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Chapter 14: Problem 40

(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?

Short Answer

Expert verified
(a) The equation for the half-life of a zero-order reaction is \(t_{1/2} = \frac{[A]_0}{2k}\). (b) The half-life of a zero-order reaction decreases as the reaction proceeds.

Step by step solution

01

Write the rate law for a zero-order reaction

A zero-order reaction is one in which the rate of the reaction is independent of the concentration of the reactant. The rate law for a zero-order reaction can be written as: Rate = k where Rate is the rate of reaction, and k is the rate constant. Since Rate is also equal to the change in concentration (∆[A]) over change in time (∆t): k = -∆[A]/∆t Now, let's integrate this equation to find the relationship between the concentration of the reactant and time.
02

Integrate the rate law equation

Integrating the rate law equation, we get: \(\int_{[A]_0}^{[A]} -d[A] = \int_0^t kdt\) Solving this integral: \(-([A]-[A]_0) = kt\) Rearranging the equation for [A]: [A] = [A]_0 - kt Now we can use this equation to find the half-life of the reaction.
03

Calculate the half-life equation

The half-life (t1/2) is the time it takes for the concentration of the reactant to reach half of its initial value. In our equation for [A], let [A] = [A]_0/2, and the time as t = t1/2. Substituting these values and solving for t1/2: [A]_0/2 = [A]_0 - kt1/2 Now, let's isolate t1/2 to obtain the equation for half-life: kt1/2 = [A]_0/2 t1/2 = [A]_0/(2k) This is the equation for the half-life of a zero-order reaction.
04

End Task (a)

We have found the equation for the half-life of a zero-order reaction: t1/2 = [A]_0/(2k) Now, let's move on to part (b) of the exercise.
05

Determine the relationship between half-life and reaction progress

We want to find out if the half-life of a zero-order reaction increases, decreases, or remains the same as the reaction proceeds. For this, let's consider the half-life equation we derived in step 3: t1/2 = [A]_0/(2k) Notice that t1/2 depends on the initial concentration [A]_0. As the reaction proceeds, the concentration of the reactant decreases. Therefore, as [A]_0 decreases, t1/2 will also decrease. #End Task (b)# The half-life of a zero-order reaction decreases as the reaction proceeds.

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Most popular questions from this chapter

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 $t=8000 \mathrm{~s}$.

The reaction $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$ was performed and the following data were obtained under conditions of constant \(\left[\mathrm{Cl}_{2}\right]:\) (a) Is the following mechanism consistent with the data? $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) & \rightleftharpoons \mathrm{NOCl}_{2}(g) \quad(\text { fast }) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) & \longrightarrow 2 \mathrm{NOCl}(g) &(\text { slow }) \end{aligned} $$ (b) Does the linear plot guarantee that the overall rate law is second order?

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in $0.1 \mathrm{M} \mathrm{HCl}$ occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) $$ \text { is } 8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} $$ (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after $4.00 \times 10^{3} \mathrm{~s}\( if the starting concentration is \)0.500 \mathrm{M}$ ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

(a) The activation energy for the reaction $\mathrm{A}(g) \longrightarrow \mathrm{B}(g)\( is \)100 \mathrm{~kJ} / \mathrm{mol}$. Calculate the fraction of the molecule A that has an energy equal to or greater than the activation energy at \(400 \mathrm{~K} .(\mathbf{b})\) Calculate this fraction for a temperature of \(500 \mathrm{~K}\). What is the ratio of the fraction at $500 \mathrm{~K}\( to that at \)400 \mathrm{~K}$ ?

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