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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 \([\mathrm{A}]=0.50 \mathrm{M} ?\)

Short Answer

Expert verified
The percentage of a sample of A that remains unreacted 900 s after a reaction has been started is 10%. The rate of the reaction when [A]=0.50 M is 0.001925 M s^{-1}.

Step by step solution

01

Calculate the Rate Constant from the Half-life

The relationship between the half-life (\(t_{1/2}\)) and the rate constant (k) for a first-order reaction is given by the equation \(t_{1/2} = \frac{0.693}{k}\). We can rearrange this equation to solve for k: \(k = \frac{0.693}{t_{1/2}}\). Substituting the given half-life of 180 s into this equation gives: \(k = \frac{0.693}{180 s} = 0.00385 s^{-1}\).
02

Determine the Percentage of A that Remains Unreacted After 900 s

The amount of reactant A remaining after a certain time (t) in a first-order reaction can be found using the equation \(A_t = A_0 \times e^{-kt}\), where \(A_t\) is the amount of A at time t, \(A_0\) is the initial amount of A, k is the rate constant and t is the time. However, we want to find the percentage of A that remains, so we can normalize this equation to \( \frac{A_t}{A_0} = e^{-kt}\). Substituting the calculated k = 0.00385 s^{-1} and t = 900 s into this equation gives: \( \frac{A_t}{A_0} = e^{-0.00385 s^{-1} \times 900 s}\). This equation gives us approximately 0.10, or 10%. So, 10% of the sample of A remains unreacted after 900 s.
03

Calculate the Rate of Reaction when [A]=0.50 M

The rate of a first-order reaction can be found using the equation \( \text{rate} = k[A] \), where k is the rate constant, and [A] is the concentration of reactant A. Substituting the calculated k = 0.00385 s^{-1} and [A] = 0.50 M into this equation gives: \(\text{rate} = 0.00385 s^{-1} \times 0.50 M = 0.001925 M s^{-1}\). So, the rate of the reaction when [A] = 0.50 M is 0.001925 M s^{-1}.

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Key Concepts

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

reaction kinetics
Reaction kinetics is the study of how fast or slow a reaction occurs. It provides vital insights into the rate at which reactants convert into products. For first-order reactions, the rate is directly proportional to the concentration of the reactant. This means if we were to double the concentration of the reactant, the reaction rate would also double.

In the context of first-order reactions, the rate at which the reaction occurs can be expressed by the formula: \[\text{rate} = k[A]\]where \(k\) is the rate constant, and \([A]\) signifies the concentration of the reactant A. This equation indicates that the speed of the reaction depends on both the rate constant and the concentration of A. Observing how these two factors interact helps scientists predict and control the reaction better.

Reaction kinetics helps chemists and chemical engineers optimize conditions for industrial reactions, ensuring that they run efficiently and safely. Understanding the rate also helps predict how long a reaction will take and how changing conditions will impact the overall process.
half-life calculations
Half-life is a concept often used with reactions, especially radioactive decay, but it is also crucial in other first-order reactions. It describes the time taken for half of the initial amount of reactant to be converted into products.

For first-order reactions, the half-life \(t_{1/2}\) has a unique relationship with the rate constant \(k\):\[t_{1/2} = \frac{0.693}{k}\]This formula is pivotal because it shows that the half-life is constant and does not depend on the initial concentration of the reactant. In our exercise, the half-life is given as 180 seconds, leading us to calculate \(k\) as 0.00385 s\(^{-1}\).

By knowing the half-life and the rate constant, we can predict how much of a reactant remains after any given time. For instance, after multiple half-lives, the quantity of the reactant continues to halve until it becomes negligible. Understanding half-life helps in various applications like pharmacology, where it determines how often a drug should be administered to maintain its efficacy in the human body.
rate constant
The rate constant \(k\) is a fundamental component of reaction kinetics. It is a measure of the reaction speed under specific conditions. The rate constant varies depending on factors like temperature, presence of a catalyst, and the nature of the reactant and product.

In our problem, after calculating from the given half-life, the rate constant was found to be 0.00385 s\(^{-1}\). This value signifies how quickly the reaction progresses in this particular setup.

The rate constant has units that depend on the reaction order. For first-order reactions, it's usually in s\(^{-1}\), indicating the change in concentration over time. The importance of \(k\) is underscored by its ability to help chemists understand and tweak reaction conditions to achieve desired outcomes more efficiently.
  • Ensures that reactions run smoothly in industrial applications.
  • Helps predict reaction behavior under altered conditions.
  • Crucial for kinetic modeling and simulation tasks in various chemical processes.
Understanding \(k\) and its implications allows scientists to delve deeper into the molecular world and extract critical information about reaction behaviors.

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Most popular questions from this chapter

The rate of a chemical reaction generally increases rapidly, even for small increases in temperature, because of a rapid increase in (a) collision frequency; (b) fraction of reactant molecules with very high kinetic energies; (c) activation energy; (d) average kinetic energy of the reactant molecules.

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 a halflife, \(t_{1 / 2},\) of 46.2 min at \(25^{\circ} \mathrm{C}\) and \(2.6 \mathrm{min}\) at \(102^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) At what temperature would the half-life be 10.0 min?

The reaction \(A \longrightarrow\) products is second order. The initial rate of decomposition of \(A\) when \([\mathrm{A}]_{0}=0.50 \mathrm{M}\) is \((\mathrm{a})\) the same as the initial rate for any other value of \([\mathrm{A}]_{0} ;\) (b) half as great as when \([\mathrm{A}]_{0}=1.00 \mathrm{M} ;(\mathrm{c})\) five times as great as when \([\mathrm{A}]_{0}=[\mathrm{A}]_{0}=0.25 \mathrm{M}.\)

One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.

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