Question

In the first-order decomposition of acetone at \(500^{\circ} \mathrm{C}\) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}(g) \longrightarrow\) products It is found that it takes \(12.7 \mathrm{~h}\) to decompose \(0.0622 \mathrm{M}\) acetone to \(0.0133 \mathrm{M}\). (a) Find \(k\) at \(500^{\circ} \mathrm{C}\). (b) What is the rate of decomposition when \(\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]=0.0500 \mathrm{M} ?\) (c) How many minutes will it take to decompose enough acetone so that \(68 \%\) remains?

Short answer

Question: Calculate the rate constant (k) for the decomposition of acetone at \(500^{\circ} \mathrm{C}\) and the rate of decomposition when the concentration is \(0.0500 \mathrm{M}\). Also, find the time needed for the concentration to be reduced by \(32 \%\) (so that \(68 \%\) remains). Answer: The rate constant (k) at \(500^{\circ} \mathrm{C}\) is \(0.1788 \mathrm{~h}^{-1}\), the rate of decomposition when the concentration is \(0.0500 \mathrm{M}\) is \(0.00894 \mathrm{M~h}^{-1}\), and the time needed to decompose enough acetone, so that \(68\%\) remains, is approximately \(151.39\) minutes.

Step-by-step solution

Write the integrated rate law for first-order reactions.

In a first-order reaction, the integrated rate law has the following form: \(ln(\frac{[A]_0}{[A]}) = kt\), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time t (in hours), and k is the rate constant.

Calculate the rate constant, k, at \(500^{\circ} \mathrm{C}\) using the given data.

We are given that it takes \(12.7 \mathrm{~h}\) to decompose \(0.0622 \mathrm{M}\) acetone to \(0.0133 \mathrm{M}\). Plugging in these values and solve for k: \(ln(\frac{0.0622}{0.0133}) = k(12.7)\) \(k = \frac{ln(\frac{0.0622}{0.0133})}{12.7}\) \(k = 0.1788 \mathrm{~h}^{-1}\)

Find the rate of decomposition when \(\left[\mathrm{CH}_{3}\mathrm{COCH}_{3}\right]=0.0500 \mathrm{M}.\)

To find the rate of decomposition, we can use the rate expression for the first-order reaction, \(rate = k[A]\). Using the rate constant found in step 2 and the given concentration, we can calculate the rate: \(rate = 0.1788 \mathrm{~h}^{-1}\times 0.0500 \mathrm{M}\) \(rate = 0.00894 \mathrm{M~h}^{-1}\)

Calculate the time required to decompose enough acetone so that \(68\%\) remains.

To solve this, we can use the integrated rate law for first-order reactions, as in step 1. We need to find the time t when \([A] = 0.68 [A]_0\): \(ln(\frac{[A]_0}{0.68[A]_0}) = kt\) Since \([A]_0\) is in both the denominator and the numerator, it cancels out. Thus: \(ln(\frac{1}{0.68}) = 0.1788 \mathrm{~h}^{-1} \times t\) Now, to find the value of t: \(t = \frac{ln(\frac{1}{0.68})}{0.1788 \mathrm{~h}^{-1}}\) \(t = 2.5231 \mathrm{~h}\) To convert hours to minutes, we multiply by 60: \(t = 2.5231 \mathrm{~h} \times 60 = 151.39 \mathrm{~min}\) So, it would take roughly 151.39 minutes to decompose enough acetone so that \(68\%\) remains.