Question

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 \(500 .-\mathrm{mL}\) bulb, which is then heated to \(600 .\) 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 \(\mathrm{PCl}_{5}\) at equilibrium?

Short answer

a) The pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate would be approximately \(0.02353\, \text{atm}\). b) The partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium is approximately \(0.00093\, \text{atm}\). c) The total pressure in the bulb at equilibrium is approximately \(0.0461\, \text{atm}\). d) The percent dissociation of \(\mathrm{PCl}_{5}\) at equilibrium is approximately \(96.04 \%\).

Step-by-step solution

First, we need to calculate the initial moles of \(\mathrm{PCl}_{5}\) using the given mass and its molar mass: \(2.450 \,\text{g} \mathrm{PCl}_{5} \times (\text{1 mole} / 208.24\, \text{g}) = 0.01176 \,\text{moles} \,\mathrm{PCl}_{5}\) Now we'll use the ideal gas law to calculate the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate. The ideal gas law is \(\text{PV}=\text{nRT}\), where \(\text{P}\) is the pressure, \(\text{V}\) is the volume, \(\text{n}\) is moles, \(\text{R}\) is the ideal gas constant, and \(\text{T}\) is the temperature. Simplify the equation for pressure: \(P = \frac{nRT}{V}\). Plug in the given values and solve for the pressure \(P=\frac{(0.01176\, \text{moles})(0.0821\,\text{L atm} /(\text{K mol})(600\, \text{K})}{0.5\, \text{L}}\). #Step 2: Set up the reaction table and find expressions relating the partial pressures to the equilibrium constant#

To understand how the system changes when it reaches equilibrium, we'll set up a reaction table, where 'I' stands for Initial, 'C' for Change, and 'E' for Equilibrium: $$ \begin{array}{c|c c c} & \mathrm{PCl}_{5}(g) & \mathrm{PCl}_{3}(g) & \mathrm{Cl}_{2}(g)\\ \hline I & 0.02353\, \text{atm} & 0\, \text{atm} & 0 \, \text{atm} \\ C & -x& +x & +x \\ E & 0.02353 - x & x &x \end{array} $$ The equilibrium constant, \(K_\text{p}\), is given by: \(K_\text{p} = \frac{[PCl_3][Cl_2]}{[PCl_5]}\). Then the expressions at equilibrium are: \(K_p = \frac{x * x}{0.02353-x} = 11.5\) #Step 3: Solving for x and finding the partial pressures at equilibrium#

Next, solve the equation from Step 2 for \(x\) which represents the pressures at equilibrium: \(11.5 = \frac{x^2}{0.02353-x}\) Rearrange into a quadratic equation: \(x^2 + 11.5x - 0.27016 \approx 0\) Solve the quadratic equation for \(x\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) \(x = \frac{-11.5 \pm \sqrt{11.5^2-4(1)(-0.27016)}}{2(1)}\) Taking only the positive root estimate since pressures must be positive, \(x \approx 0.0226\, \text{atm}\) Now we can find the partial pressures at equilibrium: \(\mathrm{PCl}_{5}(g): 0.02353 - x \approx 0.00093 \, \text{atm}\) \(\mathrm{PCl}_{3}(g): x \approx 0.0226 \,\text{atm}\) \(\mathrm{Cl}_{2}(g): x \approx 0.0226\, \text{atm}\) #Step 4: Calculate total pressure and percent dissociation at equilibrium#

The total pressure at equilibrium can be found by adding the partial pressures of all the gases: \(\text{Total pressure}= 0.00093\, \text{atm} + 0.0226\, \text{atm} + 0.0226\, \text{atm} \approx 0.0461\, \text{atm}\). Next, calculate the percent dissociation of \(\mathrm{PCl}_{5}\) at equilibrium: \(\text{Percent dissociation} = \frac{0.02353\, \text{atm} - 0.00093\, \text{atm}}{0.02353\, \text{atm}} \times 100 \% \approx 96.04 \%\) Now, we have the answers for each part of the problem: a) The pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate would be approximately \(0.02353\, \text{atm}\). b) The partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium is approximately \(0.00093\, \text{atm}\). c) The total pressure in the bulb at equilibrium is approximately \(0.0461\, \text{atm}\). d) The percent dissociation of \(\mathrm{PCl}_{5}\) at equilibrium is approximately \(96.04 \%\).

Key concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant, represented as \(K_p\) for gas-phase reactions, is a crucial concept that helps quantify the balance between products and reactants.

• For the reaction \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g)\), the equilibrium constant \(K_p\) is given as 11.5 at 600 K.
• It is expressed as \(K_p = \frac{{[\mathrm{PCl_3}][\mathrm{Cl_2}]}}{[\mathrm{PCl_5}]}\), where each term represents the partial pressure of the respective gases at equilibrium.

Such constants help predict the position of equilibrium. A large \(K_p\) (greater than 1) implies that products are favored at equilibrium, as evident by the higher pressures of \(\mathrm{PCl_3}\) and \(\mathrm{Cl_2}\) compared to \(\mathrm{PCl_5}\). Conversely, a small \(K_p\) would indicate a reactant-favored equilibrium position. This constant remains unchanged for a given reaction at a specific temperature unless external conditions like temperature are altered.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases, and it directly relates to the gas's concentration in terms of mole fraction. For the specified reaction, understanding partial pressures is key to evaluating equilibria:

• The initial partial pressure of \(\mathrm{PCl}_{5}\) is calculated assuming no dissociation, using \(n = 0.01176\) moles and the ideal gas law, resulting in \(0.02353\, \text{atm}\).
• At equilibrium, changes occur where \(\mathrm{PCl}_{5}(g)\) reduces by \(x\), and \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g)\) increase by the same value \(x = 0.0226\, \text{atm}\), computed by solving the quadratic equation derived from the \(K_p\) expression.
• The partial pressure across each component of the reaction describes the equilibrium state, which, when summed, should equal the total pressure in the system.

Thus, partial pressures are directly linked to reaction equilibria and reflective of the actual concentrations of each species in a gaseous mix.

Percent Dissociation

Percent dissociation is the percentage of an initial compound that breaks down or changes into other substances. It's essential in quantifying how much of a reactant like \(\mathrm{PCl}_{5}\) actually dissociates under the given conditions.

• This can be calculated using the formula: \( \text{Percent Dissociation} = \left( \frac{\text{initial amount} - \text{equilibrium amount}}{\text{initial amount}}\right) \times 100 \% \)
• For \(\mathrm{PCl}_{5}\), it is \(\left( \frac{0.02353 - 0.00093}{0.02353} \right) \times 100 \approx 96.04\%\).

In this context, the high percent dissociation reflects a significant shift towards product formation, confirming the equilibrium constant's prediction. This measure helps assess the degree to which a chemical equilibrium process is complete, providing insights into the efficiency and extent of reaction conversions.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle that relates the pressure, volume, temperature, and amount of a gas. The equation is \(PV = nRT\), where:

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant (0.0821 L atm/K mol), and
• \(T\) is the temperature in Kelvin.

For example, knowing the initial moles of \(\mathrm{PCl}_{5}\) and the conditions of the reaction environment (0.5 L bulb at 600 K), the law allows us to determine \(\mathrm{PCl}_{5}\)'s potential pressure, assuming no dissociation:\[P = \frac{(0.01176 \text{ moles})(0.0821 \text{ L atm/K mol})(600 \text{ K})}{0.5 \text{ L}} \approx 0.02353 \text{ atm}\]The Ideal Gas Law thus provides a foundational tool for calculating theoretical pressures or moles of a gas before any adjustments due to changes in actual chemical reactions.