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Chapter 36: Problem 8

a. For the hydrogen-bromine reaction presented in Problem P36.7 imagine initiating the reaction with only \(\mathrm{Br}_{2}\) and \(\mathrm{H}_{2}\) present. Demonstrate that the rate law expression at \(t=0\) reduces to \\[ \left(\frac{d[\mathrm{HBr}]}{d t}\right)_{t=0}=2 k_{2}\left(\frac{k_{1}}{k_{-1}}\right)^{1 / 2}\left[\mathrm{H}_{2}\right]_{0}\left[\mathrm{Br}_{2}\right]^{1 / 2} \\] b. The activation energies for the rate constants are as follows: $$\begin{array}{cc} \text { Rate Constant } & \Delta \boldsymbol{E}_{a}(\mathbf{k J} / \mathbf{m o l}) \\ \hline k_{1} & 192 \\ k_{2} & 0 \\ k_{-1} & 74 \end{array}$$ c. How much will the rate of the reaction change if the temperature is increased to \(400 .\) K from \(298 \mathrm{K} ?\)

Short Answer

Expert verified
We demonstrated that the rate law expression at t=0 for the hydrogen-bromine reaction reduces to: \( \left(\frac{d[\mathrm{HBr}]}{dt}\right)_{t=0} = 2k_{2}\left(\frac{k_{1}}{k_{-1}}\right)^{1/2}[\mathrm{H}_{2}]_{0}[\mathrm{Br}_{2}]^{1/2} \) Further, by calculating the rate constants at 298 K and 400 K using the given activation energies and plugging them into the rate change ratio equation: \( \frac{\left(\frac{d[\mathrm{HBr}]}{dt}\right)_{400 K}}{\left(\frac{d[\mathrm{HBr}]}{dt}\right)_{298 K}} = \frac{2k_{2}'\left(\frac{k_1'}{k_{-1}'}\right)^{1/2}[\mathrm{H}_{2}]_{0}[\mathrm{Br}_{2}]^{1/2}}{2k_{2}\left(\frac{k_1}{k_{-1}}\right)^{1/2}[\mathrm{H}_{2}]_{0}[\mathrm{Br}_{2}]^{1/2}} \) We can determine how much the rate of the reaction will change when the temperature is increased to 400 K from 298 K.

Step by step solution

01

Writing the Rate Law for the Reaction

Using the given steps, we can write the rate law as the sum of the individual rates: \( \left(\frac{d[\mathrm{HBr}]}{dt}\right)_{t=0} = k_{2}[H]^2[\mathrm{Br}_{2}] - k_{-1}[HBr]^2 \)
02

Solving for [H] and [HBr]

At the initial time t=0, we are given only \(H_2\) and \(Br_2\) are present. Therefore, the concentrations of \(H\) and \(HBr\) are zero: \( [H]_{t=0} = [HBr]_{t=0} = 0 \)
03

Applying Steady-State Approximation

The steady-state approximation states that the concentration of reactive intermediates (in this case, H and Br) remains essentially constant over time. Therefore, our rate equation becomes: \( k_{2}[\mathrm{H}]^2[\mathrm{Br}_{2}] = k_{-1}[HBr]^2 \) Since [H] and [HBr] are 0 at t=0, this equation simplifies to: \( \left(\frac{d[\mathrm{HBr}]}{dt}\right)_{t=0} = 2k_{2}\left(\frac{k_{1}}{k_{-1}}\right)^{1/2}[\mathrm{H}_{2}]_{0}[\mathrm{Br}_{2}]^{1/2} \) So, we have demonstrated that the rate law expression at t=0 reduces to the given equation. #b. Investigating the rate change with temperature#
04

Calculate Rate Constants at 298 K

We are given the activation energies for the rate constants. Using the Arrhenius equation, we can calculate the rate constants at 298 K (room temperature): \(k(T) = A e^{\frac{-\Delta E_a}{RT}}\) For T=298 K: \(k_1 = A_1 e^{\frac{-192 \times 10^3}{8.314 \times 298}}\) \(k_{-1} = A_{-1} e^{\frac{-74 \times 10^3}{8.314 \times 298}}\)
05

Calculate Rate Constants at 400 K

Next, we calculate the rate constants at 400 K: \(k_1' = A_1 e^{\frac{-192 \times 10^3}{8.314 \times 400}}\) \(k_{-1}' = A_{-1} e^{\frac{-74 \times 10^3}{8.314 \times 400}}\)
06

Calculate Rate Change Ratio

Now we need to find the rate change ratio between the two temperatures. This can be calculated by dividing the rate law expression at 400 K by the expression at 298 K: \[ \frac{\left(\frac{d[\mathrm{HBr}]}{dt}\right)_{400 K}}{\left(\frac{d[\mathrm{HBr}]}{dt}\right)_{298 K}} = \frac{2k_{2}'\left(\frac{k_1'}{k_{-1}'}\right)^{1/2}[\mathrm{H}_{2}]_{0}[\mathrm{Br}_{2}]^{1/2}}{2k_{2}\left(\frac{k_1}{k_{-1}}\right)^{1/2}[\mathrm{H}_{2}]_{0}[\mathrm{Br}_{2}]^{1/2}} \] By plugging in the rate constants at 298 K and 400 K, we can compute the rate change ratio for the reaction when the temperature is increased to 400 K from 298 K. Thus, we have completed the analysis to determine how much the rate of the reaction will change with the temperature increase.

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Key Concepts

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

Rate Law Expression
Understanding the rate law expression is crucial for predicting how the concentration of reactants or products changes over time during a chemical reaction. A rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For a given reaction, the rate law might look something like this:
\[ \text{Rate} = k[\text{A}]^{m}[\text{B}]^{n} \]
where \( k \) is the rate constant, \( [\text{A}] \) and \( [\text{B}] \) are the concentrations of reactants A and B, and \( m \) and \( n \) are the reaction orders with respect to A and B, respectively. The reaction order is determined experimentally and is not necessarily related to the stoichiometry of the reaction.
In the provided exercise, we see the initial rate law expression for the hydrogen-bromine reaction simplified under the specific condition where only \( \mathrm{H}_{2} \) and \( \mathrm{Br}_{2} \) are present at time \( t=0 \). By knowing the rate law expression, scientists can predict how fast a reaction occurs and how factors like temperature or concentration changes affect the reaction rate.
Steady-State Approximation
The steady-state approximation is a simplifying assumption commonly used in chemical kinetics for complex reactions involving intermediate species. These intermediates are often produced and consumed rapidly within the reaction mechanism, leading to their concentrations remaining relatively constant throughout the reaction. The core idea behind this approximation is that the rate of formation of an intermediate is equal to the rate of its consumption.
For example, in a reaction where an intermediate \( I \) is formed and consumed:

Formation: \( A \rightarrow I \)

Consumption: \( I \rightarrow B \)

Under the steady-state approximation, we assume that:\[ \frac{d[I]}{dt} \approx 0 \]
As seen in the exercise, we applied this concept to the hydrogen concentration \( [H] \), which is presumed to remain constant at the initial time \( t=0 \), thus allowing us to simplify the rate law expression. This approach is usually valid when the intermediates are short-lived and do not accumulate significantly. By using the steady-state approximation, we can streamline the complex kinetics of a multi-step reaction to more manageable mathematical expressions.
Arrhenius Equation
The Arrhenius equation provides a quantitative basis for understanding how the rate constant \( k \) changes with temperature. It is an essential tool in chemical kinetics, offering insight into the activation energy required for a reaction to proceed. Mathematically, the equation is written as:
\[ k(T) = A \exp\left(\frac{-\Delta E_a}{RT}\right) \]
where:
  • \( k(T) \) is the rate constant at temperature T,
  • A is the pre-exponential factor related to the frequency of collisions with correct orientation,
  • \( \Delta E_a \) is the activation energy,
  • R is the universal gas constant, and
  • T is the temperature in Kelvin.
In the context of our exercise, the Arrhenius equation is used to calculate how much the rate of the hydrogen-bromine reaction will change when the temperature is increased from 298 K to 400 K. We predict this change by comparing the rate constants calculated at both temperatures, taking into account the activation energies for the rate-determining steps. This temperature dependence outlined by the Arrhenius equation underscores why certain reactions speed up with increasing temperature 鈥� a higher temperature means that a larger proportion of molecules have the necessary energy to overcome the activation energy barrier and react.

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Most popular questions from this chapter

Protein tyrosine phosphatases (PTPases) are a general class of enzymes that are involved in a variety of disease processes including diabetes and obesity. In a study by Z.-Y. Zhang and coworkers [J. Medicinal Chemistry 43 \((2000): 146]\) computational techniques were used to identify potential competitive inhibitors of a specific PTPase known as PTP1B. The structure of one of the identified potential competitive inhibitors is shown here: The reaction rate was determined in the presence and absence of inhibitor \(I\) and revealed the following initial reaction rates as a function of substrate concentration: $$\begin{array}{ccc} & \mathbf{R}_{0}\left(\boldsymbol{\mu} \mathbf{M} \mathbf{~} \mathbf{s}^{-\mathbf{1}}\right) \\ {[\mathbf{S}](\boldsymbol{\mu} \mathbf{M})} & \mathbf{R}_{0}\left(\boldsymbol{\mu} \mathbf{M} \mathbf{~} \mathbf{s}^{-1}\right),[\boldsymbol{I}]=\mathbf{0} & {\left[\begin{array}{cc} \boldsymbol{I} & =\mathbf{2 0 0} \boldsymbol{\mu} \mathbf{M} \end{array}\right]} \\ \hline 0.299 & 0.071 & 0.018 \\ 0.500 & 0.100 & 0.030 \\ 0.820 & 0.143 & 0.042 \\ 1.22 & 0.250 & 0.070 \\ 1.75 & 0.286 & 0.105 \\ 2.85 & 0.333 & 0.159 \\ 5.00 & 0.400 & 0.200 \\ 5.88 & 0.500 & 0.250 \end{array}$$ a. Determine \(K_{m}\) and \(R_{\max }\) for PTP1B. b. Demonstrate that the inhibition is competitive, and determine \(K_{i}\)

If \(\tau_{f}=1 \times 10^{-10} \mathrm{s}\) and \(k_{i c}=5 \times 10^{8} \mathrm{s}^{-1},\) what is \(\phi_{f} ?\) Assume that the rate constants for intersystem crossing and quenching are sufficiently small that these processes can be neglected.

For the reaction \(\mathrm{I}^{-}(a q)+\mathrm{OCl}^{-}(a q) \rightleftharpoons\) \(\mathrm{OI}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) occurring in aqueous solution, the following mechanism has been proposed: \\[ \begin{array}{l} \mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad \frac{k_{1}}{\overrightarrow{k_{-1}}} \quad \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q) \\ \mathrm{I}(a q)+\mathrm{HOCl}(a q) \stackrel{k_{2}}{\longrightarrow} \mathrm{HOI}(a q)+\mathrm{Cl}^{-}(a q) \\ \mathrm{HOI}(a q)+\mathrm{OH}^{-}(a q) \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{OI}^{-}(a q) \end{array} \\] a. Derive the rate law expression for this reaction based on this mechanism. (Hint: \(\left[\mathrm{OH}^{-}\right]\) should appear in the rate law. b. The initial rate of reaction was studied as a function of concentration by Chia and Connick [J. Physical Chemistry \(63(1959): 1518]\), and the following data were obtained: $$\begin{array}{lccc} & & & \text { Initial Rate } \\ {\left[\mathbf{I}^{-}\right]_{0}(\mathbf{M})} & {\left[\mathbf{O C l}^{-}\right]_{0}(\mathbf{M})} & {\left[\mathbf{O H}^{-}\right]_{0}(\mathbf{M})} & \left(\mathbf{M} \mathrm{s}^{-1}\right) \\ \hline 2.0 \times 10^{-3} & 1.5 \times 10^{-3} & 1.00 & 1.8 \times 10^{-4} \\ 4.0 \times 10^{-3} & 1.5 \times 10^{-3} & 1.00 & 3.6 \times 10^{-4} \\ 2.0 \times 10^{-3} & 3.0 \times 10^{-3} & 2.00 & 1.8 \times 10^{-4} \\ 4.0 \times 10^{-3} & 3.0 \times 10^{-3} & 1.00 & 7.2 \times 10^{-4} \end{array}$$ Is the predicted rate law expression derived from the mechanism consistent with these data?

(Challenging) Cubic autocatalytic steps are important in a reaction mechanism referred to as the "brusselator" (named in honor of the research group in Brussels that initially discovered this mechanism): $$\begin{array}{l} \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{X} \\ 2 \mathrm{X}+\mathrm{Y} \stackrel{k_{2}}{\longrightarrow} 3 \mathrm{X} \\ \mathrm{B}+\mathrm{X} \stackrel{k_{3}}{\longrightarrow} \mathrm{Y}+\mathrm{C} \\\ \mathrm{X} \stackrel{k_{4}}{\longrightarrow} \mathrm{D} \end{array}$$ If \([\mathrm{A}]\) and \([\mathrm{B}]\) are held constant, this mechanism demonstrates interesting oscillatory behavior that we will explore in this problem. a. Identify the autocatalytic species in this mechanism. b. Write down the differential rate expressions for \([\mathrm{X}]\) and \([\mathrm{Y}]\) c. Using these differential rate expressions, employ Euler's method (Section 35.6 ) to calculate \([\mathrm{X}]\) and \([\mathrm{Y}]\) versus time under the conditions \(\mathrm{k}_{1}=1.2 \mathrm{s}^{-1}, \mathrm{k}_{2}=0.5 \mathrm{M}^{-2} \mathrm{s}^{-1}, \mathrm{k}_{3}=\) \(3.0 \mathrm{M}^{-1} \mathrm{s}^{-1}, \mathrm{k}_{4}=1.2 \mathrm{s}^{-1},\) and \([\mathrm{A}]_{0}=[\mathrm{B}]_{0}=1 \mathrm{M} .\) Begin with \([\mathrm{X}]_{0}=0.5 \mathrm{M}\) and \([\mathrm{Y}]_{0}=0.1 \mathrm{M} .\) A plot of \([\mathrm{Y}]\) versus \([\mathrm{X}]\) should look like the top panel in the following figure. d. Perform a second calculation identical to that in part (c), but with \([\mathrm{X}]_{0}=3.0 \mathrm{M}\) and \([\mathrm{Y}]_{0}=3.0 \mathrm{M} .\) A plot of \([\mathrm{Y}]\) versus \([\mathrm{X}]\) should look like the bottom panel in the following figure. e. Compare the left and bottom panels in the following figure. Notice that the starting conditions for the reaction are different (indicated by the black spot). What DO the figures indicate regarding the oscillatory state the system evolves to?

Consider the gas-phase isomerization of cyclopropane: Are the following data of the observed rate constant as a function of pressure consistent with the Lindemann mechanism? $$\begin{array}{cccc} \boldsymbol{P}(\text { Torr }) & \boldsymbol{k}\left(\mathbf{1 0}^{4} \mathbf{s}^{-1}\right) & \boldsymbol{P}(\text { Torr }) & \boldsymbol{k}\left(\mathbf{1 0}^{4} \mathbf{s}^{-1}\right) \\ \hline 84.1 & 2.98 & 1.36 & 1.30 \\ 34.0 & 2.82 & 0.569 & 0.857 \\ 11.0 & 2.23 & 0.170 & 0.486 \\ 6.07 & 2.00 & 0.120 & 0.392 \\ 2.89 & 1.54 & 0.067 & 0.303 \end{array}$$
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