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Chapter 14: Problem 25

For the reaction \(A \longrightarrow\) products, the following data give \([\mathrm{A}]\) as a function of time \(t=0 \mathrm{s},[\mathrm{A}]=0.600 \mathrm{M};100 \mathrm{s}, 0.497 \mathrm{M} ; 200 \mathrm{s}, 0.413 \mathrm{M} ; 300 \mathrm{s}, 0.344 \mathrm{M} ; 400 \mathrm{s}\) \(0.285 \mathrm{M} ; 600 \mathrm{s}, 0.198 \mathrm{M} ; 1000 \mathrm{s}, 0.094 \mathrm{M}.\) (a) Show that the reaction is first order. (b) What is the value of the rate constant, \(k ?\) (c) What is \([\mathrm{A}]\) at \(t=750 \mathrm{s} ?\)

Short Answer

Expert verified
Detailed answer will be given after doing calculations. The reaction order is dependent upon the plot of the natural log of concentration versus time, the rate constant (\(k\)) is dependent upon the slope of that plot, and the concentration of A at a particular time is determined using the first order reaction equation with the derived rate constant, and initial concentration.

Step by step solution

01

Determining Reaction Order

Plot the natural logarithm of the concentration, ln([\(\mathrm{A}\])), against time. If a straight line is obtained, it's evidence that the reaction is first order. So for each given time and concentration, calculate ln([\(\mathrm{A}\)]). Then, plot these values and check if the plot is approximately a straight line.
02

Calculate Rate Constant \(k\)

Assuming that the reaction is first order, the slope of the straight line plot will equal to \(-k\). Since, based on the formula for the equation of a line \(y=mx+b\), \(m\) represents the slope of the line. Therefore, \(-k = m\). Calculate the slope of line and take its absolute value to find the value of \(k\).
03

Determining the concentration at a certain time

Use the calculated rate constant \(k\) and the first order reaction equation: \(ln[A]_t = -kt + ln[A]_0 \) to determine concentration of A at \(t=750s\). First substitute the values for \(k\) and the initial concentration \([A]_0\) into the equation. Then solve the equation to find \([A]_t\) at \(t=750s\). The solution should be converted back from a logarithm to a concentration by taking the exponent of the solution: \([A]_t = e^{(ln[A]_t)}\).

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Most popular questions from this chapter

The rate equation for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) is found to be rate \(=k[\mathrm{A}][\mathrm{B}] .\) For this reaction, we can conclude that (a) the unit of \(k=s^{-1} ;\) (b) \(t_{1 / 2}\) is constant; (c) the value of \(k\) is independent of the values of \([\mathrm{A}]\) and \([\mathrm{B}] ;\) (d) the rate of formation of \(\mathrm{C}\) is twice the rate of disappearance of A.

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ What is the approximate half-life of the first-order reaction?

The following first-order reaction occurs in \(\mathrm{CCl}_{4}(1)\) at \(45^{\circ} \mathrm{C}: \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) The rate constant is \(k=6.2 \times 10^{-4} \mathrm{s}^{-1} .\) An \(80.0 \mathrm{g}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}(\mathrm{l})\) is allowed to decompose at \(45^{\circ} \mathrm{C}.\) (a) How long does it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 \mathrm{g} ?\) (b) How many liters of \(\mathrm{O}_{2},\) measured at \(745 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C},\) are produced up to this point?

Derive a plausible mechanism for the following reaction in aqueous solution, \(\mathrm{Hg}_{2}^{2+}+\mathrm{Tl}^{3+} \longrightarrow 2 \mathrm{Hg}^{2+}+\mathrm{Tl}^{+}\) for which the observed rate law is: rate \(=k\left[\mathrm{Hg}_{2}^{2+1}\right]\) \(\left.[\mathrm{T}]^{3+}\right] /\left[\mathrm{Hg}^{2+}\right].\)

We have used the terms order of a reaction and molecularity of an elementary process (that is, unimolecular, bimolecular). What is the relationship, if any, between these two terms?
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