/*! 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 64 The decomposition of phosphine, ... [FREE SOLUTION] | 91Ó°ÊÓ

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The decomposition of phosphine, \(\mathrm{PH}_{3},\) proceeds according to the equation $$\mathrm{PH}_{3}(\mathrm{g}) \rightarrow 1 / 4 \mathrm{P}_{4}(\mathrm{g})+3 / 2 \mathrm{H}_{2}(\mathrm{g})$$ It is found that the reaction has the following rate equation: Rate \(=k\left[\mathrm{PH}_{3}\right] .\) The half-life of \(\mathrm{PH}_{3}\) is 37.9 seconds at \(120^{\circ} \mathrm{C}.\) (a) How much time is required for three fourths of the \(\mathrm{PH}_{3}\) to decompose? (b) What fraction of the original sample of \(\mathrm{PH}_{3}\) remains after 1.00 minute?

Short Answer

Expert verified
(a) 75.8 seconds, (b) \(\frac{1}{3}\) remains after 1 minute.

Step by step solution

01

Understand the Reaction Order

The rate equation given is of the form \( \text{Rate} = k[\text{PH}_3] \), which indicates a first-order reaction since the rate depends linearly on the concentration of \( \text{PH}_3 \). For first-order reactions, the rate constant \( k \) and the half-life \( t_{1/2} \) are related as \( t_{1/2} = \frac{0.693}{k} \).
02

Use Half-Life to Find Rate Constant

Given the half-life \( t_{1/2} = 37.9 \) seconds, we calculate \( k \):\[ k = \frac{0.693}{37.9} \approx 0.0183 \text{ s}^{-1} \]
03

Calculate Time for Three Fourths to Decompose

The integrated rate law for a first-order reaction is\[ \ln \left( \frac{[\text{PH}_3]_0}{[\text{PH}_3]} \right) = kt \]If three fourths of \( \text{PH}_3 \) is decomposed, \( [\text{PH}_3] = \frac{1}{4}[\text{PH}_3]_0 \). Substitute into the equation:\[ \ln \left( \frac{[\text{PH}_3]_0}{\frac{1}{4}[\text{PH}_3]_0} \right) = (0.0183 \text{ s}^{-1}) t \]\[ \ln(4) = 0.0183t \]\[ t = \frac{\ln(4)}{0.0183} \approx 75.8 \text{ seconds} \]
04

Calculate Fraction Remaining after 1 Minute

To find the fraction of \( \text{PH}_3 \) remaining after 1 minute (60 seconds), we use:\[ \ln \left( \frac{[\text{PH}_3]_0}{[\text{PH}_3]} \right) = (0.0183 \text{ s}^{-1})\times 60 \]\[ \ln \left( \frac{[\text{PH}_3]_0}{[\text{PH}_3]} \right) = 1.098 \]\( e^{1.098} \approx \frac{[\text{PH}_3]_0}{[\text{PH}_3]} = 3 \ [\text{PH}_3] = \frac{[\text{PH}_3]_0}{3} \)Therefore, \( \frac{1}{3} \) of the original sample remains.

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Key Concepts

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

First-Order Reaction
In chemical kinetics, a first-order reaction is one where the rate of reaction is directly proportional to the concentration of a single reactant. This means that the decay or transformation of the reactant follows an exponential pattern. First-order reactions are common in chemical processes, including radioactive decay and enzyme-catalyzed reactions.
The general rate equation for a first-order reaction is expressed as:
  • Rate = k[A]
Here, 'Rate' signifies the speed at which the reaction progresses, 'k' is the reaction rate constant, and '[A]' is the concentration of the reactant 'A'.
First-order reactions are characterized by their simplicity as they do not involve interactions between multiple reactants, making their study and predictions easier than complex reactions.
Rate Equation
The rate equation is crucial in understanding how a chemical reaction progresses over time. It's an equation that links the rate of reaction to the concentrations of the reactants, each raised to some power, called the order of reaction. For a first-order reaction, the rate equation takes the form:
  • Rate = k[A]
In this equation, *k* is the rate constant and is unique for every distinct reaction. It represents the speed at which a reaction occurs, showing how fast the concentration of a reactant decreases. For first-order reactions, the rate is directly proportional to the concentration of the reactant. This means a decrease in reactant concentration leads to a directly related decrease in reaction rate.
The ability of the rate equation to predict how quickly reactants convert into products is fundamental in fields like pharmacology for drug degradation and environmental science for pollutant decay.
Half-Life Calculation
Half-life is a concept often associated with radioactive decay, but it's equally important in the study of chemical reactions, particularly first-order reactions. The half-life (\(t_{1/2}\)) of a reaction is the time required for half of the reactant to be consumed or transformed.
In first-order reactions, the half-life is constant and does not depend on the initial concentration of the reactant. It's calculated with the formula:
  • \(t_{1/2} = \frac{0.693}{k}\)
where *k* is the reaction rate constant. This constancy comes from the exponential nature of first-order decay, where the time to degrade a fixed percentage of reactants remains unchanged irrespective of their starting amounts.
Understanding half-life helps in predicting how long a substance will remain active or visible in a given environment, which is critical in areas like medicine absorption and nuclear waste management.
Reaction Rate Constant
The reaction rate constant, denoted as *k*, is a numerical value that helps us understand the speed of a chemical reaction. It is important for quantifying how quick or slow a reaction proceeds. In the context of a first-order reaction, the rate constant can be determined if you know the half-life of the reaction.
For a first-order reaction, the rate constant is obtained using the half-life formula:
  • \(k = \frac{0.693}{t_{1/2}}\)
This constant is specific to each reaction and influences the rate at which reactant concentrations decrease. The larger the value of *k*, the faster the reaction proceeds, indicating a shorter duration for the reactant to be converted into product.
Knowing *k* provides insight into reaction dynamics and is pivotal for designing and controlling chemical processes, like reactors used in industry or labs for synthesizing materials.

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Most popular questions from this chapter

The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After 27 minutes at \(27^{\circ} \mathrm{C},\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)

At \(573 \mathrm{K},\) gaseous \(\mathrm{NO}_{2}(\mathrm{g})\) decomposes, forming \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) If a vessel containing \(\mathrm{NO}_{2}(\mathrm{g})\) has an initial concentration of \(2.8 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) how long will it take for \(75 \%\) of the \(\mathrm{NO}_{2}(\mathrm{g})\) to decompose? The decomposition of \(\mathrm{NO}_{2}(\mathrm{g})\) is second order in the reactant and the rate constant for this reaction, at \(573 \mathrm{K},\) is \(1.1 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}.\)

The decomposition of nitrogen dioxide at a high temperature $$\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ is second order in this reactant. The rate constant for this reaction is \(3.40 \mathrm{L} / \mathrm{mol} \cdot\) min. Determine the time needed for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(2.00 \mathrm{mol} / \mathrm{L}\) to \(1.50 \mathrm{mol} / \mathrm{L}\)

We want to study the hydrolysis of the beautiful green, cobalt-based complex called trans-dichlorobis(ethylenediamine) cobalt(III) ion, (Check your book to see figure) In this hydrolysis reaction, the green complex ion trans\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) ). The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. Changes in color with time as \(\mathrm{Cl}^{-}\) ion is replaced by \(\mathrm{H}_{2} \mathrm{O}\) in a cobalt(III) complex. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker. Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: $$\begin{aligned}\operatorname{trans}-\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\end{aligned}$$ Fast: $$\begin{aligned}\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})\end{aligned}$$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k .\) Use the data below to find the activation energy. $$\begin{array}{cc}\text { Temperature }^{\circ} \mathrm{C} & \text { (for the Same Initial Concentration) } \\\\\hline 56 & 156 \mathrm{s} \\\60 & 114 \mathrm{s} \\\65 & 88 \mathrm{s} \\\75 & 47 \mathrm{s} \\\\\hline\end{array}$$

The gas-phase reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has an activation energy of \(103 \mathrm{kJ} / \mathrm{mol},\) and the rate constant is \(0.0900 \min ^{-1}\) at \(328.0 \mathrm{K}\). Find the rate constant at \(318.0 \mathrm{K}.\)

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