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A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

Short Answer

Expert verified
The time required for the reaction to be 75% complete will depend on the order of the reaction. For a first order reaction, it will be twice the time taken for 50% completion; for a zero order reaction, it will be 1.5 times the time taken for 50% completion, assuming that the concentration is proportional to the percentage of reaction completion.

Step by step solution

01

Calculate the Rate Constant 'k' for 50% reaction completion

As 'k' is a constant, irrespective of its being either zero order or first order reaction, it will remain the same at 50% or 75% completion or any other percentage of completion. First, as the reaction is 50% complete serving as our base-status, the ratio \([A_0]/[A]\) should equal 2 (100%/50%). Now, reinterpret the formula for each order to solve the rate constant 'k'. For the first order reaction: \( k = ln(2) / t \) For the zero order reaction: \( k = ([A_0]-[A]) / t = (1-0.5)[A_0]/t \)
02

Calculate Time 't' for 75% Reaction Completion

Given 75% completion, the ratio \([A_0]/[A]\) should equal 4 (100%/25%). This ratio, along with the 'k' value calculated in previous step, can be used to calculate 't' using the formulas for first and zero order reactions. For the first order reaction: \( t = ln(4) / k \) And for zero order reaction: \( t = (1-0.25) [A_0]/k = 0.75 [A_0]/k \)
03

Provide the Final Answer

Plug in the 'k' values you calculated in step 1 into the formulas for 't' in step 2, paying attention to whether it's a first order or zero order reaction.

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Key Concepts

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

First Order Reaction
In chemical kinetics, a first order reaction is characterized by the rate of reaction being directly proportional to the concentration of one of the reactants. In other words, if you were to double the concentration of the reactant, the rate of the reaction would also double.

This type of reaction has a unique mathematical expression for the rate law, represented as \(\text{Rate} = k[A]\), where \([A]\) is the concentration of the reactant and \(k\) is the first order rate constant.

To fully grasp a first order reaction, let's consider the case where a reaction is 50% complete. Using the natural logarithm (ln), the rate constant \(k\) can be calculated as \(k = \frac{\ln(2)}{t}\), where \(t\) is the elapsed time. Knowing \(k\), one can determine the time it will take for the reaction to reach 75% completion by rearranging the formula: \(t = \frac{\ln(4)}{k}\). This showcases the logarithmic relationship between the concentration of reactants and the time in first order kinetics.
Zero Order Reaction
A zero order reaction is identified by its rate being independent of the concentration of reactants. This means if you were to alter the concentration of the reactant, the rate of reaction would stay unchanged. This is in contrast to first order reactions, for which rate changes proportionally with concentration.

The rate law for a zero order reaction is given by \(\text{Rate} = k[A]^0 = k\), indicating that the rate is constant regardless of the concentration of reactant \([A]\).

When dealing with exercises involving zero order reactions, it's key to understand that after a certain time period, the concentration of the reactant has decreased linearly with time. For instance, after 50% completion in a given time, to calculate \(k\) you would use the equation \(k = \frac{([A_0]-[A])}{t}\). If we want to find out the time necessary to reach 75% completion, we apply the formula \(t = \frac{0.75 [A_0]}{k}\) using the previously calculated rate constant \(k\). It illustrates how the concentration decreases in a straight-line manner over time in a zero order reaction.
Reaction Rate Constant
The reaction rate constant, denoted by \(k\), is a pivotal component in the study of chemical kinetics. It is a proportionality factor that connects the rate of a chemical reaction with the concentrations of its reactants.

For any chemical reaction, the rate constant is determined experimentally and depends on several factors such as temperature, the presence of a catalyst, and the physical state of the reactants. It provides crucial information about the reaction's speed and can help predict the time required for a certain level of completion.

In practice, the rate constant is found by observing the time taken for a reaction to reach a certain conversion percentage – like 50%. As seen in the exercise, one could determine \(k\) for both first-order and zero-order reactions and then use it to predict further aspects of the reaction progression, such as how long it would take to reach 75% completion. Regardless of the reaction order, the rate constant remains the same for a given reaction under constant conditions, making it a fundamental aspect of predicting and understanding chemical reaction behavior.

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Most popular questions from this chapter

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.

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.

Ammonia decomposes on the surface of a hot tungsten wire. Following are the half-lives that were obtained at \(1100^{\circ} \mathrm{C}\) for different initial concentrations of \(\mathrm{NH}_{3}:\left[\mathrm{NH}_{3}\right]_{0}=0.0031 \mathrm{M}, t_{1 / 2}=7.6 \mathrm{min} ; 0.0015 \mathrm{M}\) \(3.7 \mathrm{min} ; 0.00068 \mathrm{M}, 1.7 \mathrm{min.}\) For this decomposition reaction, what is (a) the order of the reaction; (b) the rate constant, \(k ?\)

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

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