/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Ethane, \(\mathrm{C}_{2} \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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Ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}\), forms \(\cdot \mathrm{CH}_{3}\) radicals at \(700 .^{\circ} \mathrm{C}\) in a firstorder reaction, for which \(k=1.98 \mathrm{~h}^{-1}\). (a) What is the half-life for the reaction? (b) Calculate the time needed for the amount of ethane to fall from \(1.15 \times 10^{-3} \mathrm{~mol}\) to \(2.35 \times 10^{-4} \mathrm{~mol}\) in a 500.-mL reaction vessel at \(700 .{ }^{\circ} \mathrm{C}\). (c) How much of a 6.88-mg sample of ethane in a \(500 .-\mathrm{mL}\) reaction vessel at \(700 .{ }^{\circ} \mathrm{C}\) will remain after \(45 \mathrm{~min}\) ?

Short Answer

Expert verified
a) Half-life is 0.350 h. b) Time needed is 1.386 h. c) After 45 min, 1.93 mg of ethane will remain.

Step by step solution

01

Determine the half-life for a first-order reaction

The half-life (\( t_{1/2} \) of a first-order reaction is determined using the equation \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. Plug in the given rate constant \( k = 1.98 \, \text{h}^{-1} \) into the equation to calculate the half-life.
02

Calculate the time for concentration decrease

For a first-order reaction, the time \( t \) it takes for the concentration of a reactant to decrease from \( [A]_0 \) to \( [A] \) is given by the equation \( \ln \left( \frac{[A]}{[A]_0} \right) = -kt \). Solve for \( t \) using the initial concentration \( [A]_0 = 1.15 \times 10^{-3} \, \text{mol} \) and final concentration \( [A] = 2.35 \times 10^{-4} \, \text{mol} \), along with the rate constant \( k = 1.98 \, \text{h}^{-1} \).
03

Determine the remaining ethane after a given time

We can use the first-order kinetics equation \( [A] = [A]_0 e^{-kt} \) to find the remaining concentration of ethane, \( [A] \), after 45 min. First, convert the mass of ethane to moles, then use the rate constant \( k \) and time \( t = 45 \, \text{min} = 0.75 \, \text{h} \) to find \( [A] \).

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Key Concepts

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

Half-Life Calculation
Understanding the half-life of a reaction is crucial for mastering chemical kinetics, especially for reactions that are described as first-order. The half-life, denoted as \( t_{1/2} \), represents the time required for half of the reactant to be converted into products. In the context of a first-order reaction, the half-life is independent of the initial concentration of the reactant, making it a particularly convenient measure.

For a first-order reaction, the half-life is calculated using the formula: \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. This elegant relationship shows that for any given first-order reaction, knowing the rate constant allows you to determine how quickly the concentration will decrease by half. This is especially helpful in predicting the behavior of reactions over time.
Concentration Decrease Over Time
If you're wondering how the concentration of a reactant in a first-order reaction changes as time progresses, the answer lies in a straightforward logarithmic relationship. Specifically, the time \( t \) it takes for the concentration to decrease from an initial value \( [A]_0 \) to a later value \( [A] \) can be described by the equation: \( \ln \left( \frac{[A]}{[A]_0} \right) = -kt \).

This expression is incredibly useful for calculating not just how much time has passed, but also for predicting future concentrations at a given time. By rearranging the equation to solve for either \( t \) or \( [A] \), we can track the decrease of reactant concentration over time and make accurate predictions about the reaction's progress, which is a central aspect of chemical kinetics.
Rate Constant
Key to mastering the concept of chemical kinetics, the rate constant, symbolized as \( k \), is a proportionality factor that connects the reaction rate to the reactant concentrations. It is specific to each reaction and conditions such as temperature. The rate constant determines the speed at which a chemical reaction proceeds.

For first-order reactions, the rate constant has units of \( \text{time}^{-1} \), and its value is crucial for calculating both the half-life and the time-dependent concentration of reactants. The higher the value of \( k \), the faster the reaction occurs, which is why knowing \( k \) provides valuable insights into the reaction's kinetics. Changes in temperature or the use of a catalyst can significantly alter the rate constant and thereby the reaction rate.
Chemical Kinetics Equations
The equations of chemical kinetics serve as mathematical models describing the rate at which reactants are transformed into products. In the realm of first-order reactions, the rate of the reaction is directly proportional to the concentration of the single reactant.

One of the fundamental equations for first-order reactions is: \( [A] = [A]_0 e^{-kt} \), where \( [A] \) is the concentration at time \( t \), \( [A]_0 \) is the initial concentration, and \( e \) is the base of the natural logarithm. This equation is invaluable for determining the concentration of a reactant at any point in time, as it incorporates the exponential decay pattern typical of first-order kinetics.

Understanding these equations not only allows for prediction of reaction progress but also equips students to interpret and manipulate kinetic data accurately in practical scenarios.

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Most popular questions from this chapter

Complete the following statements relating to the production of ammonia by the Haber process, for which the overall reaction is \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \cdot\) (a) The rate of consumption of \(\mathrm{N}_{2}\) is ______ times the rate of consumption of \(\mathrm{H}_{2}\). (b) The rate of formation of \(\mathrm{NH}_{3}\) is _____ times the _______times the rate of consumption of \(\mathrm{N}_{2}\).

The half-life for the first-order decomposition of A is \(355 \mathrm{~s}\). How much time must elapse for the concentration of A to decrease to (a) \(\frac{1}{8}[\mathrm{~A}]_{0} ;\) (b) one-fourth of its initial concentration; (c) \(15 \%\) of its initial concentration; (d) one-ninth of its initial concentration?

Consider the reaction \(\mathrm{A} \rightleftarrows \mathrm{B}\), which is first order in each direction with rate constants \(k\) and \(k^{\prime}\). Initially, only A is present. Show that the concentrations approach their equilibrium values at a rate that depends on \(k\) and \(k^{\prime}\).

For the reversible, one-step reaction \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C}+\mathrm{D}\), the forward rate constant is \(52.4 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\) and the rate constant for the reverse reaction is \(32.1 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\). The activation energy was found to be \(35.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the forward reaction and \(44.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the reverse reaction. (a) What is the equilibrium constant for the reaction? (b) Is the reaction exothermic or endothermic? (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes by first- order kinetics with a rate constant of \(0.15 \mathrm{~s}^{-1}\) at \(353 \mathrm{~K}\). (a) What is the half-life (in seconds) for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(353 \mathrm{~K}\) ? (b) If \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}=0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), what will be the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(2.0 \mathrm{~s}\) ? (c) How much time (in minutes) will elapse before the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration decreases from \(0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.0135 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) ?

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