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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 → 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]² 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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Most popular questions from this chapter

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?

(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?

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[\mathrm{~A}]^{2} .(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.

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 \(70{ }^{\circ} \mathrm{C}\) ?

(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?
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