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Chapter 26: Problem 35

The equilibrium constant for the reaction \\[ \mathrm{D}^{+}(\mathrm{aq})+\mathrm{OD}^{-}(\mathrm{aq}) \stackrel{k_{1}}{\rightleftharpoons} \mathrm{D}_{2} \mathrm{O}(\mathrm{l}) \\] at \(25^{\circ} \mathrm{C}\) is \(K_{c}=4.08 \times 10^{16} \mathrm{mol}^{-1} \cdot \mathrm{dm}^{3}\). The rate constant \(k_{-1}\) is independently found to be \(2.52 \times 10^{-6} \mathrm{s}^{-1} .\) What do you predict for the observed relaxation time for a temperaturejump experiment to a final temperature of \(25^{\circ} \mathrm{C} ?\) The density of \(\mathrm{D}_{2} \mathrm{O}\) is \(\rho=1.104 \mathrm{g} \cdot \mathrm{cm}^{-3}\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The relaxation time \(\tau\) is approximately \(9.74 \times 10^{-12} \ \mathrm{s}\).

Step by step solution

01

Understanding the relationship

The relaxation time in a chemical reaction when the system is briefly perturbed is given by the inverse of the sum of forward and backward rate constants, denoted as \(\tau = \frac{1}{k_1 + k_{-1}}\). To find \(\tau\), we first need to determine the forward rate constant \(k_1\).
02

Calculate the forward rate constant

Using the relationship between the equilibrium constant \(K_c\), forward rate constant \(k_1\), and backward rate constant \(k_{-1}\), we have \(K_c = \frac{k_1}{k_{-1}}\). Substitute the known values to find \(k_1\): \(k_1 = K_c \times k_{-1} = (4.08 \times 10^{16} \ \mathrm{mol}^{-1} \cdot \mathrm{dm}^{3}) \times (2.52 \times 10^{-6} \ \mathrm{s}^{-1}) = 1.027 \times 10^{11} \ \mathrm{mol}^{-1} \cdot \mathrm{dm}^{3} \ \mathrm{s}^{-1}\).
03

Calculate the relaxation time

Now that we have \(k_1 = 1.027 \times 10^{11} \ \mathrm{mol}^{-1} \cdot \mathrm{dm}^{3} \ \mathrm{s}^{-1}\) and \(k_{-1} = 2.52 \times 10^{-6} \ \mathrm{s}^{-1}\), we use these values to find the relaxation time using \(\tau = \frac{1}{k_1 + k_{-1}}\). After substituting the values, we get \(\tau = \frac{1}{1.027 \times 10^{11} + 2.52 \times 10^{-6}} \approx \frac{1}{1.027 \times 10^{11}} \approx 9.74 \times 10^{-12} \ \mathrm{s}\).

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Key Concepts

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

Relaxation Time
Relaxation time is an important concept in the world of chemical kinetics. It refers to the time it takes for a system to return to equilibrium after it has been disturbed. When you perform a temperature jump experiment, the system is temporarily taken out of its equilibrium state by suddenly changing the temperature. This disturbance causes a brief period where the concentrations of molecules in a reaction change.In biochemical reactions, relaxation time helps scientists understand how quickly reactions respond to changes in conditions. The concept of relaxation time is helpful in monitoring how fast equilibrium is re-established after a sudden change. Particularly in reversible reactions, the relaxation time \( \tau \) is calculated by the equation:\[ \tau = \frac{1}{k_1 + k_{-1}}\]where \( k_1 \) is the forward rate constant and \( k_{-1} \) is the backward rate constant. The smaller the relaxation time, the faster the reaction reaches equilibrium after being disturbed.
Rate Constant
The rate constant is a key player in defining the speed at which a reaction occurs. In chemical kinetics, the rate constant can be either the forward rate constant \( k_1 \) or the backward rate constant \( k_{-1} \).The forward rate constant \( k_1 \) describes the rate at which reactants transform into products, while the backward rate constant \( k_{-1} \) does the opposite, describing the rate at which products revert to reactants.The relationship between equilibrium constant \( K_c \), forward rate constant \( k_1 \), and backward rate constant \( k_{-1} \) is given by:\[ K_c = \frac{k_1}{k_{-1}}\]This equation illustrates how the rate constants influence the position of equilibrium. When \( K_c \) is very large, it suggests that the forward reaction is much more favorable, meaning \( k_1 \) is substantially greater than \( k_{-1} \).
Temperature Jump Experiment
A temperature jump experiment is an intriguing technique used to gain insights into reaction mechanisms and rate constants. This experiment involves rapidly changing the temperature of a reacting system. The sudden alteration in temperature triggers a disturbance, moving the system out of equilibrium. Here's why scientists use temperature jump experiments:
  • To understand reaction dynamics: They help in studying how reaction rates change with temperature.
  • To explore equilibria: They reveal how systems approach equilibrium after a sudden temperature change.
  • To calculate relaxation times:
In such experiments, the fast temperature change allows researchers to observe the system's response by monitoring changes in concentrations over a brief time period. This data enables the calculation of reaction rate constants and further prediction of how the system behaves in different conditions.

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Most popular questions from this chapter

The kinetics of a chemical reaction can be followed by a variety of experimental techniques, including optical spectroscopy, NMR spectroscopy, conductivity, resistivity, pressure changes, and volume changes. When using these techniques, we do not measure the concentration itself but we know that the observed signal is proportional to the concentration; the exact proportionality constant depends on the experimental technique and the species present in the chemical system. Consider the general reaction given by $$ v_{\mathrm{A}} \mathrm{A}+v_{\mathrm{B}} \mathrm{B} \longrightarrow v_{\mathrm{Y}} \mathrm{Y}+v_{\mathrm{Z}} \mathrm{Z} $$ where we assume that \(\mathrm{A}\) is the limiting reagent so that \([\mathrm{A}] \rightarrow 0\) as \(t \rightarrow \infty\). Let \(p_{i}\) be the proportionality constant for the contribution of species \(i\) to \(S\), the measured signal from the instrument. Explain why at any time \(t\) during the reaction, \(S\) is given by $$ S(t)=p_{\mathrm{A}}[\mathrm{A}]+p_{\mathrm{B}}[\mathrm{B}]+p_{\mathrm{Y}}[\mathrm{Y}]+p_{\mathrm{Z}}[\mathrm{Z}] $$ Show that the initial and final readings from the instrument are given by $$ S(0)=p_{\mathrm{A}}[\mathrm{A}]_{0}+p_{\mathrm{B}}[\mathrm{B}]_{0}+p_{\mathrm{Y}}[\mathrm{Y}]_{0}+p_{\mathrm{Z}}[\mathrm{Z}]_{0} $$ tand $$ S(\infty)=p_{\mathrm{B}}\left([\mathrm{B}]_{0}-\frac{v_{\mathrm{B}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right)+p_{\mathrm{Y}}\left([\mathrm{Y}]_{0}+\frac{v_{\mathrm{Y}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right)+p_{\mathrm{Z}}\left([\mathrm{Z}]_{0}+\frac{v_{\mathrm{Z}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right) $$ Combine Equations 1 through 3 to show that $$ [\mathrm{A}]=[\mathrm{A}]_{0} \frac{S(t)-S(\infty)}{S(0)-S(\infty)} $$ (Hint: Be sure to express [B], [Y], and [Z] in terms of their initial values, [A] and \([\mathrm{A}]_{0}\).)

The nucleophilic substitution reaction \\[ \mathrm{PhSO}_{2} \mathrm{SO}_{2} \mathrm{Ph}(\mathrm{sln})+\mathrm{N}_{2} \mathrm{H}_{2}(\mathrm{sin}) \longrightarrow \mathrm{PhSO}_{2} \mathrm{NHNH}_{2}(\mathrm{sln})+\mathrm{PhSO}_{2} \mathrm{H}(\mathrm{sln}) \\] was studied in cyclohexane solution at \(300 \mathrm{K}\). The rate law was found to be first order in \(\left.\mathrm{PhSO}_{2} \mathrm{SO}_{2} \mathrm{Ph} \text { . For an initial concentration of } \mathrm{IPhSO}_{2} \mathrm{SO}_{2} \mathrm{Ph}\right]_{0}=3.15 \times 10^{-5} \mathrm{mol} \cdot \mathrm{dm}^{-3}\) the following rate data were observed. Determine the rate law and the rate constant for this reaction \\[ \begin{array}{c|cccc} {\left[\mathrm{N}_{2} \mathrm{H}_{2}\right]_{0} / 10^{-2} \mathrm{mol} \cdot \mathrm{dm}^{-3}} & 0.5 & 1.0 & 2.4 & 5.6 \\ \hline v / \mathrm{mol} \cdot \mathrm{dm}^{-3} \cdot \mathrm{s}^{-1} & 0.085 & 0.17 & 0.41 & 0.95 \end{array} \\]

The Arrhenius parameters for the reaction described by \\[ \mathrm{HO}_{2}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \\] are \(A=5.01 \times 10^{10} \mathrm{dm}^{3} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}\) and \(E_{\mathrm{a}}=4.18 \mathrm{kJ} \cdot \mathrm{mol}^{-1}\). Determine the value of the rate constant for this reaction at \(298 \mathrm{K}\).

In this problem, we will derive Equation 26.32 from the rate law (Equation 26.31 ) \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}][\mathrm{B}] \\] Use the reaction stoichiometry of Equation 26.30 to show that \([\mathrm{B}]=[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\) Use this result to show that Equation 1 can be written as \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}]\left|[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\right| \\] Now separate the variables and then integrate the resulting equation subject to its initial conditions to obtain the desired result, Equation 26.32 \\[ k t=\frac{1}{[\mathbf{A}]_{0}-[\mathbf{B}]_{0}} \ln \frac{[\mathbf{A}][\mathbf{B}]_{0}}{[\mathbf{B}][\mathbf{A}]_{0}} \\]

The rate constant for the chemical reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) $$ doubles from \(22.50^{\circ} \mathrm{C}\) to \(27.47^{\circ} \mathrm{C}\). Determine the activation energy of the reaction. Assume the pre-exponential factor is independent of temperature.
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