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Chapter 14: Problem 26

The decomposition of dimethyl ether at \(504^{\circ} \mathrm{C}\) is $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$$ The following data are partial pressures of dimethyl ether (DME) as a function of time: \(t=0\) s, \(P_{\text {DME }}=\) \(312 \mathrm{mmHg} ; 390 \mathrm{s}, 264 \mathrm{mmHg} ; 777 \mathrm{s}, 224 \mathrm{mmHg} ; 1195 \mathrm{s},187 \mathrm{mmHg} ; 3155 \mathrm{s}, 78.5 \mathrm{mmHg}.\) (a) Show that the reaction is first order. (b) What is the value of the rate constant, \(k ?\) (c) What is the total gas pressure at 390 s? (d) What is the total gas pressure when the reaction has gone to completion? (e) What is the total gas pressure at \(t=1000\) s?

Short Answer

Expert verified
The reaction follows first order kinetics with a certain value of rate constant \(k\). The total gas pressure at 390 s, at the completion of the reaction and at 1000 s can be calculated using the relevant equations and stoichiometry.

Step by step solution

01

Verify the order of the reaction

According to the reaction, partial pressure of dimethyl ether (DME) decreases with time, and it can be observed that the reaction is following first order kinetics. This can be verified by plotting the natural logarithm of partial pressure of DME against time. If a straight line is obtained, it confirms that the reaction is first order.
02

Calculate the rate constant, k

In a first order reaction, the rate constant (\(k\)) can be calculated by the equation: \[ k = - \frac{2.303}{t} \times \log \frac{P_{\text {final}}}{P_{\text {initial}}} \] Applying the given data points, calculate the average of k using the above formula and thus derive the value of the rate constant.
03

Calculate total gas pressure at \(t=390\) s

At \(t=390\) s, partial pressure of DME is given as 264 mmHg. For the total pressure at \(t=390\) s, apply the stoichiometry of the reaction and the decrease in pressure of DME to calculate the individual pressures of the resultant gases and then add them together including the remaining DME pressure.
04

Calculate the total gas pressure after the reaction has gone to completion

When the reaction has gone to completion, the total gas pressure will be the sum of the pressures of the final product gases. This can be found by applying the ideal gas law and stoichiometric calculations.
05

Calculate the total gas pressure at \(t=1000\) s

At \(t=1000\) s, find the pressure of DME using the earlier derived rate constant (\(k\)) and stoichiometry of the reaction. This is done by applying the equation for the pressure of DME at time \(t = k.P_{\text {DME}}\), and then calculate the pressures of the resultant gases and add them together.

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Most popular questions from this chapter

The reaction \(A+B \longrightarrow C+D\) is second order in \(A\) and zero order in B. The value of \(k\) is \(0.0103 \mathrm{M}^{-1} \mathrm{min}^{-1}.\) What is the rate of this reaction when \([\mathrm{A}]=0.116 \mathrm{M}\) and \([\mathrm{B}]=3.83 \mathrm{M} ?\)

For the reaction \(A+B \longrightarrow C+D\), the following initial rates of reaction were found. What is the rate law for this reaction? $$\begin{array}{llll} \hline & & & \text { Initial Rate, } \\ \text { Expt } & \text { [A], M } & \text { [B], M } & \text { M min }^{-1} \\\ \hline 1 & 0.50 & 1.50 & 4.2 \times 10^{-3} \\ 2 & 1.50 & 1.50 & 1.3 \times 10^{-2} \\ 3 & 3.00 & 3.00 & 5.2 \times 10^{-2} \\ \hline \end{array}$$

The decomposition of nitric oxide occurs through two parallel reactions: $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad k_{1}=25.7 \mathrm{s}^{-1}$$ $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{4} \mathrm{O}_{2}(\mathrm{g}) \quad k_{2}=18.2 \mathrm{s}^{-1}$$ (a) What is the reaction order for these reactions? (b) Which reaction is the slow reaction? (c) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(2.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2}(\mathrm{g})\) after 0.1 seconds? (d) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(4.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) after 0.025 seconds?

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

For the reaction \(A \longrightarrow\) products, the data tabulated below are obtained. (a) Determine the initial rate of reaction (that is, \(-\Delta[\mathrm{A}] / \Delta t)\) in each of the two experiments. (b) Determine the order of the reaction. $$\begin{array}{ll} \hline \text { First Experiment } & \\ \hline[\mathrm{A}]=1.512 \mathrm{M} & t=0 \mathrm{min} \\ \begin{array}{l} | \mathrm{A}\rfloor=1.490 \mathrm{M} \\ {[\mathrm{A}]=1.469 \mathrm{M}} \end{array} & \begin{array}{l} t=1.0 \mathrm{min} \\ t=2.0 \mathrm{min} \end{array} \\ \hline & \\ \hline \text { Second Experiment } & \\ \hline[\mathrm{A}]=3.024 \mathrm{M} & t=0 \mathrm{min} \\ {[\mathrm{A}]=2.935 \mathrm{M}} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=2.852 \mathrm{M}} & t=2.0 \mathrm{min} \\ \hline \end{array}$$
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