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 \(\mathrm{K}\) . Concentration versus time data were collected for this reaction, and a plot of [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 \(1.25 \times 10^{-2} \mathrm{M}\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose?

The given reaction: \[C_{2}H_{5}OH(g) \longrightarrow C_{2}H_{4}(g)+H_{2}O(g)\] The concentration vs. time graph is a straight line with a slope, which means the reaction is a first-order reaction. Thus, the rate law is: Rate Law: \[r = k[C_{2}H_{5}OH]\] Now, let's find the integrated rate law. For a first-order reaction, the integrated rate law is: Integrated Rate Law: \[\ln{\dfrac{[C_{2}H_{5}OH]_t}{[C_{2}H_{5}OH]_0}} = -kt\]

We are given that the slope of the concentration vs. time graph is \(-4.00 \times 10^{-5} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) . Since the reaction is first-order, the rate constant (k) is equal to the negative slope. Rate Constant: \[k = -(-4.00 \times 10^{-5} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) = 4.00 \times 10^{-5} \mathrm{L /mol \cdot s}\]

To calculate the half-life of a first-order reaction, we use the formula: Half-Life: \[t_{1/2} = \dfrac{\ln 2}{k}\] Using the rate constant obtained previously, we have: \[t_{1/2} = \dfrac{\ln 2}{4.00 \times 10^{-5} \mathrm{L/mol\cdot s}}\] Calculating the half-life: \[t_{1/2} = 1.73 \times 10^{4}s\]

Given the initial concentration of ethanol \([C_{2}H_{5}OH]_0 = 1.25 \times 10^{-2} M\) and the integrated rate law earlier, we want to find the time when the concentration \([C_{2}H_{5}OH]_t = 0\) . Plugging in the values into the integrated rate law, we have: \[\ln{\dfrac{0}{1.25 \times 10^{-2} M}} = -kt\] However, the above equation has a zero in the numerator, and the natural logarithm of zero is undefined. In this context, complete decomposition implies that the concentration of ethanol is close enough to zero that it is practically negligible. So we can compute the time required for, say, 99.99% of \(C_{2}H_{5}OH\) to decompose: \[\ln{\dfrac{(1 - 0.9999) \times 1.25 \times 10^{-2} M}{1.25 \times 10^{-2} M}} = -kt\] \[t = \dfrac{\ln{0.0001}}{-4.00 \times 10^{-5} \mathrm{L/mol\cdot s}}\] Calculating the time: \[t = 9.21 \times 10^4 s\] Thus, the time required for the concentration of ethanol to reach a negligible level is \(9.21 \times 10^4 s\).