Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\), \([\mathrm{B}]_{0}=3.0 \mathrm{M}\), and \([\mathrm{C}]_{0}=2.0 \mathrm{M} .\) The reaction is started, and after \(8.0\) seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\). a. Calculate the value of \(k\) for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after \(13.0\) seconds. d. Calculate the concentration of \(\mathrm{C}\) after \(13.0\) seconds.

Short Answer

Expert verified
a. The value of k is approximately \(4.17\times10^{-3}\,\mathrm{M}^{-1}\mathrm{s}^{-1}\). b. The half-life for this experiment is approximately \(23.9\, \mathrm{s}\). c. The concentration of A after 13.0 seconds is approximately \(2.16\times10^{-3}\,\mathrm{M}\). d. The concentration of C after 13.0 seconds is approximately \(1.32\,\mathrm{M}\).

Step by step solution

01

Find the initial rate of the reaction

Using the rate law given, we can find the initial rate of the reaction: Rate \( = k[A]_0[B]_0^2 = k(1.0\times10^{-2}\,\mathrm{M})(3.0\,\mathrm{M})^2\)
02

Find the change in concentration of A after 8.0 seconds

After 8.0 seconds, the concentration of A is given \([A]_t = 3.8\times10^{-3}\,\mathrm{M}\). We need to find the change in concentration of A, so we subtract the initial concentration: \(\Delta[A] = [A]_t - [A]_0 = 3.8\times10^{-3}\,\mathrm{M} - 1.0\times10^{-2}\,\mathrm{M}\) Now, we can find the average rate of the reaction over these 8.0 seconds: \(-\frac{\Delta[A]}{\Delta t} = -\frac{3.8\times10^{-3}\,\mathrm{M} - 1.0\times10^{-2}\,\mathrm{M}}{8.0\,\mathrm{s}}\)
03

Calculate the value of k

Equate the initial rate (step 1) with the average rate (step 2) to find the value of k: \(k(1.0\times10^{-2}\,\mathrm{M})(3.0\,\mathrm{M})^2 = -\frac{3.8\times10^{-3}\,\mathrm{M} - 1.0\times10^{-2}\,\mathrm{M}}{8.0\,\mathrm{s}}\) Solve for k.
04

Calculate the half-life for this experiment (using k)

For a second-order reaction, the half-life can be determined using the equation: \(t_{1/2} = \frac{1}{k[A]_0}\) Substitute the value of k found in step 3 and the initial concentration of A, and solve for \(t_{1/2}\).
05

Calculate the concentration of A after 13.0 seconds

To find the concentration of A after 13.0 seconds, we can use the integrated rate law for a second-order reaction: \(\frac{1}{[A]_{t}} = kt + \frac{1}{[A]_{0}}\) Plug in the value of k found in step 3, the initial concentration of A, and the time (13.0 seconds), then solve for \([A]_{t}\).
06

Calculate the concentration of C after 13.0 seconds

Use the stoichiometry of the reaction to find the concentration of C after 13.0 seconds: Since 2 moles of C are used for every 1 mole of A, the change in the concentration of C will be twice the change in the concentration of A, and in the opposite direction: \(\Delta[C] = -2\Delta[A]\) Substitute the change in concentration of A found in step 2, and calculate the change in concentration of C. Then, subtract this from the initial concentration of C to find the concentration of C after 13.0 seconds: \([C]_{t} = [C]_{0} + \Delta[C]\)

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Key Concepts

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

Reaction Rate Calculation
Understanding the speed of chemical reactions is essential in the study of chemical kinetics. The reaction rate refers to the rate at which the reactants are consumed or the products are formed over time. In the given hypothetical reaction \[ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} \], the reaction rate can be calculated by measuring the change in concentration of a reactant or product over time. The rate law for this reaction follows \[\text{Rate} = k[\mathrm{~A}][\mathrm{B}]^{2}\]where \(k\) is the rate constant, and the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are raised to the powers corresponding to their stoichiometric coefficients in the rate-determining step. By plugging the concentrations of the reactants into the rate law equation and measuring the change in concentration over time, we can calculate the reaction rate.

For instance, to find the initial rate, we would use the initial concentrations and the rate law equation while to find the average rate over 8 seconds, we would use the change in concentration of \(\mathrm{A}\). These calculations provide a basis to characterize the kinetics of the reaction and determine the rate constant \(k\), which is crucial for predicting the reaction's behavior under different conditions.
Determining Reaction Order
The reaction order is a critical concept in kinetics as it provides insight into the relationship between the concentration of reactants and the rate of reaction. It is often not apparent from the chemical equation and must be determined experimentally. The order of reaction with respect to each reactant shows the power to which its concentration term is raised in the rate law. In our example, the rate law is given by \[\text{Rate}=k[\mathrm{A}][\mathrm{B}]^{2}\]which suggests that the reaction is first-order with respect to [\(\mathrm{A}\)] and second-order with respect to [\(\mathrm{B}\)]. The overall reaction order is the sum of these individual orders, thus making it a third-order reaction.

To confirm these orders experimentally, one would typically perform a series of reactions varying the concentration of one reactant while holding others constant and observing the effect on the reaction rate. The data obtained helps establish the rate law's exponents empirically and define the kinetics of the reaction.
Integrated Rate Law Application
Integrated rate laws are algebraic expressions that relate concentrations of reactants to time and are derived from the differential rate laws for a given order of reaction. For a second-order reaction, the integrated rate law expression is \[\frac{1}{[A]_{t}} = kt + \frac{1}{[A]_{0}}\]where \([A]_{t}\) is the concentration of reactant A at time \(t\), \([A]_{0}\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time. By applying this formula, we can predict future concentrations of reactants or determine how the concentration has changed over a given time.

In solving the exercise problem, we can calculate the concentration of A after 13.0 seconds by first finding the rate constant \(k\) using initial rate and average rate calculations. Once \(k\) is determined, we can manipulate the integrated rate law to find \([A]_{t}\) at any point in time, providing valuable insight into how concentrations vary throughout the course of the reaction.
Half-Life of a Reaction
Half-life, symbolized as \(t_{1/2}\), is the amount of time required for half of the reactant to be consumed in a chemical reaction. For different orders of reactions, the half-life formulas vary. In a second-order reaction, the half-life is directly dependent on the initial concentration and inversely proportional to the rate constant. The half-life expression for a second-order reaction is given by \[t_{1/2} = \frac{1}{k[A]_0}\]where \(k\) is the rate constant, and \([A]_0\) is the initial concentration of the reactant.

Calculating half-life is crucial in understanding how quickly a reaction proceeds and can be especially important in pharmaceuticals, environmental engineering, and other fields where reaction timing is critical. In the case of our particular problem, knowing \(k\) and \([A]_0\), we can determine the half-life of the reaction which essentially provides a timescale for the reaction's progress under the given conditions.
Second-Order Reaction
A second-order reaction is characterized by a rate that is proportional to the square of the concentration of one reactant or the product of the concentrations of two reactants. In our textbook case, we deal with the second-order kinetics since the rate law is \[\text{Rate}= -\frac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}][\mathrm{B}]^{2}\]suggesting the reaction rate is proportional to the concentration of A and the square of the concentration of B. For second-order reactions, the integrated rate law and half-life expressions are unique as previously discussed.

These reactions exemplify kinetic behavior where interactions between two reactant molecules or two reactive sites within a single molecule are necessary for the reaction to proceed. The nuances of second-order reactions are important to master as they often appear in biochemical processes, such as enzyme kinetics, and have different characteristics and mathematical treatments compared to zero or first-order kinetics.

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Most popular questions from this chapter

Consider the following initial rate data for the decomposition of compound \(\mathrm{AB}\) to give \(\mathrm{A}\) and \(\mathrm{B}\) : Determine the half-life for the decomposition reaction initially having \(1.00 M \mathrm{AB}\) present.

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{~s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\), respectively. What is the value of the activation energy?

The decomposition of hydrogen iodide on finely divided gold at \(150^{\circ} \mathrm{C}\) is zero order with respect to HI. The rate defined below is constant at \(1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) $$ \begin{array}{r} 2 \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t}=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{array} $$ a. If the initial HI concentration was \(0.250 \mathrm{~mol} / \mathrm{L}\), calculate the concentration of HI at 25 minutes after the start of the reaction. b. How long will it take for all of the \(0.250 M\) HI to decompose?

The rate law for the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) is $$ \text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right] $$ It takes \(120 .\) s for \(1.00 M \mathrm{PH}_{3}\) to decrease to \(0.250 \mathrm{M}\). How much time is required for \(2.00 \mathrm{M} \mathrm{PH}_{3}\) to decrease to a concentration of \(0.350 \mathrm{M} ?\)

In an effort to become more environmentally friendly, you have decided that your next vehicle will run on biodiesel that you will produce yourself. You have researched how to make biodiesel in your own home and have decided that your best bet is to use the following chemical reaction: $$ \mathrm{Oil}+\mathrm{NaOH} \text { (in methanol) } \longrightarrow \text { biodiesel }+\text { glycerin } $$ You performed a test reaction in your kitchen to study the kinetics of this process. You were able to monitor the concentration of the oil and found that the concentration dropped from \(0.500 M\) to \(0.250 \mathrm{M}\) in \(20.0\) minutes. It took an additional \(40.0\) minutes for the concentration of the oil to further drop to \(0.125 M\). How long will it take for you to convert \(97.0 \%\) of the oil to biodiesel?

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