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Chapter 14: Problem 64

In the presence of excess thiocyanate ion, \(\mathrm{SCN}^{-}\), the following reaction is first order in iron(III) ion, \(\mathrm{Fe}^{3+}\); the rate constant is \(1.27 / \mathrm{s}\) $$ \mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Fe}(\mathrm{SCN})^{2+}(a q) $$ If \(90.0 \%\) reaction is required to obtain a noticeable color from the formation of the \(\mathrm{Fe}(\mathrm{SCN})^{2+}\) ion, how many seconds are required?

Short Answer

Expert verified
It takes approximately 1.813 seconds for 90% of the reaction to occur.

Step by step solution

01

Identify the Order and Rate Constant

The problem states that the reaction is first order in \( \mathrm{Fe}^{3+} \). Hence, the rate law for the reaction is \( \text{Rate} = k [\mathrm{Fe}^{3+}] \), where \( k \) is the rate constant given as \( 1.27 \, \mathrm{s}^{-1} \).
02

Use First Order Kinetics Formula

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

Set Up the Calculation for 90% Reaction

When 90% of the reaction has occurred, 10% of \( \mathrm{Fe}^{3+} \) ions remain. Assuming the initial concentration of \( \mathrm{Fe}^{3+} \) is \( [\mathrm{Fe}^{3+}]_0 \), the remaining concentration is \( 0.1[\mathrm{Fe}^{3+}]_0 \). The equation becomes: \( 0.1[\mathrm{Fe}^{3+}]_0 = [\mathrm{Fe}^{3+}]_0 e^{-1.27t} \).
04

Simplify and Solve for Time \( t \)

By cancelling \([\mathrm{Fe}^{3+}]_0\) from both sides, we have: \( 0.1 = e^{-1.27t} \). Taking the natural logarithm of both sides gives: \( \ln(0.1) = -1.27t \).
05

Solve for \( t \)

Calculate \( t \) by rearranging the equation from Step 4: \( t = \frac{\ln(0.1)}{-1.27} \). By computing the value: \( t \approx \frac{-2.3026}{-1.27} \approx 1.813 \) seconds.

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Key Concepts

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

First-order reactions
First-order reactions are a class of reactions where the rate is directly proportional to the concentration of one reactant. This means that as the concentration of the reactant decreases, the rate of the reaction slows down. First-order kinetics are common in chemical reactions, from radioactive decay to the decomposition of certain substances.
  • The rate of the reaction depends on the concentration of only one reactant, which simplifies calculations.
  • This characteristic is particularly helpful in predicting how long a reaction will take to reach a certain point, like when 90% of reactants are converted to products.
In the given problem, the reaction is first order in iron(III) ion, \[\text{Rate} = k [\mathrm{Fe}^{3+}]\]where \(k\) is constant. Understanding these basics is crucial in solving problems involving kinetic equations.
Rate constant
The rate constant, often denoted as \(k\), is a crucial part of the rate equation for a first-order reaction, \(\text{Rate} = k [\text{Reactant}]\). It provides insight into the speed of a reaction at a given temperature, acting as the proportionality factor that relates the rate of reaction to the concentration of reactants.
  • The unit of the rate constant for a first-order reaction is inverse seconds, \(\text{s}^{-1}\), which signifies how quickly the reactant concentration decreases.
  • In our example, \(k\) is given as \(1.27 \text{/s}\), meaning that each second, the concentration of reactant, \([\mathrm{Fe}^{3+}]\), changes at this rate.
The value of \(k\) is determined experimentally and may vary with temperature. A higher rate constant usually indicates a faster reaction.
Integrated rate law
The integrated rate law for first-order reactions provides a direct relationship between the concentrations of the reactant at different times. This mathematical tool is very helpful for calculating how long it takes for a reaction to reach a certain state. The integrated rate law for first-order reactions is written as:\[ [A] = [A]_0 e^{-kt} \]where:
  • \([A]_0\) is the initial concentration of the reactant,
  • \([A]\) is the concentration of the reactant at time \(t\), and
  • \(k\) is the rate constant.
In the example, to find how many seconds are needed for 90% of the reaction to be completed, we use this integrated rate law equation. When 90% of \([\mathrm{Fe}^{3+}]\) has reacted, only 10% remains, making it possible to set up the equation to solve for time \(t\) by simplifying terms. From our solution steps, we took the natural logarithm to isolate \(t\), a common strategy in solving these kinds of problems.

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Most popular questions from this chapter

Chlorine dioxide oxidizes iodide ion in aqueous solution to iodine; chlorine dioxide is reduced to chlorite ion. $$ 2 \mathrm{ClO}_{2}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+\mathrm{I}_{2}(a q) $$ The order of the reaction with respect to \(\mathrm{ClO}_{2}\) was determined by starting with a large excess of \(\mathrm{I}^{-}\), so that its concentration was essentially constant. Then $$ \text { Rate }=k\left[\mathrm{ClO}_{2}\right]^{m}\left[\mathrm{I}^{-}\right]^{n}=k^{\prime}\left[\mathrm{ClO}_{2}\right]^{m} $$ where \(k^{\prime}=k\left[\mathrm{I}^{-}\right]^{n} .\) Determine the order with respect to \(\mathrm{ClO}_{2}\) and the rate constant \(k^{\prime}\) by plotting the following data assuming first- and then second-order kinetics. [Data from \(\mathrm{H}\). Fukutomi and G. Gordon, J. Am. Chem. Soc., 89, 1362 \((1967) .]\) $$ \begin{array}{ll} \text { Time (s) } & {\left[\mathrm{ClO}_{2}\right](\mathrm{mol} / \mathrm{L})} \\ 0.00 & 4.77 \times 10^{-4} \\ 1.00 & 4.31 \times 10^{-4} \\ 2.00 & 3.91 \times 10^{-4} \\ 3.00 & 3.53 \times 10^{-4} \\ 5.00 & 2.89 \times 10^{-4} \\ 10.00 & 1.76 \times 10^{-4} \\ 30.00 & 2.4 \times 10^{-5} \\ 50.00 & 3.2 \times 10^{-6} \end{array} $$

The reaction of methylacetate with water is shown by the equation below: $$ \mathrm{CH}_{3} \mathrm{COOCH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow $$ The rate of the reaction is given by the rate law: $$ \text { Rate }=k\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right] $$ Consider one liter of solution that is \(0.15 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\) and \(0.015 \mathrm{M}\) in \(\mathrm{H}_{3} \mathrm{O}^{+}\) 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. Some concentrated sulfuric acid is added to the solution. ii. 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. Some concentrated sulfuric acid is added to the solution. ii. The reaction is carried out at \(35^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C}\).

The reaction \(3 \mathrm{I}^{-}(a q)+\mathrm{H}_{3} \mathrm{As} \mathrm{O}_{4}(a q)+2 \mathrm{H}^{+}(a q)\) $$ \mathrm{I}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(I) $$ is found to be first order with respect to each of the reactants. Write the rate law. What is the overall order?

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), decomposes when heated. $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ In an experiment, the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was \(0.0248 \mathrm{~mol} / \mathrm{L}\). If the rate constant is \(2.2 \times 10^{-5} / \mathrm{s}\), what is the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after \(2.0 \mathrm{hr}\) ? The reaction is first order.

A study of the decomposition of azomethane, \(\mathrm{CH}_{3} \mathrm{NNCH}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{N}_{2}(g)\) gave the following concentrations of azomethane at various times: Time \(\left[\mathbf{C H}_{3} \mathbf{N N C H}_{3}\right]\) \(\begin{array}{rl}0 \mathrm{~min} & 1.50 \times 10^{-2} M \\ 10 \mathrm{~min} & 1.29 \times 10^{-2} M \\ 20 \mathrm{~min} & 1.10 \times 10^{-2} M \\ 30 \mathrm{~min} & 0.95 \times 10^{-2} M\end{array}\) Obtain the average rate of decomposition in units of \(M / \mathrm{s}\) for each time interval.
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