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Chapter 14: Problem 40

For the reaction \(A \longrightarrow 2 B+C\), the following data are obtained for \([\mathrm{A}]\) as a function of time: \(t=0 \mathrm{min}\) \([\mathrm{A}]=0.80 \mathrm{M} ; 8 \mathrm{min}, 0.60 \mathrm{M} ; 24 \mathrm{min}, 0.35 \mathrm{M} ; 40 \mathrm{min}\) \(0.20 \mathrm{M}\) (a) By suitable means, establish the order of the reaction. (b) What is the value of the rate constant, \(k ?\) (c) Calculate the rate of formation of \(\mathrm{B}\) at \(t=30 \mathrm{min}\).

Short Answer

Expert verified
By plotting the \(ln[A]\) against time, a straight line indicates first order reaction. The absolute value of the slope of the line provides the rate constant \(k\). Then use the rate equation for the formation of \(B\), \(Rate = 2k[A]\) to find the rate at \(t=30 \mathrm{min}\).

Step by step solution

01

Establish the order of the reaction

In the given data, the concentration of \(A\) is reducing over time, so the reaction order can be determined by plotting the log of concentration of \(A\) against time (a linear relationship suggests a first order reaction). A semi-log plot can be drawn, plotting the natural logarithm (ln) of concentration of \(A\) against time \(t\). If the plot yields a straight line, the reaction is first-order.
02

Determine the rate constant, \(k\)

The slope of the semi-log plot \(ln[A]\) vs \(t\) is equal to \(-k\) (negative because it is a reactant). Calculate the slope of the line and the rate constant \(k\) can be determined by taking absolute value of the slope.
03

Calculate rate of formation of \(B\)

For a first-order reaction like \(A \longrightarrow 2 B+C\), the rate of formation of \(B\) is twice the rate constant times the concentration of \(A\). So at \(t=30 \mathrm{min}\), interpolate the concentration of \(A\) from the given data, multiply it by twice the rate constant to get the rate of formation of \(B\) at \(t=30 \mathrm{min}\).

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Key Concepts

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

Reaction Order
The reaction order is an essential concept in reaction kinetics that helps us understand how the concentration of reactants influences the rate of a chemical reaction. To determine the order of a reaction, we need to observe how the concentration of the reactant changes over time. If the concentration of the reactant and time data produce a straight line when plotted using a logarithmic scale, we can confirm that the reaction is of first-order.

In this exercise, we have a reaction involving a reactant \( A \) gradually converting to products over time. By taking the natural logarithm of the concentration of \( A \) and plotting these values against time, a linear relationship indicates a first-order reaction.

  • A first-order reaction means that the rate of reaction is directly proportional to the concentration of the reactant.
  • For first-order reactions, the half-life is independent of the initial concentration of the reactant.
  • The slope of the plot on a semi-log scale provides further insight into the reaction kinetics.
Rate Constant
The rate constant, \( k \), is a crucial parameter that provides insights into the speed of a chemical reaction. It is a unique value for every reaction at a given temperature and can be influenced by factors like temperature and catalyst presence. For a first-order reaction, the rate constant is determined from the slope of the natural log of the concentration of the reactant plotted against time.

In the example given, \( k\) is calculated from the slope of the line obtained in the semi-logarithmic plot of \( \ln[A] \) versus time. Since the line is straight, indicating a first-order reaction, the slope of the line is equal to \( -k \). Taking the absolute value of the slope gives us the rate constant.

  • The rate constant provides information on how fast or slow a reaction proceeds.
  • For first-order reactions, the unit of \( k \) is \( \text{time}^{-1} \).
  • The value of \( k \) remains constant as long as the conditions (such as temperature) remain unchanged.
Rate of Formation
The rate of formation is an important concept that tells us how quickly products are generated in a chemical reaction. Specifically, for the formation of a product like \( B \) in a reaction, it is crucial to understand how it relates to the reactant consumption.

In a reaction such as \( A \rightarrow 2B + C \), the appearance of \( B \) happens twice as fast as the disappearance of \( A \), since two moles of \( B \) are formed for every mole of \( A \) that reacts. To calculate the formation rate of \( B \), we need to know the concentration of \( A \) at the time in question and the rate constant. The rate of formation at any given time, such as 30 minutes in this case, can be calculated with the formula: Rate = \( 2k[A] \).

  • The rate of formation tells us the speed at which a specific product is produced.
  • For first-order reactions, the rate of formation can be easily estimated once \( k \) and concentration of the reactant are known.
  • The factor of "2" in this context depends on the stoichiometry of the reaction, which shows how many units of \( B \) are formed per unit of \( A \) reacted.

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Most popular questions from this chapter

The object is to study the kinetics of the reaction between peroxodisulfate and iodide ions. $$\begin{aligned} &\text { (a) } \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \end{aligned}$$ The \(I_{3}^{-}\) formed in reaction (a) is actually a complex of iodine, \(\mathrm{I}_{2},\) and iodide ion, \(\mathrm{I}^{-}\). Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) also present in the reaction mixture, reacts with \(\mathrm{I}_{3}^{-}\) just as fast as it is formed. $$\text { (b) } 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}+3 \mathrm{I}^{-}(\mathrm{aq})$$ When all of the thiosulfate ion present initially has been consumed by reaction (b), a third reaction occurs between \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) and starch, which is also present in the reaction mixture. $$\text { (c) } \mathrm{I}_{3}^{-}(\mathrm{aq})+\operatorname{starch} \longrightarrow \text { blue complex }$$ The rate of reaction (a) is inversely related to the time required for the blue color of the starch-iodine complex to appear. That is, the faster reaction (a) proceeds, the more quickly the thiosulfate ion is consumed in reaction (b), and the sooner the blue color appears in reaction (c). One of the photographs shows the initial colorless solution and an electronic timer set at \(t=0 ;\) the other photograph shows the very first appearance of the blue complex (after 49.89 s). Tables I and II list some actual student data obtained in this study. $$\begin{array}{l} \hline\text { TABLE I } \\ \text { Reaction conditions at } 24^{\circ} \mathrm{C}: 25.0 \mathrm{mL} \text { of the } \\ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8}(\text { aq) listed, } 25.0 \mathrm{mL} \text { of the } \mathrm{KI}(\mathrm{aq}) \\ \text { listed, } 10.0 \mathrm{mL} \text { of } 0.010 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}), \text { and } 5.0 \mathrm{mL} \\ \text { starch solution are mixed. The time is that of the } \\ \text { first appearance of the starch-iodine complex. } \\ \hline & \text { Initial Concentrations, } \mathrm{M} \\ \hline \text { Experiment } & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8} & \mathrm{KI} & \text { Time, s } \\ \hline 1 & 0.20 & 0.20 & 21 \\ 2 & 0.10 & 0.20 & 42 \\ 3 & 0.050 & 0.20 & 81 \\ 4 & 0.20 & 0.10 & 42 \\ 5 & 0.20 & 0.050 & 79 \\ \hline \end{array}$$ $$\begin{array}{l} \hline \text { TABLE II } \\ \text { Reaction conditions: those listed in Table I for } \\ \text { Experiment } 4, \text { but at the temperatures listed. } \\ \hline \text { Experiment } & \text { Temperature, }^{\circ} \mathrm{C} & \text { Time, } \mathrm{s} \\ \hline 6 & 3 & 189 \\ 7 & 13 & 88 \\ 8 & 24 & 42 \\ 9 & 33 & 21 \\ \hline \end{array}$$ (a) Use the data in Table I to establish the order of reaction (a) with respect to \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and to I \(^{-}\). What is the overall reaction order? [Hint: How are the times required for the blue complex to appear related to the actual rates of reaction? (b) Calculate the initial rate of reaction in Experiment 1 expressed in \(\mathrm{M} \mathrm{s}^{-1} .\) [Hint: You must take into account the dilution that occurs when the various solutions are mixed, as well as the reaction stoichiometry indicated by equations \((a),(b), \text { and }(c) .]\) (c) Calculate the value of the rate constant, \(k,\) based on experiments 1 and 2 (d) Calculate the rate constant, \(k\), for the four different temperatures in Table II. (e) Determine the activation energy, \(E_{\mathrm{a}}\), of the peroxodisulfate- iodide ion reaction. (f) The following mechanism has been proposed for reaction (a). The first step is slow, and the others are fast. $$\begin{array}{c} \mathrm{I}^{-}+\mathrm{S}_{2} \mathrm{O}_{8}^{2-} \longrightarrow \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \\ \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}^{+} \\ \mathrm{I}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2} \\ \mathrm{I}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-} \end{array}$$ Show that this mechanism is consistent with both the stoichiometry and the rate law of reaction (a). Explain why it is reasonable to expect the first step in the mechanism to be slower than the others.

The first-order reaction \(A \longrightarrow\) products has \(t_{1 / 2}=180 \mathrm{s}\) (a) What percent of a sample of A remains unreacted \(900 \mathrm{s}\) after a reaction has been started? (b) What is the rate of reaction when \([\mathrm{A}]=0.50 \mathrm{M} ?\)

The half-life for the first-order decomposition of nitramide, \(\mathrm{NH}_{2} \mathrm{NO}_{2}(\mathrm{aq}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1),\) is \(123 \min\) at \(15^{\circ} \mathrm{C} .\) If \(165 \mathrm{mL}\) of a \(0.105 \mathrm{M} \mathrm{NH}_{2} \mathrm{NO}_{2}\) solution is allowed to decompose, how long must the reaction proceed to yield \(50.0 \mathrm{mL}\) of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) collected over water at \(15^{\circ} \mathrm{C}\) and a barometric pressure of \(756 \mathrm{mm} \mathrm{Hg} ?\) (The vapor pressure of water at \(15^{\circ} \mathrm{C}\) is \(12.8 \mathrm{mmHg} .)\)

Explain why (a) A reaction rate cannot be calculated from the collision frequency alone. (b) The rate of a chemical reaction may increase dramatically with temperature, whereas the collision frequency increases much more slowly. (c) The addition of a catalyst to a reaction mixture can have such a pronounced effect on the rate of a reaction, even if the temperature is held constant.

Hydroxide ion is involved in the mechanism of the following reaction but is not consumed in the overall reaction. $$\mathrm{OCI}^{-}+\mathrm{I}^{-} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{OI}^{-}+\mathrm{Cl}^{-}$$ (a) From the data given, determine the order of the reaction with respect to \(\mathrm{OCl}^{-}, \mathrm{I}^{-},\) and \(\mathrm{OH}^{-}\) (b) What is the overall reaction order? (c) Write the rate equation, and determine the value of the rate constant, \(k.\) $$\begin{array}{lccl} \hline & & & \text { Rate Formation } \\ {\left[\mathrm{OC}^{-}\right], \mathrm{M}} & {\left[\mathrm{l}^{-}\right], \mathrm{M}} & {\left[\mathrm{OH}^{-}\right], \mathrm{M}} & \mathrm{O}^{-}, \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.0040 & 0.0020 & 1.00 & 4.8 \times 10^{-4} \\ 0.0020 & 0.0040 & 1.00 & 5.0 \times 10^{-4} \\ 0.0020 & 0.0020 & 1.00 & 2.4 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.50 & 4.6 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.25 & 9.4 \times 10^{-4} \\ \hline \end{array}$$
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