The decomposition of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) on an alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) surface$$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$was studied at 600 K. Concentration versus time data were collected for this reaction, and a plot of \([\mathrm{A}]\) versus time resulted in a straight line with a slope of \(-4.00 \times 10^{-5} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. If the initial concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) was \(1.25 \times 10^{-2}\) \(M,\) calculate the half-life for this reaction. c. How much time is required for all the \(1.25 \times 10^{-2} M\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose?

Step by step solution

01

Write an expression for the rate of reaction

Since the dependence of the concentration of ethanol on time is given as a straight line, this means that the rate law is of the form: Rate = k[\(C_{2}H_{5}OH]^{n}\), where k is the rate constant and n is the order of the reaction.

02

Determine the order of reaction

Since the plot of [A] versus time results in a straight line, the reaction is first order. So, n=1 and the rate law is: Rate = k[\(C_{2}H_{5}OH]\).

03

Write the integrated rate law

As the reaction is a first-order reaction, the integrated rate law is: \(\ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}] = \ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_{0} - kt\), where [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)] and [\(C_{2}H_{5}OH]_{0}\) are the concentrations of ethanol at time t and initial time respectively.

04

Calculate the rate constant

The slope of the plot of [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)] vs time is -k. Therefore, k = \(4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}\). b) Find the half-life of the reaction

05

Write the expression for half-life of a first-order reaction

The half-life (t₁/₂) of a first-order reaction is given by: t₁/₂ = \(\frac{0.693}{k}\).

06

Calculate the half-life

Using the rate constant obtained in part a, we can find the half-life. t₁/₂ = \(\frac{0.693}{4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}}\) = 17325 s. c) Find the time required for all the \(1.25 \times 10^{-2} \mathrm{M}\) ethanol to decompose

07

Use the integrated rate law

We will use the integrated rate law obtained in part a. \(\ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}] = \ln [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_{0} - kt\). Here, [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)] = 0 (as all ethanol has decomposed), and [\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)]₀ = \(1.25 \times 10^{-2}\) M.

08

Calculate the time required for complete decomposition

Now, we can solve for time t: \(\ln (0) = \ln (1.25 \times 10^{-2}\mathrm{M}) - 4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}} \cdot t\). Divide both sides by -4.00 \(\times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}\): t = \(\frac{\ln (1.25 \times 10^{-2}\mathrm{M})}{4.00 \times 10^{-5} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{s}}}\) = 69396 s.

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