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Chapter 35: Problem 13

A certain reaction is first order, and 540 s after initiation of the reaction, \(32.5 \%\) of the reactant remains. a. What is the rate constant for this reaction? b. At what time after initiation of the reaction will \(10 \%\) of the reactant remain?

Short Answer

Expert verified
a. The rate constant for this reaction is approximately \(k \approx 0.00264\,s^{-1}\). b. It will take approximately 943.79 seconds after the initiation of the reaction for 10% of the reactant to remain.

Step by step solution

01

Find the rate constant (k) using the given information

From the problem, we have \(A(t) = 0.325A_0\) (since 32.5% remains) and \(t = 540s\). To find the rate constant k, we will first solve the first-order reaction formula for k: \[ k = -\frac{1}{t} \log(\frac{A(t)}{A_0}) \] Now, substitute the given values into the formula to find k: \[ k = -\frac{1}{540} \log(\frac{0.325A_0}{A_0}) \] Simplify and calculate k: \[ k = -\frac{1}{540} \log(0.325) \approx 0.00264\,s^{-1} \] So, the rate constant for this reaction is approximately 0.00264 s^{-1}.
02

Find the time when 10% of the reactant remains

Now, we need to find the time (t) at which \(A(t) = 0.1A_0\) (since 10% remains). Use the first-order reaction formula and the value of k found in Step 1: \[ 0.1A_0 = A_0 \exp(-0.00264t) \] Divide both sides of the equation by \(A_0\): \[ 0.1 = \exp(-0.00264t) \] To solve for t, take the natural logarithm of both sides of the equation: \[ \log(0.1) = -0.00264t \] Now, divide both sides of the equation by -0.00264: \[ t = \frac{\log(0.1)}{-0.00264} \approx 943.79\,s \] So, it will take approximately 943.79 seconds after the initiation of the reaction for 10% of the reactant to remain.

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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 chemistry that deals with the rates of chemical reactions, the factors that affect these rates, and the mechanisms by which the reactions proceed. Understanding the kinetics of a chemical reaction is crucial for controlling reactions in manufacturing, determining shelf life of products, and more. It involves the study of how different conditions, such as temperature, pressure, and concentration of reactants, influence the speed of a chemical reaction.

In the realm of chemical kinetics, we explore how substances transform over time and through what pathways they interact. The rate of a chemical reaction is a centerpiece in this study, leading to the concept of the reaction rate constant, which is instrumental in quantifying the speed of a reaction.
Reaction Rate
The reaction rate refers to the speed at which the concentrations of reactants or products in a chemical reaction change over time. It's typically expressed as the change in concentration of a reactant or product per unit time. In a practical sense, knowing the reaction rate helps predict how long a chemical reaction will take to reach a certain point or how it can be controlled to optimize the desired product yield.

For example, if you know that a reactant is being consumed quickly, you can infer that the reaction rate is high. However, if the concentration of this reactant barely changes over an extended period, the reaction is likely to be slower, indicating a low reaction rate.
First-Order Reaction
A first-order reaction is a type of chemical reaction where the rate depends on the concentration of a single reactant raised to the first power. This means that the rate of reaction is directly proportional to the concentration of the reactant. A characteristic feature of first-order reactions is that they have a constant half-life, which means the time it takes for half of the reactant to be consumed is the same, regardless of the starting concentration.

In a mathematical sense, a first-order reaction is described by the linear equation,
\( \text{rate} = k[A]^1 \) where \( k \) is the rate constant, and \( [A] \) is the concentration of the reactant. In this equation, the '1' as an exponent is typically omitted since it indicates first-order kinetics.
Rate Constant Calculation
For first-order reactions, the rate constant (\( k \)) is a measure of how quickly a reaction proceeds. It is a unique value for each chemical reaction at a given temperature and can be calculated using the formula:
\[ k = -\frac{1}{t} \times \text{ln}\bigg(\frac{[A](t)}{[A]_0}\bigg) \]
In this formula, \( [A](t) \) is the concentration of reactant at time \( t \), and \( [A]_0 \) is the initial concentration of the reactant. By measuring how the concentration of a reactant changes over time, we can determine the rate constant \( k \), which in turn enables us to predict reaction behavior under various conditions.
Exponential Decay
Exponential decay is a fundamental concept describing how the quantity of a reactant decreases over time in a first-order reaction. In a chemical context, it signifies that the amount of a substance decreases at a rate proportional to its current value. This is mathematically represented as:
\( [A](t) = [A]_0 \times e^{-kt} \)
where \( e \) is the base of the natural logarithm, \( k \) is the rate constant, and \( t \) is the time. Exponential decay is not just limited to chemical reactions; it is a widespread phenomenon seen in many natural processes, such as radioactive decay and cooling of substances. Understanding exponential decay is essential when studying the dynamics of first-order reactions, as it provides a clear model of how reactant concentrations diminish over time.

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Most popular questions from this chapter

For the sequential reaction \(\mathrm{A} \stackrel{k_{A}}{\rightarrow} \mathrm{B} \stackrel{k_{B}}{\rightarrow} \mathrm{C}$$k_{A}=1.00 \times 10^{-3} \mathrm{s}^{-1} .\) Using a computer spreadsheet program such as Excel, plot the concentration of each species for cases where \(k_{B}=10 k_{A}, k_{B}=1.5 k_{A},\) and \(k_{B}=0.1 k_{A} .\) Assume that only the reactant is present when the reaction is initiated.

Hydrogen abstraction from hydrocarbons by atomic chlorine is a mechanism for \(\mathrm{Cl} \cdot\) loss in the atmosphere. Consider the reaction of \(\mathrm{Cl} \cdot\) with ethane: $$\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{Cl} \cdot(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \cdot(g)+\mathrm{HCl}(g)$$ This reaction was studied in the laboratory, and the following data were obtained: $$\begin{array}{cc}\mathbf{T}(\boldsymbol{K}) & \mathbf{k}\left(\times \mathbf{1 0}^{-\mathbf{1 0}} \mathbf{M}^{-\mathbf{2}} \mathbf{s}^{-\mathbf{1}}\right) \\\\\hline 270 & 3.43 \\\370 & 3.77 \\ 470 & 3.99 \\\570 & 4.13 \\\670 & 4.23\end{array}$$ a. Determine the Arrhenius parameters for this reaction. b. At the tropopause (the boundary between the troposphere and stratosphere located approximately \(11 \mathrm{km}\) above the surface of Earth \(),\) the temperature is roughly \(220 \mathrm{K}\). What do you expect the rate constant to be at this temperature? c. Using the Arrhenius parameters obtained in part (a), determine the Eyring parameters \(\Delta H^{\dagger}\) and \(\Delta S^{\frac{1}{r}}\) for this reaction at \(220 \mathrm{K}\)

The conversion of \(\mathrm{NO}_{2}(g)\) to \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g)\) can occur through the following reaction: $$\mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ The activation energy for this reaction is \(111 \mathrm{kJ} \mathrm{mol}^{-1}\) and the pre-exponential factor is \(2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1}\). Assume that these quantities are temperature independent. a. What is the rate constant for this reaction at \(298 \mathrm{K} ?\) b. What is the rate constant for this reaction at the tropopause where \(\mathrm{T}=225 \mathrm{K}\).

In the limit where the diffusion coefficients and radii of two reactants are equivalent, demonstrate that the rate constant for a diffusion controlled reaction can be written as $$k_{d}=\frac{8 R T}{3 \eta}$$

At \(552.3 \mathrm{K}\), the rate constant for the thermal decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.02 \times 10^{-6} \mathrm{s}^{-1} .\) If the activation energy is \(210 . \mathrm{kJ} \mathrm{mol}^{-1},\) calculate the Arrhenius preexponential factor and determine the rate constant at \(600 .\) K.
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