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Chapter 14: Problem 35

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

Short Answer

Expert verified
\(a) ([\mathrm{A}]_{0}\): Initial concentration of reactant A (M). \(t_{1/2}\): Half-life of reactant A. \([\mathrm{A}]_{t}\): Concentration of reactant A at time t (M). \(k\): Rate constant of the reaction. \(b\) For a first-order reaction, the quantity that will yield a straight line when graphed versus time is: \(\ln \dfrac{[\mathrm{A}]_{t}}{[\mathrm{A}]_{0}}\)

Step by step solution

01

Definition of symbols

In rate equations, the symbols represent the following: 1. \([\mathrm{A}]_{0}\): Initial concentration of reactant A at the beginning of the reaction (t = 0). This is usually measured in moles per liter (M). 2. \(t_{1/2}\): Half-life of the reactant A, which is the time required for the concentration of A to reduce by half its initial concentration. 3. \([\mathrm{A}]_{t}\): Concentration of reactant A at a specific time, t. Like the initial concentration, this is measured in moles per liter (M). 4. \(k\): Rate constant of the reaction, which depends on temperature and has different units based on the order of the reaction.
02

First-order reactions

For a first-order reaction, the rate law has the form: \[rate = k[\mathrm{A}]\] To find the quantity that, when graphed against time, will yield a straight line for a first-order reaction, we need to work with the integrated rate law. For a first-order reaction, the integrated rate law is: \(\ln [\mathrm{A}]_{t} = -kt + \ln [\mathrm{A}]_{0}\)
03

Graphing the first-order reaction

Rearranging the integrated rate law for a first-order reaction, we can obtain a relationship between the natural logarithm of concentration and time: \(\ln \dfrac{[\mathrm{A}]_{t}}{[\mathrm{A}]_{0}} = -kt\) Therefore, for a first-order reaction, the quantity that will yield a straight line when graphed versus time is: \(\ln \dfrac{[\mathrm{A}]_{t}}{[\mathrm{A}]_{0}}\) This is because when plotting \(\ln \dfrac{[\mathrm{A}]_{t}}{[\mathrm{A}]_{0}}\) against time (\(t\)), the slope will be equal to the negative rate constant \(-k\), and the line will be linear.

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Key Concepts

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

First-order Reaction
In chemistry, understanding the concept of a first-order reaction is essential. A first-order reaction is characterized by the fact that the rate at which the reaction proceeds is directly proportional to the concentration of one reactant. This means that if you double the concentration of the reactant, the reaction rate also doubles. This is expressed mathematically with the rate law:
  • rate = k[A]
Here, k is the rate constant, and [A] is the concentration of reactant A. Such reactions often involve a single reactant whose decomposition or transformation drives the overall reaction rate. Examples include radioactive decay and some enzyme-catalyzed reactions.
Rate Constant
The rate constant, often denoted by k, is a crucial parameter in reaction kinetics. It provides insight into the reaction's speed under specific conditions. The units of k depend on the reaction order, making it a unique identifier for reaction types. For a first-order reaction, the units are reciprocal seconds (s-1).
The rate constant is influenced by factors such as temperature, which often causes it to increase with rising temperatures due to faster molecular movements. Chemists use the Arrhenius equation to describe this temperature dependence, showing the exponential relationship between k and temperature. Understanding k assists in predicting how quickly a reaction will occur under various conditions.
Half-life
Half-life is an essential concept in understanding reaction progress. It refers to the time required for half of the initial concentration of a reactant to be converted into product. In a first-order reaction, the half-life (
  • t1/2
) remains constant and is independent of the initial concentration. This characteristic distinguishes first-order processes from other reaction orders. The formula for calculating the half-life of a first-order reaction is:
  • t1/2 = 0.693/k
This property is particularly useful in fields like pharmacokinetics and nuclear physics, where understanding how substances degrade over time is vital.
Integrated Rate Law
The integrated rate law provides a clear and mathematical way to express how the concentration of a reactant changes over time in a reaction. For first-order reactions, it is expressed as:
  • \(\ln [A]_t = -kt + \ln [A]_0\)
In this equation, \([A]_t\) is the concentration of reactant A at any time t, and \([A]_0\) is the initial concentration. The formula shows a linear relationship between the natural logarithm of concentration and time, with a slope of -k. This makes it possible to derive the rate constant from experimental data by plotting \(\ln [A]_t\) against time. Such a plot yields a straight line, with the slope representing the rate constant and the intercept providing the initial concentration details. This methodology aids significantly in analyzing reaction kinetics and understanding the underlying mechanisms involved.

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Most popular questions from this chapter

(a) Two reactions have identical values for \(E_{a}\). Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a higher rate constant than the other. Account for these observations.

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c)\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c)\), has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) \(\mathrm{A}\) solution of KOH in ethanol is made up by dissolving \(0.335\) g KOH in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Similarly, \(1.453 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C} ?(\mathrm{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) From your everyday experience, give two examples of the effects of temperature on the rates of reactions. (c) What is the difference between average rate and instantaneous rate?

Explain why rate laws generally cannot be written from balanced equations. Under what circumstance is the rate law related directly to the balanced equation for a reaction?
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