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(a) Define the following symbols that are encountered in rate equations: \([\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k .(\mathrm{b})\) What quantity, when graphed versus time, will yield a straight line for a firstorder reaction?

Short Answer

Expert verified
The symbols in rate equations are defined as follows: 1. \([\mathrm{A}]_{0}\) represents the initial concentration of reactant A at time \(t=0\). 2. \(t_{1 / 2}\) represents the half-life of a reactant, the time for its concentration to decrease by half. 3. \([\mathrm{A}]_{t}\) represents the concentration of reactant A at a specific time \(t\). 4. \(k\) represents the rate constant of the reaction. For a first-order reaction, the quantity that produces a straight line when graphed versus time is the natural logarithm of the concentration of the reactant A, \(\ln [\mathrm{A}]_{t}\).

Step by step solution

01

Define \([\mathrm{A}]_{0}\)

\([\mathrm{A}]_{0}\) represents the initial concentration of the reactant A at time \(t=0\). It is the starting point from which we measure the changes in concentration as the reaction progresses.
02

Define \(t_{1 / 2}\)

\(t_{1 / 2}\) represents the half-life of a reactant, which is the time it takes for the concentration of the reactant to decrease by half. In other words, it's the time required for the concentration of the reactant to reach half of its initial value \([\mathrm{A}]_{0}\).
03

Define \([\mathrm{A}]_{t}\)

\([\mathrm{A}]_{t}\) represents the concentration of the reactant A at a specific time \(t\). It shows the changes in the concentration of the reactant A as the reaction progresses.
04

Define \(k\)

\(k\) represents the rate constant of the reaction. It is a proportionality constant that relates the reaction rate with the concentrations of the reactants raised to the power of their stoichiometric coefficients. The rate constant's values and units depend on the order of the reaction.
05

Determine the quantity that produces a straight line when graphed versus time for a first-order reaction

For a first-order reaction, the rate equation can be expressed as: \( -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}] \) Integrating from \(t = 0\) to \(t = t\) yields: \( \ln \frac{[\mathrm{A}]_{0}}{[\mathrm{A}]_{t}} = kt \) Rearranging the equation, we get: \( \ln [\mathrm{A}]_{t} = -kt + \ln [\mathrm{A}]_{0} \) In this equation, the quantity that yields a straight line when graphed against time is the natural logarithm of the concentration of the reactant A, \(\ln [\mathrm{A}]_{t}\). This is because the equation above has the linear form \(y = mx + b\), where \(y = \ln [\mathrm{A}]_{t}\), \(m = -k\), \(x = t\), and \(b = \ln [\mathrm{A}]_{0}\). The slope of the line will equal the negative rate constant, \(-k\).

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Key Concepts

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

Initial Reactant Concentration
Understanding the concept of initial reactant concentration is pivotal in the study of chemical kinetics. It refers to the amount of a substance present in the system at the start of the reaction, denoted by \( [\mathrm{A}]_{0} \). This value serves as a baseline for chemists to track the progress of the reaction as the reactant is consumed over time.

When a chemical reaction begins, the initial reactant concentration is often known or set by the experimenter. Throughout the reaction, this concentration diminishes as reactants are transformed into products. Measuring how this concentration changes gives valuable information about the reaction rate and the reaction order. It is particularly critical when dealing with first-order reactions, as the reaction rate is directly proportional to the concentration of the reactant at any given time.
Half-life
The term half-life, represented by \( t_{1/2} \), is commonly used in chemistry to describe the amount of time it takes for the concentration of a reactant to reduce to half its original value. In a first-order reaction, the half-life is constant and independent of the initial concentration.

This is highly useful for predicting how long a reaction will take to reach a certain point or for determining the relative stability of a reactant. Given a first-order reaction with a known rate constant, the half-life can be calculated using the formula: \( t_{1/2} = \frac{\ln 2}{k} \). This relationship shows that the half-life is inversely proportional to the rate constant — the larger the rate constant, the quicker the reactant concentration halves.
Rate Constant
The rate constant, noted as \( k \), is a crucial component in the calculation of reaction rates. In the equation \( -\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}] \), the rate constant connects the rate of reaction to the reactant's concentration. It is determined experimentally and varies depending on various factors such as temperature and the presence of a catalyst.

For a given reaction at a fixed temperature, the rate constant is a measure of the inherent 'speed' at which the reaction occurs. In first-order reactions, the rate constant is used in conjunction with the initial concentration to predict the concentrations of reactants at any future time point. The units of the rate constant depend on the order of the reaction, and for first-order reactions, it is typically expressed in reciprocal seconds \( s^{-1} \).
First-order Reaction
A first-order reaction is characterized by a reaction rate that is directly proportional to the concentration of a single reactant. These reactions have a signature rate law of the form \( \text{rate} = k[\mathrm{A}] \). The integrated rate law, which comes from manipulating the differential rate law, allows us to connect the concentration of reactants at any time point with the rate constant and the initial concentration.

In the context of graphing data to analyze reaction order, plotting the natural logarithm of the reactant concentration against time \( (\ln[\mathrm{A}]_{t} \text{ vs. } t) \) yields a straight line for a first-order reaction. This linear relationship is invaluable for chemists as it helps determine the rate constant and other kinetic parameters through the slope and intercept of the plot. Understanding and identifying first-order kinetics are essential in fields ranging from pharmaceuticals to environmental science.

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Most popular questions from this chapter

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbonic acid with carbon dioxide and water. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\), which is commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) : At \(23^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{cl} \hline \text { Time }(\mathrm{min}) & \left.\mathrm{IC}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M) \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?\) (b) What is the value of the rate constant?

(a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction? \(\mathrm{NO}+\mathrm{O} \longrightarrow \mathrm{NO}_{2}\) or \(\mathrm{H}+\mathrm{Cl} \longrightarrow \mathrm{HCl}\) ? (b) How does the kinetic-molecular theory help us understand the temperature dependence of chemical reactions?

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4} \quad\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M}\) \(\mathrm{HCl}\) occurs according to the reaction \(\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+\underset{2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)}\) The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the value for the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M} ?(\mathrm{c})\) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

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