Question

The decomposition of azomethane, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2},\) to nitrogen and ethane gases is a first-order reaction, $$ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{6}(g) $$ At a certain temperature, a 29 -mg sample of azomethane is reduced to \(12 \mathrm{mg}\) in \(1.4 \mathrm{~s}\). (a) What is the rate constant \(k\) for the decomposition at that temperature? (b) What is the half-life of the decomposition? (c) How long will it take to decompose \(78 \%\) of the azomethane?

Short answer

b) What is the half-life of the decomposition? c) How much time is required to decompose 78% of the azomethane? a) The rate constant (k) is approximately 0.638 s⁻¹. b) The half-life of the decomposition is approximately 1.087 seconds. c) It takes approximately 2.104 seconds to decompose 78% of the azomethane.

Step-by-step solution

a) Finding the rate constant k

Using the given data and the first-order reaction equation, we can solve for the rate constant k: $$ \ln \frac{[A]_{0}}{[A]}=kt $$ We know the initial amount of azomethane is 29 mg, and after 1.4 s, there are 12 mg remaining. Thus: $$ \ln \frac{29 \mathrm{mg}}{12 \mathrm{mg}} = k(1.4s) $$ Now, we can solve for k: $$ k=\frac{\ln \frac{29 \mathrm{mg}}{12 \mathrm{mg}}}{1.4s} $$ $$ k \approx 0.638 \,\mathrm{s}^{-1} $$ So, the rate constant k for the decomposition of azomethane at that temperature is approximately \(0.638 \,\mathrm{s}^{-1}\).

b) Finding the half-life of the decomposition

For first-order reactions, the half-life (\(t_{1/2}\)) can be determined using the following equation: $$ t_{1/2} = \frac{\ln(2)}{k} $$ We have calculated the value of k (\(0.638 \,\mathrm{s}^{-1}\)), so now we can determine the half-life: $$ t_{1/2} = \frac{\ln(2)}{0.638 \,\mathrm{s}^{-1}} $$ $$ t_{1/2} \approx 1.087\, \mathrm{s} $$ Therefore, the half-life of the decomposition is approximately \(1.087\, \mathrm{s}\).

c) Finding the time to decompose 78% of azomethane

To determine the time required to decompose 78% of azomethane, we can use the first-order reaction equation again: $$ \ln \frac{[A]_{0}}{[A]}=kt $$ This time, we know that \([A] = [A]_0 - 0.78[A]_0\). So, the equation becomes: $$ \ln \frac{[A]_{0}}{[A]_0 - 0.78[A]_0}=kt $$ Let's simplify the equation by letting \([A]_0 = 1\) (as a simplification, there is no actual change in the problem because the concentration will be the same proportion): $$ \ln \frac{1}{1 - 0.78} = kt $$ We already know the value of k (\(0.638\,\mathrm{s}^{-1}\)), so now we can solve for t: $$ t = \frac{\ln \frac{1}{1 - 0.78}}{0.638\,\mathrm{s}^{-1}} $$ $$ t \approx 2.104\,\mathrm{s} $$ Therefore, it will take approximately \(2.104\,\mathrm{s}\) to decompose 78% of the azomethane.