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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

The addition of NO accelerates the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

In solution, chemical species as simple as \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) can serve as catalysts for reactions. Imagine you could measure the \(\left[\mathrm{H}^{+}\right]\) of a solution containing an acidcatalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the \(\left[\mathrm{H}^{+}\right]\) concentration profile you would measure as a function of time for the reaction, assuming \(t=0\) is when you add a drop of acid to the reaction.

From the following data for the second-order gas-phase decomposition of HI at \(430^{\circ} \mathrm{C}\), calculate the second-order rate constant and half- life for the reaction: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { [HI]/mol } \mathrm{dm}^{-3} \\ \hline 0 & 1 \\ 100 & 0.89 \\ \hline 200 & 0.8 \\ \hline 300 & 0.72 \\ \hline 400 & 0.66 \\ \hline \end{array} $$

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(392.4 \mathrm{pm} .(\mathbf{a})\) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .(\mathbf{b})\) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(280 \mathrm{pm}\) (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a 5.0-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right) .(\mathbf{a})\) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?
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