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Chapter 14: Problem 89

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\), is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C}\), \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) .\) If the reac- tion is first order with a half-life of \(56.3 \mathrm{~min}\) at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).

Short Answer

Expert verified
(a) The half-life of the reaction at room temperature is 990 seconds. (b) The rate constant for the reaction at 415掳C is approximately \(2.05 \times 10^{-4} s^{-1}\).

Step by step solution

01

Identify the given rate constant

We are given the rate constant as \(k = 7.0 \times 10^{-4} s^{-1}\).
02

Use the half-life formula

We need to calculate the half-life using the formula: \(t_{1/2} = \frac{0.693}{k}\) Plug in the given value for \(k\): \(t_{1/2} = \frac{0.693}{7.0 \times 10^{-4} s^{-1}}\)
03

Calculate the half-life

Calculate the half-life: \(t_{1/2} = 990 s \) So the half-life of the reaction is 990 seconds. (b) Calculate the rate constant in \(s^{-1}\)
04

Identify the given half-life

We are given the half-life as \(t_{1/2} = 56.3 min\). First, we need to convert minutes to seconds: \(56.3 min \times 60 \frac{s}{min} = 3378 s\).
05

Rearrange the half-life formula

We need to calculate the rate constant from the given half-life. Rearrange the formula to solve for k: \(k = \frac{0.693}{t_{1/2}}\)
06

Plug in the given half-life

Plug in the value for the half-life \(t_{1/2} = 3378 s\): \(k = \frac{0.693}{3378 s}\)
07

Calculate the rate constant

Calculate the rate constant: \(k \approx 2.05 \times 10^{-4} s^{-1}\) So the rate constant for the reaction at 415掳C is approximately \(2.05 \times 10^{-4} s^{-1}\).

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Key Concepts

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

Rate Constant
Understanding the rate constant is crucial in the study of first order reactions. The rate constant, often represented by the symbol \(k\), is a proportionality constant in the rate law equation. It dictates how fast a reaction occurs under certain conditions. For first order reactions, the rate law is typically written as:
  • Rate = \(k imes [ ext{A}]\)
The concentration of the reactant \(\text{A}\) decreases over time, and the rate constant \(k\) determines the speed of this decrease.
In the given exercise, we use the rate constant to calculate the half-life of a reaction and vice versa. Remember, unlike 0-order reactions, the rate for a first order reaction depends linearly on reactant concentration, and \(k\) provides insights into the time-dependent behavior of a chemical system. This makes the rate constant a pivotal parameter in reaction kinetics.
Half-Life Calculation
The half-life of a reaction, denoted as \(t_{1/2}\), is the time it takes for half of the reactant to be consumed. For first order reactions, the half-life is independent of the initial concentration of the reactant. This unique property simplifies the calculation considerably. The formula used is:
  • \(t_{1/2} = \frac{0.693}{k}\)
This equation implies that for first order reactions, the half-life can be directly calculated if the rate constant \(k\) is known.
In part (a) of the exercise, the half-life was calculated using this relationship and the provided rate constant, illustrating how a simple formula can elucidate the decay rate of a reaction. Conversely, if the half-life is known, the rate constant can also be determined.
This interchange adds flexibility to analyzing reaction kinetics, allowing scientists to predict how much time is left for a reaction state to halve itself.
Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical processes occur. It provides valuable insights into how a reaction progresses from reactants to products.
Within the context of this exercise, understanding first order reaction kinetics involves analyzing the exponential decay of reactants over time. For first order reactions:
  • The rate of reaction only depends on the concentration of one reactant.
  • It follows the equation: Rate = \(-\frac{d[ ext{A}]}{dt} = k[ ext{A}]\)
Studying kinetics can help predict the speed and mechanism of a reaction. In practical applications, such as drug metabolism or industrial processes, knowing the rate of a reaction allows for optimized conditions and improved efficiency.
Reactions can be modeled using different mathematical techniques, but understanding the basic principles, such as determining the rate constant and half-life, is fundamental for analyzing the kinetics of any chemical reaction. This knowledge bridges the gap between theoretical chemistry and real-world applications.

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Most popular questions from this chapter

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is catalyzed by \(\mathrm{NO}_{2}\). Thereaction proceeds as follows: $$ \begin{aligned} &\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Show that the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (Hint: The top reaction must be multiplied by a factor so the \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) cancel out.) (b) Why do we consider \(\mathrm{NO}_{2}\) a catalyst and not an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

You study the effect of temperature on the rate of two reactions and graph the natural logarithm of the rate constant for each reaction as a function of \(1 / T\). How do the two graphs compare (a) if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor, and (b) if the frequency factor of the second reaction is higher than the frequency factor of the first reaction but the two reactions have the same activation energy? [Section \(14.5\)

(a) Explain the importance of enzymes in biological systems. (b) What chemical transformations are catalyzed (i) by the enzyme catalase, (ii) by nitrogenase?

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) Reaction 4: \(\mathrm{CH}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g)\) The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & E_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(\mathrm{~g})\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.
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