/*! 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 58 OBJECTIVE. Calculate the concent... [FREE SOLUTION] | 91Ó°ÊÓ

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OBJECTIVE. Calculate the concentration-time behavior for a second-order reaction from the rate law and the rate constant. A Consider the second-order decomposition of nitrosyl chloride. $$ 2 \mathrm{NOCl}(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ At \(450 \mathrm{~K},\) the rate constant is \(15.4 \mathrm{~atm}^{-1} \mathrm{~s}^{-1}\) (a) How much time is needed for \(\mathrm{NOCl}\) originally at a partial pressure of 44 torr to decay to 22 torr? (b) How much time is needed for \(\mathrm{NOCl}\) originally at a concentration of \(0.0044 M\) to decay to \(0.0022 M ?\)

Short Answer

Expert verified
Part (a): \( t \approx 1.44 \) s; Part (b): \( t \approx 44.2 \) s.

Step by step solution

01

Write the rate equation for second-order

The rate law for a second-order reaction is generally given by \( \text{-}\frac{d[A]}{dt} = k[A]^2 \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant. For this reaction, \( \mathrm{NOCl} \) is the reactant, so the equation becomes \( \text{-}\frac{d[P]}{dt} = k[P]^2 \), where \( P \) is the partial pressure of \( \mathrm{NOCl} \).
02

Integrate the rate equation

For a second-order reaction, the integrated rate law is \( \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \). In terms of partial pressure, this becomes \( \frac{1}{P_t} = \frac{1}{P_0} + kt \). \( P_0 \) is the initial partial pressure and \( P_t \) is the partial pressure at time \( t \).
03

Solve part (a) for time using pressure

Given \( P_0 = 44 \) torr, \( P_t = 22 \) torr, and \( k = 15.4 \) atm\(^{-1}\)s\(^{-1}\), convert pressures from torr to atm by dividing by 760. Thus, \( P_0 = \frac{44}{760} \) atm and \( P_t = \frac{22}{760} \) atm. Substitute these values in the integrated rate law: \( \frac{1}{\frac{22}{760}} = \frac{1}{\frac{44}{760}} + 15.4t \). Rearrange and solve for \( t \).
04

Calculate time for part (a)

First, calculate \( \frac{1}{P_0} = \frac{760}{44} \) and \( \frac{1}{P_t} = \frac{760}{22} \). Substitute into the integrated equation to solve: \[ \frac{760}{22} = \frac{760}{44} + 15.4t \]. Hence, solving for \( t \), we get \( t = \frac{760}{15.4} \times \left( \frac{1}{22} - \frac{1}{44} \right) \).
05

Solve part (b) for time using molar concentration

For part (b), use the same integrated rate equation \( \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \). Here, \( [A]_0 = 0.0044 \) M and \( [A]_t = 0.0022 \) M. Substitute these values into the integrated equation: \( \frac{1}{0.0022} = \frac{1}{0.0044} + 15.4t \). Solve for \( t \).
06

Calculate time for part (b)

Calculate \( \frac{1}{[A]_0} = \frac{1}{0.0044} \) and \( \frac{1}{[A]_t} = \frac{1}{0.0022} \). Then solve: \[ \frac{1}{0.0022} = \frac{1}{0.0044} + 15.4t \]. Hence, \( t = \frac{1}{15.4} \times \left( \frac{1}{0.0022} - \frac{1}{0.0044} \right) \).

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Key Concepts

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

Rate Law
In the world of chemistry, understanding how reactions occur and at what speed is crucial. The rate law helps us determine the relationship between the concentration of reactants and the reaction rate. For second-order reactions, the rate law is given by the formula: \( -\frac{d[A]}{dt} = k[A]^2 \). Here, \( k \) represents the rate constant, and \( [A] \) is the concentration of the reactant.
In practical terms, this equation tells us that the rate of a reaction depends on the square of the concentration of one reactant. This is typical of second-order reactions, where the rate decreases as the concentration decreases because the reactant molecules are less likely to collide.
  • This type of reaction is commonly seen in chemical processes involving the break down or decomposition of substances.
  • It implies that as the concentration of the reactant dwindles, the reaction rate slows proportionally to the concentration squared.
This understanding is essential in predicting how fast a reaction will proceed, which is important in industries and laboratories alike.
Integrated Rate Equation
For a second-order reaction, the integrated rate law provides a tool to predict how long a reaction will take based on initial and final concentrations. The integrated rate equation is: \( \frac{1}{[A]_t} = \frac{1}{[A]_0} + kt \), where \( [A]_0 \) is the initial concentration and \( [A]_t \) is the concentration at time \( t \).
This equation can be rearranged and solved to find the time \( t \) it takes for a concentration to change from \( [A]_0 \) to \( [A]_t \).
  • The integrated rate equation is derived from the rate equation by integrating over time.
  • It simplifies the process of calculating the duration required for a specific change in concentration.
Interestingly, when integrated, the rate law moves from describing an instantaneous rate of reaction to providing a way to calculate the time over which reactions occur. This makes it extremely useful in practical chemistry applications.
Reaction Kinetics
Reaction kinetics involves studying the rates of chemical reactions and understanding the factors that influence them. It's a broad field that addresses how temperature, concentration, and catalysts affect how fast reactions occur.
For second-order reactions, kinetics can be studied using both the rate law and the integrated rate equation.
  • By examining the rate constant \( k \), we can determine the inherent speed characteristics of a reaction.
  • For second-order kinetics, an increase in temperature typically increases \( k \), making reactions faster.
  • Additionally, reaction rates are affected by the physical state of reactants, and how they are mixed.
Overall, reaction kinetics provides a theoretical framework that helps chemists understand how reactions can be controlled and optimized in real-world scenarios like drug design and food preservation.
Partial Pressure
When dealing with gases, reactions are often discussed in terms of partial pressures rather than concentrations. Partial pressure is the pressure exerted by a single type of gas in a mixture of gases.
For the decomposition of gas reactants, such as \( \text{NOCl} \), the concept of partial pressure replaces concentration in calculations. In a reaction, using the pressure equivalent of the rate law, \( \text{-}\frac{d[P]}{dt} = k[P]^2 \), is essential.
  • Partial pressure allows chemists to calculate changes in the gaseous reactant's concentration over time using the integrated rate law adjusted for pressure.
  • It's important in predicting how long it will take for a gas to decompose, just as concentration is for solutions.
  • This concept is fundamental in industries that rely on gas-phase reactions, such as the production of ammonia or refinements of crude oil.
Ultimately, understanding partial pressure is crucial for accurately applying chemical principles in practical and industrial contexts, providing valuable insight into gaseous reactions.

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Most popular questions from this chapter

What is meant by the mechanism of a reaction? How does the mechanism relate to the order of the reaction?

Nitramide decomposes to water and dinitrogen monoxide. $$ \mathrm{H}_{2} \mathrm{NNO}_{2}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{N}_{2} \mathrm{O}(\mathrm{g}) $$ This reaction was studied by J. N. Brønsted in 1924 as part of research into the fundamental nature of acids and bases. If \(1.00 \mathrm{~L}\) of \(0.440 \mathrm{M}\) nitramide is placed in a reactor at \(20^{\circ} \mathrm{C}\), the following results are expected. (The experiment was actually performed to measure the effect of malate ion on the rate of reaction.) $$ \begin{array}{cc} \mathrm{T} \text { (min) } & P \text { (torr) } \\ \hline 0.00 & 17.54 \\ 0.50 & 18.98 \\ 1.00 & 20.09 \\ 2.00 & 21.65 \\ 4.00 & 23.46 \\ 6.00 & 24.48 \\ 8.00 & 25.14 \\ 10.00 & 25.59 \\ \text { Completion } & 27.54 \end{array} $$ What is the rate law for this reaction? (Hint: The data show the increase in concentration of a product, which differs from the other problems in this book. The problem looks more familiar if you create a decay curve. Note that the initial point represents a small amount of product, and thus a large amount of reactant. The final point represents a large amount of product and no reactant. The problem looks like any other kinetics problem if you first calculate ([final pressure \(-\) current pressure \(]\) to see the data as a decay curve). Graph (final pressure \(-\) current pressure), rather than the current pressure, as a function of time to see the data. If this is not a straight line, you can graph \(\ln (\) final pressure current pressure) and \(1 /\) (final pressure \(-\) current pressure) to determine the order of the reaction. Scientists frequently transform their data to a familiar form. These operations make it easier to interpret the data.

OBJECTIVE. Relate half-life and rate constant, and calculate concentration- time behavior from the half-life of a first-order reaction. A Calculate the half-life of a first-order reaction if the concentration of the reactant is \(0.0451 \mathrm{M}\) at 30.5 seconds after the reaction starts and is \(0.0321 \mathrm{M}\) at 45.0 seconds after the reaction starts. How many seconds after the start of the reaction does it take for the reactant concentration to decrease to \(0.0100 \mathrm{M} ?\)

Derive an expression for the half-life of a reaction with the following third- order rate law: $$ \frac{1}{[\mathrm{R}]_{\mathrm{t}}^{2}}-\frac{1}{[\mathrm{R}]_{0}^{2}}=2 k t $$

OBJECTIVE. Describe a chemical reaction as a sequence of elementary processes. Sum the following elementary steps to determine the overall stoichiometry of the reaction. $$ \begin{array}{l} \mathrm{Cl}_{2} \rightarrow 2 \mathrm{Cl} \\ \mathrm{Cl}+\mathrm{CO} \rightarrow \mathrm{COCl} \\ \mathrm{COCl}+\mathrm{Cl} \rightarrow \mathrm{COCl}_{2} \end{array} $$

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