Sulfuryl chloride undergoes first-order decomposition at \(320 .{ }^{\circ} \mathrm{C}\) with a half-life of \(8.75 \mathrm{~h}\). $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ What is the value of the rate constant, \(k\), in \(\mathrm{s}^{-1}\) ? If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is 791 torr and the decomposition occurs in a \(1.25-\mathrm{L}\) container, how many molecules of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) remain after \(12.5 \mathrm{~h}\) ?

To find the rate constant k for a first-order reaction, we can use the formula for half-life: $$ t_{1/2} = \frac{0.693}{k} $$ Where \(t_{1/2}\) is the half-life and k is the rate constant. We are given the half-life to be 8.75 hours. To find the value of k in \(\mathrm{s}^{-1}\), we must first convert the half-life to seconds: $$ t_{1/2} = 8.75 \, \mathrm{h} \times \frac{3600 \, \mathrm{s}}{1 \, \mathrm{h}} = 31,500 \, \mathrm{s} $$ Now, we can find k by rearranging the formula and plugging in the value of \(t_{1/2}\): $$ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{31,500 \, \mathrm{s}} = 2.2 \times 10^{-5} \, \mathrm{s^{-1}} $$

To find the number of remaining SO2Cl2 molecules after 12.5 hours, we can use the first-order reaction formula: $$ \ln{\frac{[A]_0}{[A]}} = kt $$ Where \([A]_0\) is the initial concentration of SO2Cl2, \([A]\) is the concentration at time t, k is the rate constant, and t is the time elapsed. We are given the initial pressure of SO2Cl2 as 791 torr, and thus we can find the initial concentration by using the Ideal Gas Law: $$ PV = nRT \Rightarrow n = \frac{PV}{RT} $$ Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Note that we must first convert the pressure to atm: $$ P = 791 \, \mathrm{torr} \times \frac{1 \, \mathrm{atm}}{760 \, \mathrm{torr}} = 1.04 \, \mathrm{atm} $$ We are given the volume as 1.25 L and the temperature as 320 °C, which must be converted to K: $$ T = 320 \, \mathrm{^{\circ} C} + 273.15 = 593.15 \, \mathrm{K} $$ Now, we can find the initial concentration using the Ideal Gas Law and converting it to molecules per L: $$ n = \frac{(1.04 \, \mathrm{atm})(1.25 \, \mathrm{L})}{(0.0821 \, \mathrm{L\cdot atm/mol\cdot K})(593.15 \, \mathrm{K})} $$ $$ n = 0.0219 \, \mathrm{mol} $$ $$ [A]_0 = \frac{0.0219 \, \mathrm{mol}}{1.25 \, \mathrm{L}} = 0.0175 \, \mathrm{M} $$ Next, we'll calculate the amount of time in seconds: $$ t = 12.5 \, \mathrm{h} \times \frac{3600 \, \mathrm{s}}{1 \, \mathrm{h}} = 45,000 \, \mathrm{s} $$ Now, plug in the given values to the first-order reaction formula: $$ \ln{\frac{0.0175}{[A]}} = (2.2 \times 10^{-5} \, \mathrm{s^{-1}})(45,000 \, \mathrm{s}) $$ Solve for [A]: $$ [A] = 0.00854 \, \mathrm{M} $$ Finally, multiply [A] by the given volume and Avogadro's constant to find the number of remaining SO2Cl2 molecules: $$ \mathrm{Remaining \, molecules} = (0.00854 \, \mathrm{M})(1.25 \, \mathrm{L})(6.022 \times 10^{23} \, \mathrm{molecules/mol}) $$ $$ \mathrm{Remaining \, molecules} = 6.41 \times 10^{21} \, \mathrm{molecules} $$ Therefore, after 12.5 hours, there will be approximately \(6.41 \times 10^{21}\) SO2Cl2 molecules remaining in the container.