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Chapter 14: Problem 96

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

Short Answer

Expert verified
In conclusion, both Reaction (1) and Reaction (2) follow a 2nd order rate law based on their half-life trends. For Reaction (1), the half-life getting longer as the reaction proceeds indicates a 2nd order rate law, while for Reaction (2), the half-life getting shorter as the reaction proceeds also represents a 2nd order rate law, but with a decreasing concentration of another reactant, leading to an overall faster rate as the reaction progresses.

Step by step solution

01

Overview of Rate Laws and Half-Lives

Rate laws can be described as: rate = k [A]^n [B]^m, where k is the rate constant and n and m are the reaction orders for reactants A and B, respectively. Half-life is the time it takes for the concentration of a reactant to decrease to half its initial value. Now, let's study the half-life behaviors of 0th, 1st, and 2nd order reactions.
02

0th Order Reactions

For a 0th order reaction, the rate law is: rate = k [A]^0 = k. The half-life can be calculated using the equation: t(1/2) = [A]0 / (2k), where [A]0 is the initial concentration of reactant A. Since the half-life depends on the initial concentration and is independent of the reaction progress, this does not match any of the given reaction characteristics.
03

1st Order Reactions

For a 1st order reaction, the rate law is: rate = k [A]^1 = k[A]. The half-life can be calculated using the equation: t(1/2) = ln(2) / k. The half-life is constant and does not change as the reaction proceeds, so this does not match any of the given reaction characteristics either.
04

2nd Order Reactions

For a 2nd order reaction, the rate law is: rate = k [A]^2. The half-life can be calculated using the equation: t(1/2) = 1 / (k[A]0), where [A]0 is the initial concentration of reactant A. Here, the half-life is inversely proportional to the initial concentration of the reactant. Thus, the half-life increases as the concentration decreases, and vice versa.
05

Conclusion

By comparing the half-life behaviors of 0th, 1st, and 2nd order reactions, we can conclude that: Reaction (1): The half-life getting longer as the reaction proceeds indicates that this reaction follows a 2nd order rate law. Reaction (2): The half-life getting shorter as the reaction proceeds also indicates a 2nd order rate law, but with a decreasing concentration of the other reactant in addition to reactant A, leading to an overall faster rate as the reaction progresses. Therefore, both reactions are 2nd order based on their half-life trends.

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Most popular questions from this chapter

Consider a hypothetical reaction between \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) that is zero order in A, second order in B, and first order in C. (a) Write the rate law for the reaction. (b) How does the rate change when [A] is tripled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is doubled and the other reactant concentrations are held constant? (d) How does the rate change when [C] is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are doubled? (f) By what factor does the rate change when the concentrations of all three reactants are cut in half?

For a first order reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C},\) if the half-life of \(\mathrm{A}\) at \(25^{\circ} \mathrm{C}\) is $3.05 \times 10^{4} \mathrm{~s},\( what is the rate constant \)k$ at this temperature? What percentage of A will not have reacted after one day?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: $\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .$ This rapid reaction gives the following rate data: $$ \begin{array}{ccc} \hline\left[\mathrm{OCI}^{-}\right](M) & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when $\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\( and \)\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}$

The following mechanism has been proposed for the reaction of \(\mathrm{NO}\) with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and $\mathrm{H}_{2} \mathrm{O}$ : $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{NO}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g) \\ \mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism.(c) Identify anyintermediatesin the mechanism. (d) The observed rate law is rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\). If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions?

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?
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