Question

The decomposition of ethane, \(\mathrm{C}_{2} \mathrm{H}_{6}\), is a first-order reaction. It is found that it takes 212 s to decompose \(0.00839 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{6}\) to \(0.00768 \mathrm{M}\). (a) What is the rate constant for the reaction? (b) What is the rate of decomposition (in \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}\) ) when \(\left[\mathrm{C}_{2} \mathrm{H}_{6}\right]=\) \(0.00422 \mathrm{M} ?\) (c) How long (in minutes) will it take to decompose \(\mathrm{C}_{2} \mathrm{H}_{6}\) so that \(27 \%\) remains? (d) What percentage of \(\mathrm{C}_{2} \mathrm{H}_{6}\) is decomposed after \(22 \mathrm{~min}\) ?

Short answer

Answer: The rate of decomposition at a concentration of 0.00422 M is 0.0225 mol/L∙h.

Step-by-step solution

Determine the first-order rate law equation

For a first-order reaction, the rate law is: \(\frac{-d[\mathrm{C}_{2} \mathrm{H}_{6}]}{dt}=k[\mathrm{C}_{2} \mathrm{H}_{6}]\) Integrating this equation from the initial concentration \([\mathrm{C}_{2} \mathrm{H}_{6}]_0\) to the final concentration after time \(t\), we get the integrated rate law: \(ln\frac{[\mathrm{C}_{2} \mathrm{H}_{6}]}{[\mathrm{C}_{2} \mathrm{H}_{6}]_0}=-kt\)

Calculate the rate constant, k

Using the initial and final concentrations and integration time, we can solve for k: \(ln\frac{0.00768}{0.00839} = -k(212s)\) \(k = \frac{ln\frac{0.00768}{0.00839}}{-212s} = 1.48 \times 10^{-3} s^{-1}\) (b)

Calculate the rate of decomposition

Now we will use the reaction rate law to find the rate of decomposition at \(\left[\mathrm{C}_{2} \mathrm{H}_{6}\right] = 0.00422 \mathrm{M}\) \(rate = -\frac{d[\mathrm{C}_{2} \mathrm{H}_{6}]}{dt} = k[\mathrm{C}_{2} \mathrm{H}_{6}] = (1.48 \times 10^{-3} s^{-1})(0.00422 \mathrm{M}) = 6.25 \times 10^{-6} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) To convert to \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}\), simply multiply by the number of seconds in an hour: \(6.25 \times 10^{-6} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \cdot \frac{3600 \mathrm{s}}{1 \mathrm{hr}} = 0.0225 \mathrm{mol} / \mathrm{L} \cdot \mathrm{h}\) (c)

Determine the time needed for 27% to remain

We want to know how long it will take for 73% to decompose (100% - 27%). We can use the integrated rate law to find the correct time: \(ln\frac{0.27[\mathrm{C}_2 \mathrm{H}_6]_0}{[\mathrm{C}_2 \mathrm{H}_6]_0}=-kt\) \(ln(0.27)=-kt\) \(t = \frac{ln(0.27)}{-k} = \frac{ln(0.27)}{-1.48 \times 10^{-3} s^{-1}} = 790.7 \mathrm{s}\) Converting from seconds to minutes, we get: \(t = \frac{790.7 \mathrm{s}}{60 \mathrm{min/s}} = 13.2 \mathrm{min}\) (d)

Calculate the percentage decomposed after 22 minutes

To determine the percentage decomposed after 22 minutes, we will use the integrated rate law to find the concentration of \(\mathrm{C}_{2} \mathrm{H}_{6}\) at that time. First, convert 22 min to seconds: \(22 \mathrm{min} \cdot \frac{60 \mathrm{s}}{1 \mathrm{min}} = 1320 \mathrm{s}\) Now, use the integrated rate law to find the final concentration: \(\ln\frac{[\mathrm{C}_2 \mathrm{H}_6]}{[\mathrm{C}_2 \mathrm{H}_6]_0} = -k(1320s)\) \([\mathrm{C}_2 \mathrm{H}_6] = [\mathrm{C}_2 \mathrm{H}_6]_0 e^{-k(1320s)} = [\mathrm{C}_2 \mathrm{H}_6]_0 e^{-1.48 \times 10^{-3} s^{-1}(1320s)} = 0.314[\mathrm{C}_2 \mathrm{H}_6]_0\) To find the percentage decomposed, we would subtract the remaining percentage from 100%: \(100\% - 31.4\%= 68.6\%\)

Key concepts

Decomposition rate

The decomposition rate in a chemical reaction provides insight into how fast a reactant is consumed over a period of time. For first-order reactions like the decomposition of ethane (\(\mathrm{C}_{2}\mathrm{H}_{6}\)), the rate of decomposition is directly proportional to the concentration of the reactant at any given time. This means that as the concentration of ethane decreases, the rate at which it decomposes also decreases.

Here's where the calculation comes in. To find the rate at a specific concentration, you use the formula: \[\text{rate} = -\frac{d[\mathrm{C}_{2}\mathrm{H}_{6}]}{dt} = k[\mathrm{C}_{2}\mathrm{H}_{6}]\]Where:

• "rate" refers to the change in concentration over time (\(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\))
• \(k\) is the rate constant unique to each reaction at a specific temperature
• \([\mathrm{C}_{2}\mathrm{H}_{6}]\) is the concentration of ethane at that time

When you know these values, you can calculate the rate, helping you understand how swiftly or slowly a reaction is unfolding under specified conditions. This is crucial for predicting how much of a reactant is left after a certain time period.

Rate constant calculation

Calculating the rate constant \(k\) for a first-order reaction is a fundamental step in understanding the kinetics of the reaction. It provides a measure of the reaction's speed and is essential for determining how long you need to wait for a reaction to reach a desired extent.

Use the integrated rate law for a first-order reaction to calculate \(k\):\(\ln\frac{[\mathrm{C}_{2}\mathrm{H}_{6}]}{[\mathrm{C}_{2}\mathrm{H}_{6}]_0} = -kt\) To rearrange the formula and solve for \(k\), we get:\(k = \frac{\ln\left(\frac{[\mathrm{C}_{2}\mathrm{H}_{6}]}{[\mathrm{C}_{2}\mathrm{H}_{6}]_0}\right)}{-t}\)In this formula:

• \([\mathrm{C}_{2}\mathrm{H}_{6}]\) is the concentration at time \(t\)
• \([\mathrm{C}_{2}\mathrm{H}_{6}]_0\) is the initial concentration
• \(t\) is the time elapsed

In practice, you'd start by determining the concentrations at different time points, and use them to calculate \(k\). In our scenario, we found using the initial and final concentrations that \(k = 1.48 \times 10^{-3} \mathrm{s}^{-1}\). This constant helps predict concentrations at future times and assess how changing conditions might influence reaction speed.

Integrated rate law

The integrated rate law plays a crucial role in understanding the kinetics of first-order reactions. This equation allows us to relate the concentrations of reactants over time, providing a means to calculate how much reactant remains at any given moment, or how long it might take to reach a certain level of conversion.

For first-order reactions, the integrated rate law is expressed as:\(\ln\left(\frac{[\mathrm{C}_{2}\mathrm{H}_{6}]}{[\mathrm{C}_{2}\mathrm{H}_{6}]_0}\right) = -kt\)

• \([\mathrm{C}_{2}\mathrm{H}_{6}]\) is the concentration textat time \(t\)
• \([\mathrm{C}_{2}\mathrm{H}_{6}]_0\) is the initial concentration
• \(k\) is the rate constant
• \(t\) is time

The elegance of this formula lies in its linear relationship after taking the natural logarithm of the concentrations. This results in a straight line when plotting \(\ln[\mathrm{C}_{2}\mathrm{H}_{6}]\) against time, with a slope of \(-k\).

This transforms the problem of predicting concentrations into one of calculating straight line features—a much simpler and intuitive task. By using this linear relationship, you can quickly and accurately predict how much time it takes for a reaction to reach a desired percentage of completion or to determine what percentage remains after a given period, such as after 22 minutes in this problem.