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Chapter 14: Problem 8

Which of the following linear plots do you expect for a reaction \(\mathrm{A} \longrightarrow\) products if the kinetics are (a) zero order, (b) first order, or (c) second order?

Short Answer

Expert verified
For a reaction A 鈫� products with different kinetics, we can expect the following linear plots: (a) Zero order: Plot [A] vs t (b) First order: Plot \(\ln[A]\) vs t (c) Second order: Plot \(\frac{1}{[A]}\) vs t

Step by step solution

01

(a) Analyzing zero order kinetics

For a zero order reaction, the rate law is given as: \(Rate = k\) Integrating the rate equation, we obtain the concentration of reactant A as a function of time: \([A] = [A]_0 - kt\) Where [A] is the concentration of reactant A at a given time, [A]_0 is the initial concentration, k is the rate constant, and t is the time. Since this equation is already in the form of a linear equation (y = mx + c), we can directly plot [A] vs t for a zero-order reaction.
02

(b) Analyzing first order kinetics

For a first order reaction, the rate law is given as: \(Rate = k[A]\) Dividing both sides by [A] and integrating, we have: \(\ln[A] = \ln[A]_0 - kt\) This equation also represents a linear relationship between the natural logarithm of the concentration of reactant A and time (y = mx + c). Hence, we can plot \(\ln[A]\) vs t for a first-order reaction.
03

(c) Analyzing second order kinetics

For a second order reaction, the rate law is given as: \(Rate = k[A]^2\) Dividing both sides by [A]虏 and integrating, we get: \(\frac{1}{[A]} = \frac{1}{[A]_0} + kt\) Similar to the previous cases, this equation represents a linear relationship between the reciprocal of the concentration of reactant A and time. Hence, we can plot \(\frac{1}{[A]}\) vs t for a second-order reaction. #Conclusion# To summarize, for the given reactions with different kinetics, we can expect the following linear plots: (a) Zero order: [A] vs t (b) First order: \(\ln[A]\) vs t (c) Second order: \(\frac{1}{[A]}\) vs t

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Key Concepts

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

Rate Laws
In chemical kinetics, rate laws define how the rate of a reaction depends on the concentration of reactants. They provide a mathematical relationship between the rate of a reaction and the concentration of reactants. Understanding rate laws helps us predict how changes in concentration affect reaction speed.

Rate laws have the general form:
  • Zero Order: The rate is constant and does not depend on the concentration of reactants.
  • First Order: The rate is directly proportional to the concentration of one reactant.
  • Second Order: The rate is proportional to the square of the concentration of one reactant or the product of two reactants.
Rate constants ( $k$ ) in these equations provide crucial information about the speed of the reaction and are determined experimentally.
Zero Order Reactions
Zero order reactions are unique because the rate of reaction is constant regardless of the concentration of reactants. This means that even as the reactant is consumed, the speed of the reaction remains unchanged.

For zero order reactions, the rate law is: \(Rate = k\)
If you plot the concentration of reactant \([A]\) against time \(t\), the curve will be a straight line with a negative slope, given by the formula:
\[ [A] = [A]_0 - kt \]
  • \([A]_0\) is the initial concentration.
  • \(k\) is the rate constant.
  • \(t\) is the time.
The linear aspect makes zero order reactions predictable and simple to analyze.
First Order Reactions
First order reactions have a rate that is directly proportional to the concentration of a single reactant. As the reactant concentration decreases, the reaction slows down at a rate proportional to the current concentration.

The rate law for first order reactions is:
\(Rate = k[A]\)
When integrated, it provides the following equation:
\[ \ln[A] = \ln[A]_0 - kt \]Here, plotting \(\ln[A]\) versus time \(t\) will give a straight line, with the slope equal to \(-k\). This implies a steady decrease in natural log concentration, showing exponential decay of reactant over time.

First order kinetics are common in reactions involving radioactive decay and unimolecular processes.
Second Order Reactions
Second order reactions involve rates that depend on the square of the concentration of one reactant or the product of two reactants' concentrations. These reactions are more complex due to this quadratic relationship.

The rate law is expressed as:\(Rate = k[A]^2\) or \(Rate = k[A][B]\)
Upon integrating, we derive:
\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \] This means plotting \(\frac{1}{[A]}\) against time \(t\) yields a straight line, indicating a direct relationship between inverse concentration and time.

In this case, the reaction rate changes more significantly with concentration changes. It is often seen in bimolecular reactions, where two molecules collide and react.

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Most popular questions from this chapter

(a) What is a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst? (c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both?

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) As the temperature increases, does the reaction rate increase or decrease? (c) As a reaction proceeds, does the instantaneous reaction rate increase or decrease?

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\) at \(\quad 70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .(\mathbf{a})\) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min} ?\) (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010 \mathrm{~mol}\) ? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C}\) ?

The isomerization of methyl isonitrile \(\left(\mathrm{CH}_{3} \mathrm{NC}\right)\) to acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) was studied in the gas phase at \(215^{\circ} \mathrm{C}\), and the following data were obtained: $$ \begin{array}{rc} \hline \text { Time (s) } & {\left[\mathrm{CH}_{3} \mathrm{NC}\right](M)} \\ \hline 0 & 0.0165 \\ 2000 & 0.0110 \\ 5000 & 0.00591 \\ 8000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (a) Calculate the average rate of reaction, in \(M / s\), for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from \(t=0\) to \(t=15,000 \mathrm{~s} .(\mathbf{c})\) Which is greater, the average rate between \(t=2000\) and \(t=12,000 \mathrm{~s}\), or between \(t=8000\) and \(t=15,000 \mathrm{~s} ?(\mathbf{d})\) Graph \(\left[\mathrm{CH}_{3} \mathrm{NC}\right]\) versus time and determine the instantaneous rates in \(M / \mathrm{s}\) at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).

(a) Develop an equation for the half-life of a zero-order reaction. (b) Does the half-life of a zero-order reaction increase, decrease, or remain the same as the reaction proceeds?
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