Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right),\) an extremely poisonous gas. Phosgene decomposes by the reaction $$\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$for which $K_{\mathrm{p}}=6.8 \times 10^{-9}\( at \)100^{\circ} \mathrm{C} .$ If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

Let the change in pressure during the reaction be represented by \(x\). This means that the phosgene will decrease by \(x\) atm, and both CO and Cl\(_2\) will increase by \(x\) atm. Then, the equilibrium pressures of COCl\(_2\), CO, and Cl\(_2\) will be \((1.0-x)\) atm, \(x\) atm, and \(x\) atm respectively.

Now we can write the expression for \(K_{p}\): \[K_{p} = \frac{[\mathrm{CO}][\mathrm{Cl}_{2}]}{[\mathrm{COCl}_2]}\] Since we are working with pressures, the brackets represent the partial pressures of the species: \[K_{p} = \frac{(x)(x)}{(1 - x)}\] Given the value of \(K_{p} = 6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\), we can now substitute: \[\begin{equation} 6.8 \times 10^{-9} = \frac{x^{2}}{(1 - x)} \end{equation}\] This is a quadratic equation in \(x\). We can solve it by rearranging and simplifying: \[\begin{equation} x^2 + (6.8 \times 10^{-9})(x) - 6.8 \times 10^{-9} = 0 \end{equation}\] Now we can solve for \(x\) using the Quadratic Formula: \[\begin{equation} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{equation}\] where \(a = 1\), \(b = 6.8 \times 10^{-9}\), and \(c = -6.8 \times 10^{-9}\). Plugging these values, we only consider the positive root since the pressure cannot be negative: \[x \approx 2.61 \times 10^{-5}\]

Using the value of \(x\), we can now determine the equilibrium pressures for each species in the reaction: \[\mathrm{COCl}_{2}: \: 1.0 - 2.61 \times 10^{-5} \approx 0.999974 \, \mathrm{atm}\] \[\mathrm{CO}: \: 2.61 \times 10^{-5} \, \mathrm{atm}\] \[\mathrm{Cl}_{2}: \: 2.61 \times 10^{-5} \, \mathrm{atm}\] Hence, the equilibrium pressures for COCl\(_2\), CO, and Cl\(_2\) are \(0.999974 \, \mathrm{atm}\), \(2.61 \times 10^{-5} \, \mathrm{atm}\), and \(2.61 \times 10^{-5} \, \mathrm{atm}\), respectively.