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Chapter 14: Problem 96

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

Short Answer

Expert verified
In summary, Reaction 1 has a constant half-life, indicating a zero-order rate law with the formula Rate = k. On the other hand, Reaction 2's half-life increases as the reaction proceeds, suggesting a first-order rate law with a possible formula Rate = k [A]. However, the rate law for Reaction 2 may also depend on other factors and reactants.

Step by step solution

01

Understanding reaction rates and rate laws

Reaction rates describe how fast the concentration of reactants change as a reaction proceeds, while rate laws represent the relationship between the reaction rate and the concentrations of reactants. Rate laws typically have the form: Rate = k [A]^m [B]^n where k is the rate constant, [A] and [B] are the concentrations of reactants A and B, and m and n are their respective reaction orders.
02

Considering Reaction 1's constant half-life

For Reaction 1, the half-life is constant, which means the rate of Reaction 1 does not depend on the concentration of the reactants. Hence, the rate is directly proportional to the concentration to the power of 0. This gives us the following rate law for Reaction 1: Rate = k Here, k is the rate constant.
03

Considering Reaction 2's increasing half-life

For Reaction 2, the half-life increases as the reaction proceeds, meaning the reaction slows down as it progresses. This indicates that the reaction rate is dependent on the concentration of reactants. Since the half-life increases, this is suggestive of a first-order reaction, where the rate is directly proportional to the concentration of a reactant to the power of 1. Given this information, an example rate law for Reaction 2 could be: Rate = k [A] where k is the rate constant and [A] is the concentration of reactant A. Please note that it is just an example of a possible rate law based on the given information, and it may also be dependent on other factors and reactants.
04

Concluding the rate laws of the reactions

In summary, Reaction 1 has a constant half-life, indicating that it has a zero-order rate law, which can be written as: Rate = k On the other hand, Reaction 2 has a half-life that increases as the reaction proceeds, suggesting it could have a first-order rate law, which can be written as: Rate = k [A] However, for Reaction 2, the rate law provided is an example based on the given information, and it may also be dependent on other factors and reactants.

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Most popular questions from this chapter

(a) A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol}\) ? (b) Another first-order reaction also has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\) What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol}\) ? (c) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M}\) ? (c) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ? (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to \(0.0120 \mathrm{M}\) ?

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

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). At \(23^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](\mathrm{M})} \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \\ \hline \end{array} $$

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\)-fold increase in the reaction rate?
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