Question

The decomposition of \(\mathrm{R}\) at \(33^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 128 minutes to decompose \(41.0 \%\) of an intial mass of \(739 \mathrm{mg}\) at \(33^{\circ} \mathrm{C}\). At \(33^{\circ} \mathrm{C}\) (a) what is \(k\) ? (b) what is the half-life of \(739 \mathrm{mg}\) ? (c) what is the rate of decomposition for \(739 \mathrm{mg}\) ? (d) what is the rate of decomposition if one starts with an initial amount of \(1.25 \mathrm{~g}\) ?

Short answer

Question: Identify the rate constant, half-life, and decomposition rate for a zero-order reaction of substance R, given the following information: 41% of the initial 739 mg is decomposed after 128 minutes at 33°C. Answer: (a) The rate constant, k, for the decomposition of substance R is 2.367 mg/min. (b) The half-life for 739 mg of substance R is 155.96 min. (c) The rate of decomposition for 739 mg of substance R is 2.367 mg/min. (d) The rate of decomposition when starting with an initial amount of 1.25 g is 2.367 mg/min.

Step-by-step solution

1. Calculate the initial concentration of R

First, we need to find what percentage of the initial mass wasn't decomposed: \(100 - 41.0 = 59.0\%\). Then, find the initial mass decomposed by multiplying the initial mass by the percentage decomposed: \(739\,\text{mg} \times 0.41 = 303.0\,\text{mg}\).

2. Calculate the rate constant, k

Now, use the relation \([A] = [A]_0 - kt\) to find k: \([A]_0 - [A] = kt \Rightarrow 303.0 = k \times 128 \Rightarrow k = \frac{303.0}{128} = 2.367\,\text{mg/min}\). (a) The rate constant, k, is \(2.367\,\text{mg/min}\).

3. Calculate the half-life

To find the half-life, we have to determine the time it takes for the initial concentration to be reduced by half. The relation at half-life becomes \([A]_0 - \frac{[A]_0}{2} = kt_{1/2}\), where \(t_{1/2}\) is the half-life. Solving for \(t_{1/2}\), \(t_{1/2} = \frac{[A]_0}{2k}\). Now, plug in the initial mass of \(739\,\text{mg}\) and the rate constant \(2.367\,\text{mg/min}\): \(t_{1/2} = \frac{739}{2 \times 2.367} = 155.96\,\text{min}\). (b) The half-life of \(739\,\text{mg}\) is \(155.96\,\text{min}\).

4. Calculate the rate of decomposition for 739 mg

Since it is a zero-order reaction, the rate constant k is equal to the rate of decomposition. Therefore, the decomposition rate is the same as k, which is \(2.367\,\text{mg/min}\). (c) The rate of decomposition for \(739\,\text{mg}\) is \(2.367\,\text{mg/min}\).

5. Calculate the rate of decomposition for 1.25 g

In a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. Hence, for an initial amount of \(1.25\,\text{g}\) (1250 mg), the rate of reaction will still be the same. (d) The rate of decomposition when starting with an initial amount of \(1.25\,\text{g}\) is \(2.367\,\text{mg/min}\).