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Chapter 13: Problem 69

In the presence of excess thiocyanate ion, \(\mathrm{SCN}^{-}\), the following reaction is first order in chromium(III) ion, \(\mathrm{Cr}^{3+}\); the rate constant is \(2.0 \times 10^{-6} / \mathrm{s}\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Cr}(\mathrm{SCN})^{2+}(a q)$$ If \(85.0 \%\) reaction is required to obtain a noticeable color from the formation of the \(\mathrm{Cr}(\mathrm{SCN})^{2+}\) ion, how many hours are required?

Short Answer

Expert verified
Approximately 263.47 hours are required for 85.0% completion.

Step by step solution

01

Understand the Reaction Conditions

This is a first-order reaction involving chromium(III) ion, \(\text{Cr}^{3+}\). The reaction rate is given by \( \text{Rate} = k[\text{Cr}^{3+}] \), and the rate constant \( k \) is \( 2.0 \times 10^{-6} \, \text{s}^{-1} \).
02

Use the First-Order Kinetics Equation

For a first-order reaction, the integrated rate law is \( \ln \frac{[A]_t}{[A]_0} = -kt \), where \([A]_0\) is the initial concentration, \([A]_t\) is the concentration after time \( t \), and \( k \) is the rate constant.
03

Determine the Fraction Remaining

Since 85.0% of the reaction is complete, 15.0% of the original chromium(III) ion remains. Thus, \( \frac{[\text{Cr}^{3+}]_t}{[\text{Cr}^{3+}]_0} = 0.15 \).
04

Substitute Values into the Kinetics Equation

Substitute the known values into the equation: \( \ln 0.15 = -2.0 \times 10^{-6} \, \text{s}^{-1} \times t \). Solve for \( t \).
05

Solve for Time

Rearrange and solve: \( t = \frac{\ln 0.15}{-2.0 \times 10^{-6}} \). Calculate \( \ln 0.15 \approx -1.897 \). Thus, \( t \approx \frac{-1.897}{-2.0 \times 10^{-6}} \approx 9.485 \times 10^{5} \, \text{s} \).
06

Convert Time to Hours

Convert from seconds to hours: \( t \approx 9.485 \times 10^{5} \, \text{s} \times \frac{1 \, \text{hour}}{3600 \, \text{s}} \approx 263.47 \, \text{hours} \).

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Key Concepts

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

Rate Constant
In chemical kinetics, the rate constant is a crucial parameter that provides insights into the speed of a chemical reaction. It is denoted by the symbol \( k \) and is specific to each reaction. For a first-order reaction, the rate constant has the unit of time inverse, such as \( ext{s}^{-1} \), indicating how quickly the concentration of a reactant changes over time.
For the reaction involving chromium(III) ion, the rate constant is given as \( 2.0 \times 10^{-6} \, ext{s}^{-1} \). This small value indicates that the reaction proceeds slowly. Knowing the rate constant allows us to calculate the progress of the reaction at any given time. It serves as a "speedometer" for the reaction, indicating whether it proceeds quickly or slowly.
  • The larger the rate constant, the faster the reaction.
  • For first-order reactions, the rate directly depends on the concentration of one of the reactants.
Understanding rate constants helps chemists and engineers in designing processes and predicting how long a reaction will take under certain conditions.
Integrated Rate Law
In the context of first-order reaction kinetics, the integrated rate law is a vital tool used to determine the concentration of reactants at any point in time during the reaction. The integrated rate law for a first-order reaction is expressed as \( \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \).
Here, \([A]_0\) represents the initial concentration of the reactant, while \([A]_t\) is its concentration after a given time \( t \). The negative sign in the equation indicates that the concentration of the reactant decreases with time. This equation directly relates the concentration of reactants to time, allowing us to calculate how much of the reactant remains at any point during the reaction.
Integrated rate laws are particularly useful when a specific conversion percentage of a reactant is desired. For example, if 85% of the reactant is to be converted, the remaining concentration after the desired reaction time would be 15% of its original concentration. By substituting the known values into the integrated rate law, one can easily solve for the time needed to achieve this conversion.
  • The integrated rate law makes it possible to compare reaction speeds under different conditions.
  • It provides a clear mathematical relationship between reactant concentration and time.
Chromium(III) Ion
The chromium(III) ion, \( \text{Cr}^{3+} \), is a common ion in many chemical reactions, particularly those involving oxidation and complex formation. In this particular reaction, it plays a crucial role in forming a colored complex when it reacts with thiocyanate ions, \( \text{SCN}^{-} \). This reaction follows first-order kinetics with respect to chromium(III) ion, meaning the rate of reaction is directly proportional to its concentration.
Chromium is a transition metal, known for forming various complexes due to its ability to exist in multiple oxidation states. The complex formed in the given reaction, \( \text{Cr(SCN)}^{2+} \), exhibits a distinct color, making reactions involving chromium ions useful as qualitative indicators in labs. The color change provides a visual cue to the reaction's progress, as the intensive blue color of the complex becomes apparent once a significant portion, like 85%, of the reaction has occurred.
Understanding the role of chromium(III) ions in reactions helps in fields such as analytical chemistry, where they are used in various titrations and colorimetric assays.
  • Chromium ions are versatile due to their multiple valence states.
  • They form stable complexes, which often have distinct colors used in detection and analysis.
Recognizing how ions like chromium(III) participate in reactions is crucial for effective reaction monitoring and control.

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Most popular questions from this chapter

Azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3}\), decomposes according to the following equation: $$\mathrm{CH}_{3} \mathrm{NNCH}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{N}_{2}(g)$$ The initial concentration of azomethane was \(1.50 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) After \(7.00 \mathrm{~min}\), this concentration decreased to \(1.01 \times 10^{-2}\) \(\mathrm{mol} / \mathrm{L}\). Obtain the average rate of reaction during this time interval. Express the answer in units of \(\mathrm{mol} /(\mathrm{L} \cdot \mathrm{s})\).

Methyl chloride, \(\mathrm{CH}_{3} \mathrm{Cl}\), reacts in basic solution to give methanol. $$\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{OH}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{Cl}^{-}$$ This reaction is believed to occur in a single step. If so, what should be the rate law?

Rate constants for reactions often follow the Arrhenius equation. Write this equation and then identify each term in it with the corresponding factor or factors from collision theory. Give a physical interpretation of each of those factors.

The rate law for the reaction $$2 \mathrm{NO}_{2} \mathrm{Cl}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{Cl}_{2}(g) \quad \text { (overall equation) }$$ is first order in nitryl chloride, \(\mathrm{NO}_{2} \mathrm{Cl}\). $$\text { Rate }=k\left[\mathrm{NO}_{2} \mathrm{Cl}\right]$$ Explain why the mechanism for this reaction cannot be the single elementary reaction $$2 \mathrm{NO}_{2} \mathrm{Cl} \longrightarrow 2 \mathrm{NO}_{2}+\mathrm{Cl}_{2} \quad \text { (elementary reaction) }$$

The reaction of thioacetamide with water is shown by the equation below: $$\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{CH}_{3} \mathrm{C}(\mathrm{O}) \mathrm{NH}_{2}(a q)$$ The rate of reaction is given by the rate law: $$\text { Rate }=k\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}\right]$$ Consider one liter of solution that is \(0.20 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}\) and \(0.15 \mathrm{M}\) in \(\mathrm{HCl}\) at \(25^{\circ} \mathrm{C}\). a. For each of the changes listed below, state whether the rate of reaction increases, decreases, or remains the same. Why? i. A \(4-g\) sample of \(\mathrm{NaOH}\) is added to the solution. ii. \(500 \mathrm{~mL}\) of water is added to the solution. b. For each of the changes listed below, state whether the value of \(k\) will increase, decrease, or remain the same. Why? i. A catalyst is added to the solution. ii. The reaction is carried out at \(15^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C}\).
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