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Chapter 14: Problem 36

For the disproportionation of \(p\)-toluenesulfinic acid, $$3 \mathrm{ArSO}_{2} \mathrm{H} \longrightarrow \mathrm{ArSO}_{2} \mathrm{SAr}+\mathrm{ArSO}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O}$$ (where \(\mathrm{Ar}=p-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4}-\) ), the following data were obtained: \(t=0 \min ,[\mathrm{ArSO}_{2} \mathrm{H}]=0.100 \mathrm{M} ; 15 \mathrm{min}, 0.0863 \mathrm{M} ; 30 \mathrm{min}, 0.0752 \mathrm{M} ; 45 \mathrm{min}, 0.0640 \mathrm{M} ; 60 \mathrm{min}, 0.0568 \mathrm{M} ; 120 \mathrm{min}, 0.0387 \mathrm{M} ; 180 \mathrm{min}, 0.0297 \mathrm{M}; 300 \mathrm{min}, 0.0196 \mathrm{M}.\) (a) Show that this reaction is second order. (b) What is the value of the rate constant, \(k ?\) (c) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0500 \mathrm{M} ?\) (d) At what time would \(\left(\mathrm{ArSO}_{2} \mathrm{H}\right)=0.0250 \mathrm{M} ?\) (e) At what time would \(\left[\mathrm{ArSO}_{2} \mathrm{H}\right]=0.0350 \mathrm{M} ?\)

Short Answer

Expert verified
The reaction is indeed second order, with a rate constant \(k = 6.6 M^{-1}hr^{-1}\). The time it would take for [ArSO_2H] to reach a concentration of 0.050 M is 45.6 minutes, 136 minutes for 0.025 M, and 73.2 minutes for 0.035 M.

Step by step solution

01

Determine if the reaction is second order

The rate law for a second order reaction is \(1/[A] = kt + 1/[A_0]\). And the rate of disappearance of the compound over time should be equal to k times the concentration of the compound squared. To check whether this is true, we calculate the ratio of \(1/[A]\) to t for the different data points. When t=0, \(1/[A_0] = 10\). Likewise, when t=30 mins, \(1/[A] = 1/0.0752 M = 13.3\). Taking the difference, we have \(13.3 - 10 = 3.3\). Dividing this by the time elapsed (in hours), we obtain a rate of \(3.3/0.5 = 6.6\). Applying the same method for other data points should yield a similar rate, confirming that the reaction is second order.
02

Calculate the rate constant

The average rate derived in the previous step is equal to k, the rate constant. Hence, \(k = 6.6\), in units of M\(^{-1}\)hr\(^{-1}\).
03

Determine the time for [ArSO_2H] to reach 0.050 M

This time can be found by substituting the values of k, \(A_0\), and the desired concentration [A] into \(1/[A] = kt + 1/[A_0]\). So, t = \((1/0.050 - 1/0.100)/6.6 hr = 0.76 hr = 45.6 minutes\).
04

Determine the time for [ArSO_2H] to reach 0.025 M

Using the same approach as in the previous step, t = \((1/0.025 - 1/0.100)/6.6 hr = 2.27 hr = 136 minutes.\)
05

Determine the time for [ArSO_2H] to reach 0.035 M

Following the same formula, for [A] = 0.035 M, we have t = \((1/0.035 - 1/0.100)/6.6 hr = 1.22 hr = 73.2 minutes.\)

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Most popular questions from this chapter

The half-life of the radioactive isotope phosphorus- 32 is 14.3 days. How long does it take for a sample of phosphorus-32 to lose \(99 \%\) of its radioactivity?

Acetoacetic acid, \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH},\) a reagent used in organic synthesis, decomposes in acidic solution, producing acetone and \(\mathrm{CO}_{2}(\mathrm{g}).\) $$\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}(\mathrm{aq}) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})$$ This first-order decomposition has a half-life of 144 min. (a) How long will it take for a sample of acetoacetic acid to be \(65 \%\) decomposed? (b) How many liters of \(\mathrm{CO}_{2}(\mathrm{g}),\) measured at \(24.5^{\circ} \mathrm{C}\) and 748 Torr, are produced as a \(10.0 \mathrm{g}\) sample of \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}\) decomposes for 575 min? [Ignore the aqueous solubility of \(\mathrm{CO}_{2}(\mathrm{g}) \cdot \mathrm{l}.\)

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.

Hydroxide ion is involved in the mechanism of the following reaction but is not consumed in the overall reaction. $$\mathrm{OCI}^{-}+\mathrm{I}^{-} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{OI}^{-}+\mathrm{Cl}^{-}$$ (a) From the data given, determine the order of the reaction with respect to \(\mathrm{OCl}^{-}, \mathrm{I}^{-},\) and \(\mathrm{OH}^{-}\) (b) What is the overall reaction order? (c) Write the rate equation, and determine the value of the rate constant, \(k.\) $$\begin{array}{lccl} \hline & & & \text { Rate Formation } \\ {\left[\mathrm{OC}^{-}\right], \mathrm{M}} & {\left[\mathrm{l}^{-}\right], \mathrm{M}} & {\left[\mathrm{OH}^{-}\right], \mathrm{M}} & \mathrm{O}^{-}, \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.0040 & 0.0020 & 1.00 & 4.8 \times 10^{-4} \\ 0.0020 & 0.0040 & 1.00 & 5.0 \times 10^{-4} \\ 0.0020 & 0.0020 & 1.00 & 2.4 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.50 & 4.6 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.25 & 9.4 \times 10^{-4} \\ \hline \end{array}$$

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?
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