/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 When heated, cyclobutane, \(\mat... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 116

When heated, cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8},\) decomposes to ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}\) \(\mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) The activation energy, \(E_{a},\) for this reaction is \(262 \mathrm{~kJ} / \mathrm{mol}\). (a) If the rate constant \(k=0.032 \mathrm{~s}^{-1}\) at \(800 . \mathrm{K},\) calculate the value of \(k\) at \(900 . \mathrm{K}\). (b) Calculate the cyclobutane concentration after \(2 \mathrm{~h}\) at \(850 . \mathrm{K}\) if the initial cyclobutane concentration was \(0.0427 \mathrm{M}\)

Short Answer

Expert verified
(a) \(k_{900} \approx 10.98 \, s^{-1}\). (b) Cyclobutane will decompose completely after 2 hours at 850 K.

Step by step solution

01

Use Arrhenius Equation to Find k at 900 K

To find the rate constant \(k\) at 900 K, we use the Arrhenius equation: \[ k = A e^{-E_a / (RT)} \] However, since we have \(k\) at a different temperature already, we use the modified form:\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]Given \(k_1 = 0.032 \text{s}^{-1}\) at \(T_1 = 800 \, K\), \(E_a = 262 \text{kJ/mol} = 262000 \, J/mol\), and using \(R = 8.314 \, J/mol\cdot K\), we need to find \(k_2\) at \(T_2 = 900 \, K\). Plug the values into the equation to solve.
02

Solving for k at 900 K

First, calculate the terms:\[ \frac{1}{T_2} = \frac{1}{900} \, K^{-1}, \quad \frac{1}{T_1} = \frac{1}{800} \, K^{-1} \]Find the difference:\[ \frac{1}{900} - \frac{1}{800} = -\frac{1}{720000} \, K^{-1} \]Calculate the logarithm term:\[ \ln \left( \frac{k_2}{0.032} \right) = \frac{-262000}{8.314} \cdot \left( -\frac{1}{720000} \right) \]Solve for:\[ k_2 = 0.032 \times e^{12.33} \approx 10.98 \, s^{-1} \]
03

Calculate Cyclobutane Concentration After 2 Hours

We assume first-order kinetics, so the integrated rate equation is:\[ \ln \left( \frac{[C_{4}H_{8}]_t}{[C_{4}H_{8}]_0} \right) = -kt \]Given \([C_{4}H_{8}]_0 = 0.0427 \, M\), \(t = 2 \times 3600 \, s = 7200 \, s\) at \(T = 850 \, K\), first, use the Arrhenius equation to find \(k\) at 850 K:\[ \ln \left( \frac{k_{850}}{0.032} \right) = \frac{-262000}{8.314} \left( \frac{1}{850} - \frac{1}{800} \right) \]Solve to find \(k_{850} \approx 0.221 \, s^{-1}\). Substitute this into the rate equation to find \([C_{4}H_{8}]_t\).
04

Substitution and Calculation for Concentration

Calculate:\[ k_{850} \approx 0.221 \, s^{-1}\] \[ \ln \left( \frac{[C_{4}H_{8}]_t}{0.0427} \right) = -(0.221)(7200) \]\[ \frac{[C_{4}H_{8}]_t}{0.0427} \approx e^{-1591.2} \approx 0 \]Thus, \([C_{4}H_{8}]_t\) is practically 0 M after 2 hours.

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Key Concepts

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

Activation Energy
Activation energy is the minimum energy required for a chemical reaction to occur. Think of it as a barrier the reactants need to overcome to transform into products. In our cyclobutane decomposition, the activation energy is given as 262 kJ/mol. This means each mole of cyclobutane requires this much energy to break apart into ethylene.
  • High activation energy means the reaction is slow, as few molecules have enough energy to react.
  • Low activation energy indicates a faster reaction, as more molecules can overcome the barrier.
Activation energy influences how temperature affects reaction rates. When temperature increases, more molecules gain sufficient energy, leading to a higher rate of reaction.
Rate Constant
The rate constant (k) is a crucial factor in determining how fast a reaction proceeds. It varies with temperature and is specific to a given reaction at a determined temperature, as expressed by the Arrhenius equation: \[ k = A e^{-E_a / (RT)} \] Here, A is the frequency factor, a constant.The rate constant is a measure of the speed of a reaction. For example, if at 800 K, \( k = 0.032 \, \text{s}^{-1} \), it signifies the speed at which cyclobutane decomposes at this temperature. To find \( k \) at a different temperature, like 900 K, a modified Arrhenius equation helps: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{-E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] This relationship helps calculate \( k_2 \), demonstrating the temperature's effect on reaction speed.
First-Order Kinetics
First-order kinetics refers to reactions where the rate is directly proportional to the concentration of one reactant. In our case, cyclobutane decomposes into ethylene following first-order kinetics. This means the rate of decomposition depends linearly on the concentration of cyclobutane. The rate law can be expressed as: \[ \text{Rate} = k[C_{4}H_{8}] \] For first-order reactions, the concentration over time follows the integrated rate equation: \[ \ln \left( \frac{[C_{4}H_{8}]_t}{[C_{4}H_{8}]_0} \right) = -kt \] Using this equation, we calculate how much cyclobutane remains after a certain time at a specific temperature. With initial concentration \( [C_{4}H_{8}]_0 = 0.0427 \, \text{M} \) and \( k \) known, solving it helps find the concentration left after 2 hours, showing the decay pattern typical of first-order kinetics.

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Most popular questions from this chapter

Two mechanisms are proposed for the reaction $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ Mechanism $$ \text { 1: } \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3} $$ fast $$ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} $$ slow $$ \text { Mechanism } 2: \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} $$ fast Show that each mechanism is consistent with the observed rate law: Rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\).

When phenacyl bromide and pyridine are both dissolved in methanol, they react to form phenacylpyridinium bromide. \(\mathrm{C}_{6} \mathrm{H}_{5}-\stackrel{\mathrm{O}}{\|} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Br}+\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{C}-\mathrm{CH}_{2} \mathrm{NC}_{5} \mathrm{H}_{5}^{+}+\mathrm{Br}^{-}\) When equal concentrations of reactants were mixed in methanol at \(35^{\circ} \mathrm{C}\), these data were obtained: \begin{tabular}{rc|rr} \hline Time \((\min )\) & [Reactant] \((\mathrm{mol} / \mathrm{L})\) & Time \((\mathrm{min})\) & \([\mathrm{Reactant}]\) \((\mathrm{mol} / \mathrm{L})\) \\ \hline 0 & 0.0385 & \(500 .\) & 0.0208 \\ \(100 .\) & 0.0330 & \(600 .\) & 0.0191 \\ \(200 .\) & 0.0288 & \(700 .\) & 0.0176 \\ \(300 .\) & 0.0255 & \(800 .\) & 0.0163 \\ \(400 .\) & 0.0220 & \(1000 .\) & 0.0143 \\ \hline \end{tabular} (a) Determine the rate law for this reaction. (b) Determine the overall order of this reaction. (c) Determine the rate constant for this reaction. (d) Determine the rate constant for this reaction when the concentration of each reactant is \(0.030 \mathrm{~mol} / \mathrm{L}\)

Nitrosyl bromide, NOBr, is formed from \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\). $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g}) $$ Experiment shows that the reaction is first-order in \(\mathrm{Br}_{2}\) and second-order in NO. (a) Write the rate law for the reaction. (b) If the concentration of \(\mathrm{Br}_{2}\) is tripled, determine how the reaction rate changes. (c) Determine what happens to the reaction rate when the concentration of \(\mathrm{NO}\) is doubled.

For each reaction listed with its rate law, propose a reasonable mechanism. (a) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$ \begin{array}{l} \text { CH }_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH} \text { (aq) } \\ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{H}^{+}\right] \\ \text {(b) } \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] \end{array} $$ For each reaction listed with its rate law, propose a reasonable mechanism. (a) \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow\) $$ \begin{array}{l} \text { CH }_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq}) \\ \text { Rate }=k\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]\left[\mathrm{H}^{+}\right] \\ \text {(b) } \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] \end{array} $$

Assuming that each reaction is elementary, predict the rate law. (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\) (b) \(\mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\) (d) \(2 \mathrm{HI}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\)
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