/*! 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 33 The iodide ion reacts with hypoc... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 33

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鈦籡[I鈦籡 (b) The rate constant (k) is approximately 6.08 x 10虏 M鈦宦箂鈦宦�. (c) The reaction rate when [OCl鈦籡 = 2.0 x 10鈦宦� M and [I鈦籡 = 5.0 x 10鈦烩伌 M is approximately 6.08 x 10鈦烩伒 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鈦� and I鈦�). 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鈦籡^m[I鈦籡^n Where k is the rate constant, m and n are the orders of OCl鈦� and I鈦� respectively. Consider the two sets of initial concentrations: 1. [OCl鈦籡 = 1.5 x 10鈦宦� M, [I鈦籡 = 1.5 x 10鈦宦� M, Initial rate = 1.36 x 10鈦烩伌 M/s 2. [OCl鈦籡 = 3.0 x 10鈦宦� M, [I鈦籡 = 1.5 x 10鈦宦� M, Initial rate = 2.72 x 10鈦烩伌 M/s From these two sets of data, we can find the order m as follows: (2.72 x 10鈦烩伌) / (1.36 x 10鈦烩伌) = ([3.0 x 10鈦宦砞^m[I鈦籡^n) / ([1.5 x 10鈦宦砞^m[I鈦籡^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鈦籡 = 1.5 x 10鈦宦� M, [I鈦籡 = 1.5 x 10鈦宦� M, Initial rate = 1.36 x 10鈦烩伌 M/s 3. [OCl鈦籡 = 1.5 x 10鈦宦� M, [I鈦籡 = 3.0 x 10鈦宦� M, Initial rate = 2.72 x 10鈦烩伌 M/s We find the order n as follows: (2.72 x 10鈦烩伌) / (1.36 x 10鈦烩伌) = ([OCl鈦籡^m[3.0 x 10鈦宦砞^n) / ([OCl鈦籡^m[1.5 x 10鈦宦砞^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鈦籡[I鈦籡
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鈦烩伌 M/s = k(1.5 x 10鈦宦� M)(1.5 x 10鈦宦� M) k = ((1.36 x 10鈦烩伌) M/s) / ((1.5 x 10鈦宦�)^2 M虏) k 鈮� 6.08 x 10虏 M鈦宦箂鈦宦� Therefore, the rate constant (k) is approximately 6.08 x 10虏 M鈦宦箂鈦宦�.
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鈦籡 = 2.0 x 10鈦宦� M [I鈦籡 = 5.0 x 10鈦烩伌 M Rate = k[OCl鈦籡[I鈦籡 Rate = (6.08 x 10虏 M鈦宦箂鈦宦�)(2.0 x 10鈦宦� M)(5.0 x 10鈦烩伌 M) Rate 鈮� 6.08 x 10鈦烩伒 M/s Therefore, the reaction rate when [OCl鈦籡 = 2.0 x 10鈦宦� M and [I鈦籡 = 5.0 x 10鈦烩伌 M is approximately 6.08 x 10鈦烩伒 M/s.

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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鈦� and I鈦�, 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.

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Most popular questions from this chapter

Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) : $$ \begin{array}{llll} \hline \text { Experiment } & {\left[\mathrm{HgCl}_{2}\right](M)} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.025 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? (b) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is \(10 \mathrm{kPa}\) per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

(a) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(\mathrm{Cl}_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. \((\mathbf{b})\) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathbf{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M} ?\) (d) What is the reaction rate at \(1000 \mathrm{~K}\) if \([\mathrm{NO}]\) is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M} ?\)
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