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 $$ \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ 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 \(-4.00 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). 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?

Since we are given a straight line when plotting the concentration of reactant (ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)) versus time, we can infer that this a first-order reaction. Therefore, the rate law is as follows: $$ \text{rate} = k [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}] $$ The integrated rate law for a first-order reaction is: $$ \ln\left(\frac{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_t}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_0}\right) = -kt $$

The slope of the straight line in the given plot is equal to the negative of the rate constant. So, $$ k = -(-4.00 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}) = 4.00 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} $$

The half-life of a first-order reaction is given by the formula: $$ t_{1/2} = \frac{0.693}{k} $$ Using the rate constant obtained in step 2, we calculate the half-life: $$ t_{1/2} = \frac{0.693}{4.00 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}} = 1.73 \times 10^{4} \mathrm{s} $$

To find the time required for all the \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose, we can use the integrated rate law equation from step 1: $$ \ln\left(\frac{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_t}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_0}\right) = -kt $$ Since we want all of the ethanol to decompose, the final concentration \([\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_t = 0\) (or as close to zero as possible). Using the initial concentration \([\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}]_0 = 1.25 \times 10^{-2} \mathrm{M}\), we get: $$ \ln\left(\frac{0}{1.25 \times 10^{-2}}\right) = -\left(4.00 \times 10^{-5}\right) t $$ However, this equation will result in a mathematical error since we cannot take the natural logarithm of 0. Instead, we can say that the ethanol concentration will be very close to zero once it has completely decomposed. Let's assume that after complete decomposition, the concentration is close to \(10^{-10}\) M. So our equation becomes: $$ \ln\left(\frac{10^{-10}}{1.25 \times 10^{-2}}\right) = -\left(4.00 \times 10^{-5}\right) t $$ Solving for t: $$ t = \frac{\ln\left(\frac{10^{-10}}{1.25 \times 10^{-2}}\right)}{-\left(4.00 \times 10^{-5}\right)} \approx 3.69 × 10^6 \mathrm{s} $$ So, the time required for all the \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose is approximately \(3.69 × 10^6\) seconds.