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Chapter 12: Problem 96

A certain substance, initially present at \(0.0800 M\), decomposes by zero-order kinetics with a rate constant of \(2.50 \times 10^{-2}\) \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). Calculate the time (in seconds) required for the system to reach a concentration of \(0.0210 \mathrm{M}\).

Short Answer

Expert verified
The time required for the system to reach a concentration of \(0.0210 M\) is 2360 seconds.

Step by step solution

01

List down the given values

The given values are: - Initial concentration (\(C_0\)) = \(0.0800 M\) - Rate constant (\(k\)) = \(2.50 \times 10^{-2}\) \( \frac{mol}{L \cdot s}\) - Final concentration (\(C_t\)) = \(0.0210 M\) We want to find the time (t).
02

Use the zero-order rate equation

We have the zero-order rate equation: \[C_t = C_0 - kt\] Now, plug in the given values and solve for t: \(0.0210 M = 0.0800 M - (2.50 \times 10^{-2} \frac{mol}{L \cdot s}) t\)
03

Solve for t

First, get the term with 't' on one side of the equation: \((2.50 \times 10^{-2} \frac{mol}{L \cdot s}) t = 0.0800 M - 0.0210 M\) Next, simplify the right side of the equation: \((2.50 \times 10^{-2} \frac{mol}{L \cdot s}) t = 0.0590 M\) To find the value of 't', divide both sides by the rate constant: \[t = \frac{0.0590 M}{2.50 \times 10^{-2} \frac{mol}{L \cdot s}}\]
04

Calculate the value of t

Finally, divide the values to get the result: \[t = \frac{0.0590}{2.50 \times 10^{-2}} = 2360\ seconds\] So, it will take 2360 seconds for the system to reach a concentration of \(0.0210 M\).

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of physical chemistry that deals with understanding the rates of chemical reactions. It is pivotal in understanding how reactions occur and the factors that influence their rates. Factors that can affect reaction rates include temperature, concentration of reactants, surface area, catalysts, and the presence of light.

The rate of a chemical reaction tells us how fast the reactants are converted into products. In our exercise, we are dealing with zero-order kinetics, a scenario where the rate of reaction is independent of the concentration of reactants. This is in stark contrast to more common reactions that follow first or second-order kinetics, where the rate is dependent on the concentration(s) of one or more reactants.
Reaction Rate Constant
The reaction rate constant, represented by the symbol 'k', is a key concept in chemical kinetics. It's a proportionality constant that relates the reaction rate to the reactant concentrations at a given temperature. The value of 'k' can reveal how quickly a reaction proceeds; the larger the value, the faster the reaction rate.

In our given problem, the rate constant for the zero-order reaction is provided as \(2.50 \times 10^{-2}\) \(\frac{mol}{L \cdot s}\). This value is critical for calculating how long it will take for the substance to decompose to a certain concentration. It's also worth noting that the units of the rate constant vary depending on the order of the reaction. For zero-order reactions, the unit is concentration per time, as seen in this exercise.
Concentration-Time Relationship
Understanding the relationship between concentration and time is fundamental in chemical kinetics. This relationship tells us how the concentration of a reactant or product changes as a reaction proceeds. For zero-order reactions, the concentration decreases linearly with time.

In our exercise, we have applied this principle to determine the time it takes for the concentration to change from \(0.0800 M\) to \(0.0210 M\). This straight-line relationship can be depicted on a concentration versus time graph, with the slope representing the negative of the reaction rate constant 'k'. The slope of this line is crucial because it leads directly to the mathematical solution for the time 't' in the exercise.
Rate Equations
Rate equations, also known as rate laws, express the rate of a chemical reaction in terms of the concentration of reactants. They are empirical equations derived from experimental data, which make them indispensable tools in chemical kinetics. For zero-order reactions, the rate equation is: \[ rate = k \]

In our exercise, the rate equation simplifies to \(C_t = C_0 - kt\), which shows that the concentration of the substance at any given time 't' is the initial concentration minus the product of the rate constant and time. By rearranging this equation and solving for 't', we can determine the exact time required for a reaction to reach a certain concentration, explicitly demonstrating the utility of rate equations in chemical problem-solving.

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Most popular questions from this chapter

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of the rate constant. c. The half-life for a reaction, \(\mathrm{aA} \longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0}\). d. The half-life for a second-order reaction increases as the reaction proceeds.

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t\) : $$ \begin{aligned} -\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ -\frac{\Delta \mathrm{B}}{\Delta t} &=0.0120 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{aligned} &\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \\ &\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ Overall reaction \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{a}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ is \(14.0 \mathrm{~kJ} . E_{a}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.

The decomposition of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) on an alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) surface $$ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ was studied at \(600 \mathrm{~K}\). Concentration versus time data were collected for this reaction, and a plot of \([\mathrm{A}]\) versus time resulted in a straight line with a slope of \(-4.00 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. If the initial concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) was \(1.25 \times 10^{-2}\) \(M\), calculate the half-life for this reaction. c. How much time is required for all the \(1.25 \times 10^{-2} M\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose?
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