1. 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
$$
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})
$$
was studied at $600 \mathrm{~K}$. 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 $24.00 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}$.
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 $1.25 \times 10^{-2} \mathrm{M}$, calculate the half-life for this reaction.
c) How much time is required for all the $1.25 \times 10^{-2} \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ to decompose?

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Determine the order of the reaction**: Since a plot of $[A]$ versus time yields a straight line, the reaction is first-order. ‖ ‖ 2 ⋮ **Write the rate law**: For a first-order reaction, the rate law is $rate = k[A]$, where $k$ is the rate constant and $[A]$ is the concentration of the reactant. ‖ ‖ 3 ⋮ **Calculate the rate constant $k$**: The slope of the concentration versus time plot for a first-order reaction is equal to $-k$. Thus, $k = -slope = -24.00 \times 10^{-5} \text{ s}^{-1}$. ‖ ‖ 4 ⋮ **Write the integrated rate law**: For a first-order reaction, the integrated rate law is $\ln[A] = -kt + \ln[A]_0$, where $[A]_0$ is the initial concentration. ‖ ‖ 5 ⋮ **Calculate the half-life ($t_{1/2}$)**: The half-life for a first-order reaction is given by $t_{1/2} = \frac{0.693}{k}$. Substituting the value of $k$ gives $t_{1/2} = \frac{0.693}{24.00 \times 10^{-5} \text{ s}^{-1}}$. ‖ ‖ 6 ⋮ **Determine the time for complete decomposition**: For a first-order reaction, the time required for complete decomposition is theoretically infinite. However, we can calculate the time for the concentration to decrease to a negligible amount, such as $1\%$ of the initial concentration. ‖ ∻1 Answer∻ ⚹ Rate law: $rate = k[A]$ Integrated rate law: $\ln[A] = -kt + \ln[A]_0$ Rate constant: $k = 24.00 \times 10^{-5} \text{ s}^{-1}$ ⚹ ∻2 Answer∻ ⚹ Half-life: $t_{1/2} = \frac{0.693}{24.00 \times 10^{-5} \text{ s}^{-1}}$ ⚹ ∻3 Answer∻ ⚹ Time for complete decomposition: Theoretically infinite, but can calculate time to reach a negligible concentration. ⚹ ∻Key Concept∻ ⚹ The slope of the concentration versus time plot for a first-order reaction is equal to the negative of the rate constant. ⚹ ∻Explanation∻ ⚹ Since the plot of $[A]$ versus time is linear, the reaction follows first-order kinetics, and the slope gives the rate constant. ⚹ ∻Key Concept∻ ⚹ The half-life of a first-order reaction is independent of the initial concentration and is calculated using the rate constant. ⚹ ∻Explanation∻ ⚹ The half-life for a first-order reaction is given by $t_{1/2} = \frac{0.693}{k}$, which only depends on the rate constant. ⚹ ∻Key Concept∻ ⚹ Complete decomposition in a first-order reaction takes an infinite amount of time, but practical completion can be calculated. ⚹ ∻Explanation∻ ⚹ Although complete decomposition is never reached, we can determine the time required for the concentration to fall to a very low level, such as $1\%$ of the initial◊What is the rate law for the decomposition of ethanol on an alumina surface at 600 K⍭ Generate me a similar question◊⚹

Sia