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Chapter 14: Problem 41

For the generic reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) that is zero order in \(\mathrm{A}\), what would you graph in order to obtain the rate constant?

Short Answer

Expert verified
In order to obtain the rate constant for a zero-order reaction, graph the concentration of reactant A as a function of time (\([A]\) vs \(t\)). The slope of the resulting straight line will represent the value of the rate constant (k).

Step by step solution

01

Understand the zero-order reaction rate law

A zero-order reaction is characterized by the rate law: \(Rate = k[A]^0\) Since any number raised to the power of 0 is 1, our rate law simplifies to: \(Rate = k\)
02

Express the rate in terms of reactant concentration

We can express the rate in terms of the change in reactant concentration with respect to time: \(\dfrac{-d[A]}{dt} = k\)
03

Integrating the rate expression

To find the relationship between reactant concentration and time, we need to integrate the rate expression with respect to time: \(\int_{[A]_0}^{[A]} -d[A] = \int_{0}^{t} k~dt\) Where:\([A]\) = concentration of A at time t, \([A]_0\) = initial concentration of A (at t = 0), and \(t\) = reaction time.
04

Evaluate the integrals and express concentration as a function of time

After integrating the rate expression, we get: \(-\big([A] - [A]_0\big) = kt\) Rearranging for the concentration of A, we get: \([A] = [A]_0 - kt\)
05

Determine what to graph to obtain the rate constant

As per the integrated rate law, we found that for a zero-order reaction: \([A] = [A]_0 - kt\) This has the form of a linear equation where the slope of the line is -k, and the y-intercept is \([A]_0\). Therefore, in order to obtain the rate constant (k), we should graph the concentration of the reactant A as a function of time (\([A]\) vs \(t\)). The slope of the resulting straight line will give us the value of the rate constant (k).

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Key Concepts

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

Rate Law and Zero-Order Reactions
The rate law is a mathematical expression that describes the relationship between the rate of a chemical reaction and the concentration of its reactants. In the case of a zero-order reaction, where the rate law is expressed as \(Rate = k[A]^0\), it may appear a bit counterintuitive at first glance. This is because, regardless of the concentration of reactant A, the rate remains constant. It鈥檚 a bit like setting the speed on cruise control when driving; no matter the terrain, your vehicle is set to maintain the same speed. The 'k' in the equation represents the rate constant, a unique value specific to the reaction at a given temperature.

For zero-order reactions, the rate constant can be found by monitoring the concentration of the reactant over time and then graphing this data. This concept underscores that some reactions do not depend on reactant concentrations for their speed, an important takeaway that distinguishes zero-order kinetics from other reaction orders.
Understanding Reaction Kinetics
Reaction kinetics refers to the study of rates of chemical processes and the factors that affect them. It gives us vital insight into how fast a reaction proceeds and what variables might influence this speed鈥攚hether it's temperature, catalysts, or the concentration of reactants. In the context of a zero-order reaction, the intriguing aspect is that the reactant concentration doesn't alter the reaction rate. Think of it like a factory assembly line that produces widgets at a consistent rate, no matter how many widgets are already in stock. To quantify the kinetics of such a reaction, we use the integrated rate law as shown in our step-by-step solution. This formula not only allows us to predict the concentration of reactants at any given time but also helps us to determine the lifespan of reactants in zero-order reactions.
Analyzing the Concentration-Time Graph
A concentration-time graph is one of the most effective tools for visualizing chemical reaction kinetics. By plotting reactant concentration against time, you get a clear picture of how the reaction progresses. For zero-order reactions, you'll notice a linear decrease in reactant concentration as time marches on, much like the straight path of a ball rolling down an incline with a constant slope. By examining the slope of this graph, which is essentially a straight line, we can effortlessly extract the rate constant 'k' for the reaction. This is because the slope of the line (-m, in y = mx + b format) is equal to the negative of the rate constant. Thus, in the context of a zero-order reaction, this allows us to dive directly into the heart of kinetic analysis, presenting a straightforward approach to understand the speed of our reaction.

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Most popular questions from this chapter

Consider the gas-phase reaction between nitric oxide and bromine at \(273^{\circ} \mathrm{C}: 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{NOBr}(g) .\) The following data for the initial rate of appearance of NOBr were obtained: (a) Determine the rate law. (b) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets. (c) How is the rate of appearance of NOBr related to the rate of disappearance of \(\mathrm{Br}_{2}\) ? (d) What is the rate of disappearance of \(\mathrm{Br}_{2}\) when \([\mathrm{NO}]=0.075 \mathrm{M}\) and \(\left[\mathrm{Br}_{2}\right]=0.25 \mathrm{M} ?\)

When chemists are performing kinetics experiments, the general rule of thumb is to allow the reaction to proceed for 4 half-lives. (a) Explain how you would be able to tell that the reaction has proceeded for 4 half-lives. (b) Let us suppose a reaction \(\mathrm{A} \rightarrow \mathrm{B}\) takes 6 days to proceed for 4 half-lives and is first order in A. However, when your lab partner performs this reaction for the first time, he does not realize how long it takes, and he stops taking kinetic data, monitoring the loss of A, after only 2 hours. Your lab partner concludes the reaction is zero order in A based on the data. Sketch a graph of [A] versus time to convince your lab partner the two of you need to be in the lab for a few days to obtain the proper rate law for the reaction.

The gas-phase reaction \(\mathrm{Cl}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{Br}(g)\) has an overall enthalpy change of \(-66 \mathrm{~kJ}\). The activation energy for the reaction is \(7 \mathrm{~kJ}\). (a) Sketch the energy profile for the reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

As described in Exercise \(14.43,\) the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 450 torr, what is the pressure of this substance after \(60 \mathrm{~s} ?\) (b) At what time will the pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C},\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) .\) If the reaction is first order with a half-life of 56.3 min at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).
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