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Chapter 11: Problem 39

The decomposition of dimethyl ether \(\left(\mathrm{CH}_{3} \mathrm{OCH}_{3}\right)\) to methane, carbon monoxide, and hydrogen gases is found to be first-order. At \(500^{\circ} \mathrm{C}\), a \(150.0\) -mg 35 sample of dimethyl ether is reduced to \(43.2 \mathrm{mg}\) after three quarters of an hour. Calculate (a) the rate constant. (b) the half-life at \(500^{\circ} \mathrm{C}\). (c) how long it will take to decompose \(95 \%\) of the dimethyl ether.

Short Answer

Expert verified
Question: Determine the rate constant, half-life, and the time it takes to decompose 95% of dimethyl ether (CH3OCH3) at 500°C, given that the initial mass is 150 mg and the final mass after 0.75 hours is 43.2 mg. Answer: The rate constant (k) is 2.78 h⁻¹, the half-life is 0.249 hours, and the time to decompose 95% of dimethyl ether is 1.16 hours.

Step by step solution

01

Determine the initial and final concentrations

First, convert the given masses of dimethyl ether to moles using its molar mass (46.07 g/mol). Then, determine the initial and final concentrations. Initial mass = 150.0 mg = 150.0 g / 1000 = 0.150 g Final mass = 43.2 mg = 43.2 g / 1000 = 0.0432 g Molar mass of CH3OCH3 = 46.07 g/mol Initial moles = 0.150 g / 46.07 g/mol = 0.00325 mol Final moles = 0.0432 g / 46.07 g/mol = 0.000937 mol Now, assume a volume of 1 L for calculation purposes. Therefore, we have: Initial concentration, [A]₀ = 0.00325 mol/L Final concentration, [A] = 0.000937 mol/L
02

Calculate the rate constant (k)

Using the first-order rate law and the given reaction time (0.75 h), we can calculate the rate constant (k). \(ln\frac{[A]}{[A]_0} = -kt\) \(ln\frac{0.000937}{0.00325} = -k(0.75 h)\) Now, solve for k: \(k = -\frac{ln\frac{0.000937}{0.00325}}{0.75\;h} = 2.78\;h^{-1}\) The rate constant, k, is 2.78 h⁻¹.
03

Calculate the half-life

We can determine the half-life of the reaction using the following formula: \(t_{1/2} = \frac{ln 2}{k}\) \(t_{1/2} = \frac{ln 2}{2.78 \;h^{-1}} = 0.249 \;h\) The half-life at 500°C is 0.249 h.
04

Calculate the time to decompose 95% dimethyl ether

To find the time it takes for 95% of dimethyl ether to decompose, we need to determine the concentration when only 5% is left. 95% decomposed means 5% remaining: Remaining concentration, [A] = 0.05 x [A]₀ = 0.05 x 0.00325 mol/L = 0.000163 mol/L Applying the first-order rate law again: \(ln\frac{[A]}{[A]_0} = -kt\) \(ln\frac{0.000163}{0.00325} = -2.78\;h^{-1}t\) Finally, solve for t: \(t = -\frac{ln\frac{0.000163}{0.00325}}{2.78\;h^{-1}} = 1.16\;h\) It will take 1.16 hours to decompose 95% of the dimethyl ether.

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Most popular questions from this chapter

The first-order rate constant for the decomposition of a certain drug at \(25^{\circ} \mathrm{C}\) is \(0.215 \mathrm{month}^{-1}\) (a) If \(10.0 \mathrm{~g}\) of the drug is stored at \(25^{\circ} \mathrm{C}\) for one year, how many grams of the drug will remain at the end of the year? (b) What is the half-life of the drug? (c) How long will it take to decompose \(65 \%\) of the drug?

Copper-64 is one of the metals used to study brain activity. Its decay constant is \(0.0546 \mathrm{~h}^{-1}\). If a solution containing \(5.00 \mathrm{mg}\) of \(\mathrm{Cu}-64\) is used, how many milligrams of Cu-64 remain after eight hours?

If the activation energy of a reaction is \(4.86 \mathrm{~kJ}\), then what is the percent increase in the rate constant of a reaction when the temperature is increased from \(45^{\circ}\) to \(75^{\circ} \mathrm{C}\) ?

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The following gas-phase reaction is second-order. $$2 \mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g)$$ Its half-life is \(1.51\) min when \(\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]\) is \(0.250 \mathrm{M}\) (a) What is \(k\) for the reaction? (b) How long will it take to go from \(0.187 M\) to \(0.0915 \mathrm{M}\) ? (c) What is the rate of the reaction when \(\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]\) is \(0.335 \mathrm{M} ?\)
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