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 \(22 \mathrm{~min}\) ?

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: \(ln\frac{[\mathrm{C}_{2} \mathrm{H}_{6}]}{[\mathrm{C}_{2} \mathrm{H}_{6}]_0}=-kt\)

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)

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)

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)

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%: \(100\% - 31.4\%= 68.6\%\)