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Chapter 14: Problem 97

A first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) has the rate constant \(k=3.2 \times 10^{-3} \mathrm{~s}^{-1}\). If the initial concentration of \(\mathrm{A}\) is \(2.5 \times 10^{-2} M\), what is the rate of the reaction at \(t=660 \mathrm{~s}\) ?

Short Answer

Expert verified
The rate of the reaction at \(t=660 s\) is approximately \(2.02 \times 10^{-5} M/s\).

Step by step solution

01

Write down the first-order rate law

For a first-order reaction, the rate law is given by: \[rate = k[A]\] Where, rate = rate of reaction, k = rate constant, and [A] = concentration of reactant A. We are given k and the initial concentration of A, and we need to find the rate at t=660 seconds.
02

Use the integrated rate law for a first-order reaction to find the concentration of A at t=660 seconds

The integrated rate law for a first-order reaction is given by: \[\ln{\frac{[A]}{[A]_0}} = -kt\] Where, [A] = concentration of A at time t, [A]_0 = initial concentration of A, and t = time. In this problem, we have: [A]_0 = \(2.5 \times 10^{-2} M\), k = \(3.2 \times 10^{-3} s^{-1}\), and t = \(660 s\). We need to solve this equation for [A] to find the concentration of A at t=660 seconds.
03

Calculate [A] at t=660 seconds

Substitute the given values of [A]_0, k, and t in the integrated rate law equation: \[\ln{\frac{[A]}{2.5 \times 10^{-2}}} = -(3.2 \times 10^{-3})(660)\] Now, solve for [A]: \[ [A] = 2.5 \times 10^{-2} \times e^{-(3.2 \times 10^{-3})(660)}\] \[ [A] \approx 6.32 \times 10^{-3} M\]
04

Compute the rate of reaction at t=660 seconds

Now that we have the concentration of A at t=660 seconds, we can find the rate of reaction. Use the first-order rate law: \[rate = k[A]\] Substitute the given value of k and the calculated value of [A] in the equation: \[rate = (3.2 \times 10^{-3})(6.32 \times 10^{-3})\] \[rate \approx 2.02 \times 10^{-5} M/s\] Thus, the rate of the reaction at t=660 seconds is approximately \(2.02 \times 10^{-5} M/s\).

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Key Concepts

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

Understanding the Rate Constant
The rate constant, denoted as \(k\), is a crucial component in the world of reaction kinetics, especially for first-order reactions. It is a measure that helps us understand how fast a reaction proceeds. In a first-order reaction like \(\text{A} \rightarrow \text{B}\), the rate constant determines how quickly reactant A is converted into product B.
Key features of the rate constant:
  • Units: For first-order reactions, the unit is usually \(s^{-1}\), meaning the reaction rate changes with time.
  • Dependence: The rate constant is constant at a given temperature but may change if the temperature changes.
  • Value: In our example, \(k = 3.2 \times 10^{-3} \text{s}^{-1}\), indicates the reaction proceeds at a moderate pace.
If you are given a rate constant, you can predict how changes in concentration will affect the reaction speed. It's especially vital when calculating how much reactant remains or how fast a product forms over time.
The Integrated Rate Law for First-Order Reactions
The integrated rate law is a vital equation that links concentrations of reactants with time鈥攑roviding a deeper understanding than the simple rate law. For a first-order reaction, the integrated rate law is expressed as:\[\ln{\frac{[A]}{[A]_0}} = -kt\]Here鈥檚 how it works:
  • \([A]\): concentration of A at time \(t\).
  • \([A]_0\): initial concentration of A.
  • \(k\): rate constant, conveying the reaction speed.
  • \(t\): time elapsed since the reaction began.
Using this equation, we can solve for \([A]\) at any desired time, such as the 660 seconds in our example. By solving the integrated rate law, you determine that \([A] = 6.32 \times 10^{-3} \text{M}\), the concentration of A left after 660 seconds. This equation shows the exponential relationship between concentration and time for first-order kinetics.
Exploring Reaction Kinetics
Reaction kinetics is the study of the speed or rate at which chemical reactions occur. For first-order reactions, understanding kinetics involves:
  • Determining how reactant concentration changes over time using rate laws and integrated rate laws.
  • Using rate constants to predict how quickly a reaction proceeds.
In our specific problem, reaction kinetics allows us to calculate that the rate of reaction at 660 seconds is \(2.02 \times 10^{-5} \text{M/s}\).

The reaction rate depends on both the rate constant and the current concentration of the reactant. Reaction kinetics helps chemists tailor reactions to desired speeds, crucial in industries where reaction time affects production.
By mastering the concepts of rate constants and integrated rate laws, you can predict and manipulate how quickly a reaction will reach completion. This skill is essential for chemists in both academic and industrial settings.

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Most popular questions from this chapter

The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) was performed and the following data obtained under conditions of constant \(\left[\mathrm{Cl}_{2}\right]\) : (a) Is the following mechanism consistent with the data? $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{NOCl}_{2}(g) \text { (fast) } \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) & \longrightarrow 2 \mathrm{NOCl}(g) \text { (slow) } \end{aligned} $$ (b) Does the linear plot guarantee that the overall rate law is second order?

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right)\). (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

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: (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCI}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\).

The following mechanism has been proposed for the reaction of NO with \(\mathrm{H}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}\) and $\mathrm{H}_{2} \mathrm{O} :$ $$\mathrm{NO}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)$ $\mathrm{N}_{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (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 any intermediates in 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?

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with \(\mathrm{ICl}\) : $$ \begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?
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