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Chapter 14: Problem 81

The decomposition of ethylene oxide at \(690 \mathrm{K}\) is monitored by measuring the total gas pressure as a function of time. The data obtained are \(t=10 \mathrm{min}, P_{\text {tot }}= 139.14 \mathrm{mmHg} ; 20 \mathrm{min}, 151.67 \mathrm{mmHg} ; 40 \mathrm{min}, 172.65 \mathrm{mmHg} ; 60 \mathrm{min}, 189.15 \mathrm{mmHg} ;\) \(100 \mathrm{min}, 212.34\) \(\mathrm{mmHg} ; 200 \mathrm{min}, 238.66 \mathrm{mmHg} ; \infty, 249.88 \mathrm{mmHg}\) What is the order of the reaction \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) ?\)

Short Answer

Expert verified
The exact order of the reaction will be determined by checking for linearity in the created plots in step 5. The straight line plot will correspond to the correct order of the reaction. For an accurate determination, a graphical or statistical analysis tool would be needed.

Step by step solution

01

Understand the reaction

The reaction we are analyzing is the decomposition of ethylene oxide (\(C_2H_4O\)) into methane (\(CH_4\)) and carbon monoxide (\(CO\)). No stoichiometric ratios are given, so we assume the reaction happens in a 1:1:1 ratio. That is, one molecule of \(C_2H_4O\) decomposes into one molecule of \(CH_4\) and one molecule of \(CO\).
02

Find the total pressure after complete reaction

We are given that at \(t = \infty\) the total pressure \(P_{tot}\) is 249.88 mmHg, which is the total pressure when all the \(C_2H_4O\) has decomposed.
03

Calculate the initial pressure of ethylene oxide

Since the decomposition forms two gas molecules from one, every 1 mmHg decrease in the pressure of \(C_2H_4O\) results in a 2 mmHg increase in the total pressure. So, the initial pressure of \(C_2H_4O\), \(P_{C_2H_4O}\), can be calculated by \(P_{C_2H_4O} = 2 * P_{tot}\(0 min) - P_{tot}(\infty)\)
04

Calculate the pressure of \(C_2H_4O\)|| at each time interval

We can determine the pressure of \(C_2H_4O\) at each given time interval by rearranging the formula from step 3 to \(P_{C_2H_4O} = 2 * P_{tot}(t) - P_{tot}(\infty)\)
05

Determine the order of the reaction

By observing how the pressure of \(C_2H_4O\) changes with time, we can determine the order of the reaction. We assume the generic rate law for the decomposition is \(rate = k[P_{C_2H_4O}]^n\), where \(k\) is the rate constant, \(n\) is the order of the reaction. By trying different values for n and applying the rate law, rate versus time plots are made. The plot that gives a straight line will have the correct order. For a zero-order reaction, \(P_{C_2H_4O}\) versustime would give a straight line. For a first-order reaction, \(-ln[P_{C_2H_4O}]\) versus time would give a straight line. For a second-order reaction, \(1/[P_{C_2H_4O}]\) versus time would give a straight line.

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Most popular questions from this chapter

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, \(k.\) $$\begin{array}{lccl} \hline & & & \text { Rate Formation } \\ {\left[\mathrm{OC}^{-}\right], \mathrm{M}} & {\left[\mathrm{l}^{-}\right], \mathrm{M}} & {\left[\mathrm{OH}^{-}\right], \mathrm{M}} & \mathrm{O}^{-}, \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.0040 & 0.0020 & 1.00 & 4.8 \times 10^{-4} \\ 0.0020 & 0.0040 & 1.00 & 5.0 \times 10^{-4} \\ 0.0020 & 0.0020 & 1.00 & 2.4 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.50 & 4.6 \times 10^{-4} \\ 0.0020 & 0.0020 & 0.25 & 9.4 \times 10^{-4} \\ \hline \end{array}$$

Explain why (a) A reaction rate cannot be calculated from the collision frequency alone. (b) The rate of a chemical reaction may increase dramatically with temperature, whereas the collision frequency increases much more slowly. (c) The addition of a catalyst to a reaction mixture can have such a pronounced effect on the rate of a reaction, even if the temperature is held constant.

The following rates of reaction were obtained in three experiments with the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOCl}(\mathrm{g}).\) $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial Rate of } \\ \text { Expt } & \text { [NO], M } & \text { [Cl }_{2} \text { ], M } & \text { Reaction, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 1 & 0.0125 & 0.0255 & 2.27 \times 10^{-5} \\ 2 & 0.0125 & 0.0510 & 4.55 \times 10^{-5} \\ 3 & 0.0250 & 0.0255 & 9.08 \times 10^{-5} \\ \hline \end{array}$$ What is the rate law for this reaction?

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

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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