For the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ at \(600 . \mathrm{K}\) , the equilibrium constant, \(K_{\mathrm{p}},\) is \(11.5 .\) Suppose that 2.450 \(\mathrm{g} \mathrm{PCl}_{5}\) is placed in an evacuated 500 -mL bulb, which is then heated to \(600 . \mathrm{K}\) . a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of PCl_ at equilibrium?

In order to find the pressure of \(\mathrm{PCl_5}\) without dissociation, we first need to calculate the number of moles of \(\mathrm{PCl_5}\) and then use the ideal gas law, \(PV=nRT\), to find the pressure. Given the mass of \(\mathrm{PCl_5}\) as \(2.450\, g\), we can determine the moles by: \(n=\dfrac{\text{mass}}{\text{molar mass}}\) The molar mass of \(\mathrm{PCl_5}\) is: \(\text{Molar mass} = 1 \times (\text{Atomic mass of P}) + 5 \times (\text{Atomic mass of Cl}) = 1 \times (30.97\, g/mol) + 5 \times (35.45\, g/mol) = 208.22\, g/mol\) Now we can calculate the moles of \(\mathrm{PCl_5}\): \(n=\dfrac{2.450\,g}{208.22\,g/mol} = 0.01176\,mol\) Next, we use the ideal gas law to find the pressure: \(PV = nRT\) \(P = \dfrac{nRT}{V}\) Given the temperature, \(T = 600\, K\), the volume, \(V = 500\, mL = 0.5\, L\), and the gas constant, \(R = 0.0821\, L \cdot atm/(mol \cdot K)\), we can calculate the pressure: \(P = \dfrac{(0.01176\,mol)(0.0821\, L \cdot atm/(mol \cdot K))(600\, K)}{0.5\, L} = 11.583\, atm\) So, the initial pressure of \(\mathrm{PCl_5}\) without dissociation is approximately \(11.583\, atm\).

Now we need to determine the changes in pressure when the reaction reaches equilibrium. We will use an ICE (Initial, Change, Equilibrium) table for this purpose: \(\begin{array}{c|ccc} & \mathrm{PCl_5} & & \mathrm{PCl_3} & \mathrm{Cl_2} \\ \hline \text{Initial} & 11.583 & & 0 & 0 \\ \text{Change} & -x & & +x & +x \\ \text{Equilibrium} & 11.583-x & & x & x \end{array}\) Here, \(x\) represents the change in pressure of each gas due to the reaction.

Since we have the ICE table and the equilibrium constant, \(K_p = 11.5\), we can write the expression for \(K_p\) and solve for the equilibrium pressure of \(\mathrm{PCl_5}\) (i.e.,\(11.583-x\)): \(K_p = \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}\) \(11.5 = \dfrac{x^2}{11.583-x}\) Now we can solve for \(x\), which is a quadratic equation. However, if we assume that the dissociation is not too significant (since \(K_p\) is not very large), we can approximate \(11.583-x \approx 11.583\). This simplifies the equation to: \(11.5 = \dfrac{x^2}{11.583}\) \(x^2 = 11.5 \times 11.583\) \(x = \sqrt{11.5 \times 11.583} = 3.52\) So, at equilibrium, the partial pressure of \(\mathrm{PCl_5}\) is approximately \(11.583 - 3.52 = 8.063\, atm\).

From the ICE table, we see that the partial pressures of \(\mathrm{PCl_3}\) and \(\mathrm{Cl_2}\) at equilibrium are both equal to \(x\approx 3.52\, atm\). Thus, the total pressure at equilibrium can be calculated by adding the partial pressures of all gases: \(P_{\text{total}} = [\mathrm{PCl_5}] + [\mathrm{PCl_3}] + [\mathrm{Cl_2}] = 8.063 + 3.52 + 3.52 = 15.103\, atm\) So the total pressure in the bulb at equilibrium is approximately \(15.103\, atm\).

Finally, we can calculate the percent dissociation of \(\mathrm{PCl_5}\) at equilibrium: \(\text{Percent dissociation} = \dfrac{\text{amount dissociated}}{\text{initial amount}} \times 100\%\) \(\text{Percent dissociation} = \dfrac{x}{11.583} \times 100\%\) \(\text{Percent dissociation} = \dfrac{3.52}{11.583} \times 100\% = 30.37\%\) Thus, the percent dissociation of \(\mathrm{PCl_5}\) at equilibrium is approximately \(30.37\%\).