: \[ \text{Rate} = (6.08 \times 10^2 \text{ M}^{-1}\text{s}^{-1})(2.0 \times 10^{-3} \text{ M})(5.0 \times 10^{-4} \text{ M}) \approx 6.08 \times 10^{-5} \text{ M/s} \] By understanding how to apply these calculations, students can predict changes in reaction conditions.
Rate Constant
The rate constant \(k\) is a proportionality factor in the rate law of a reaction that quantifies the speed of a reaction at a given temperature. Each reaction has its own unique rate constant, dependent on specific conditions like temperature and the presence of catalysts. To calculate the rate constant, we rearrange the rate law to solve for \(k\): \[ k = \frac{\text{Rate}}{[\text{Reactant}_1]^m [\text{Reactant}_2]^n} \] In our iodide and hypochlorite reaction, using the provided initial data: \[ 1.36 \times 10^{-4} \text{ M/s} = k (1.5 \times 10^{-3} \text{ M})(1.5 \times 10^{-3} \text{ M}) \] Solving this equation gives: \[ k \approx 6.08 \times 10^2 \text{ M}^{-1}\text{s}^{-1} \] This calculation shows the speed of converting reactants to products goes quite fast under given conditions.
Reaction Order
The reaction order refers to the power to which the concentration of a reactant is raised in the rate law equation. Identifying the reaction order for each reactant helps in understanding the influence each has on the overall reaction rate. In the iodide ion and hypochlorite ion reaction:
  • The reaction is first order with respect to both \([\text{OCl}^-]\) and \([\text{I}^-]\) as determined through experimental data comparison.
  • The overall reaction order is the sum of the individual orders, here: \[ m + n = 1 + 1 = 2 \]
This means the reaction rate is proportional to the square of the concentration of the reactants. Therefore, if the concentrations of both reactants double, the reaction rate quadruples.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?

The human body is characterized by an extremely complex system of interrelated chemical reactions. A large number of enzymes are necessary for many of these reactions to occur at suitable rates. Enzymes are very selective in the reactions they catalyze, and some are absolutely specific. Use the lock-and- key model to account for the specificity of an enzyme.

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(\) alc \()\) has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.1 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (a) Predict the rate constant for the reaction at \(30^{\circ} \mathrm{C}\). (b) A solution of KOH in ethanol is made up by dissolving \(0.500 \mathrm{~g} \mathrm{KOH}\) in ethanol to form \(500 \mathrm{~mL}\) of solution. Similarly, \(1.500 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(500 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(30^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? ((d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(40^{\circ} \mathrm{C} .\)

The reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\) \(\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) was studied with the following results: $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{CIO}_{2}\right](M)} & {\left[\mathrm{OH}^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.060 & 0.030 & 0.0248 \\ 2 & 0.020 & 0.030 & 0.00276 \\ 3 & 0.020 & 0.090 & 0.00828 \\ \hline \end{array} $$ (a) Determine the rate law for the reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{ClO}_{2}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\)

(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free