Question

Consider the reaction $$\mathbf{P}_{4}(g) \longrightarrow 2 \mathbf{P}_{2}(g)$$, where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{K},\) the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium.

Short answer

The equilibrium pressures of \(P_{4}(g)\) and \(P_{2}(g)\) are approximately 0.844 atm and 0.312 atm, respectively. The fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium is 0.156.

Step-by-step solution

Write the balanced chemical equation for the reaction

The given reaction is: \(P_4(g) \longrightarrow 2P_2(g)\)

Write the expression for the equilibrium constant

For the given reaction, the expression for the equilibrium constant (\(K_p\)) is: \[K_p = \frac{[P_2]^2}{[P_4]}\]

Set up the ICE table

Set up an ICE table to represent the initial concentrations, the change in concentrations, and the equilibrium concentrations of the reactants and products. ``` [P_4] [P_2] Initial P_4_0 0 Change -x +2x Equilibrium P_4_0 - x 2x ``` Initial pressures are unknown except for total pressure, so let the initial pressure of \(P_4(g)\) be \(P_{4_0}\) and the initial pressure of \(P_2(g)\) is 0. As the reaction progresses, the pressure of \(P_4(g)\) decreases by x, and the pressure of \(P_2(g)\) increases by 2x. At equilibrium, the pressure of \(P_4(g)\) is \(P_{4_0} - x\) and the pressure of \(P_2(g)\) is \(2x\).

Substitute the equilibrium pressures into the equilibrium constant expression

Replace the concentrations in the \(K_p\) expression by equilibrium pressures from the ICE table: \[K_p = \frac{(2x)^2}{(P_{4_0} - x)}\] It is given that the total pressure at equilibrium is 1 atm, so \((P_{4_0} - x) + 2x = 1\). Also, given that \(K_p=1.00 \times 10^{-1}\).

Solve the equations for x and the equilibrium pressures

Solve the equations obtained in step 4 for x and substitute the value of x back into the equilibrium pressure expressions to find the equilibrium pressures of the reactants and products. From the total pressure equation, we get \(P_{4_0} + x = 1\). Substituting this into the \(K_p\) equation, we get: \[\frac{(2x)^2}{(1-x)}= 1.00 \times 10^{-1}\] Solving for x, we get \(x \approx 0.156\) Now, substitute the value of x back into the equilibrium pressure expressions: \[P_{4} = P_{4_0} - x \approx 1 - 0.156 \approx 0.844\ \text{atm}\] \[P_{2} = 2x \approx 2 \times 0.156 \approx 0.312\ \text{atm}\]

Calculate the fraction of \(P_{4}(g)\) that has dissociated

To find the fraction of \(P_4(g)\) dissociated, divide the change in pressure due to dissociation by the initial pressure of \(P_4(g)\): Fraction dissociated = \(\frac{x}{P_{4_0}} = \frac{0.156}{1} = 0.156\) The equilibrium pressures of \(P_{4}(g)\) and \(P_{2}(g)\) are approximately 0.844 atm and 0.312 atm, respectively. The fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium is 0.156.

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_p\), is essential in understanding how far a reaction proceeds towards the products or the reactants in a gaseous system. For the given reaction \(P_4(g) \longrightarrow 2P_2(g)\), the equilibrium constant is given as \(K_p = 1.00 \times 10^{-1}\) at a temperature of 1325 K.
This value of \(K_p\) indicates a relatively small extent of conversion of reactants to products at this temperature. The equilibrium constant is temperature-dependent, meaning it only applies to the reaction conditions specified.
In mathematical terms, \(K_p\) is a ratio of the pressure of the products to the pressure of the reactants at equilibrium:

• Products are raised to the power of their coefficients in the balanced equation.
• In this scenario, \([P_2]^2\) denotes squared pressure of \(P_2\) due to the coefficient of 2 in the balanced equation.
• The reactant \([P_4]\) is not raised to any power because its coefficient is one.

By studying the \(K_p\), chemists can predict whether the reaction will favor the formation of reactants or products under specified conditions.

Pressure Calculation

In gas-phase reactions, calculating the equilibrium pressures is crucial for understanding the thermodynamic balance of the system. This is particularly the case when only the total pressure and the equilibrium constant are known, and individual partial pressures must be calculated.
The total pressure of the system is given as 1.00 atm. To find the equilibrium pressures of \(P_4(g)\) and \(P_2(g)\), the initial pressure of \(P_4(g)\) must be known. Given that initially, only \(P_4(g)\) is present, we set \(P_{4_0}\) as its initial pressure. As the reaction progresses, the pressures change according to the stoichiometry of the chemical equation.
The total pressure equals the sum of the partial pressures of gases at equilibrium, which leads us to the formula:

• \((P_{4_0} - x) + 2x = 1\)

where \(x\) is the change in pressure of \(P_4\). Solving the equilibrium expression yields the pressures of each component at equilibrium in the gaseous reaction.

ICE Table

An ICE (Initial, Change, Equilibrium) table is a handy tool for tracking changes in concentration or pressure in a chemical reaction as it reaches equilibrium. Using the ICE table:

• "Initial" represents the starting pressures of reactants and products.
• "Change" indicates how pressures differ as the reaction progresses towards equilibrium.
• "Equilibrium" gives the final pressures after the system has reached equilibrium.

In the reaction \(P_4(g) \to 2P_2(g)\), initially, pure \(P_4\) is placed in a container, hence its initial pressure \(P_{4_0}\) and \(P_2\) starts at 0.
As \(P_4\) decomposes, its pressure decreases by \(x\), while \(P_2\) builds up, increasing by \(2x\). The ICE table for this reaction clearly shows how these pressures are related and provide the main structure to solve the equilibrium problem using the equilibrium expression and known total pressures.

Mole Fraction

Mole fraction represents the ratio of moles of a particular gas component to the total moles of all gases present in the mixture. It's vital in evaluating how much of a particular reactant has undergone a reaction.
For the precision calculation of dissociation in the reaction \(P_4(g) \longrightarrow 2P_2(g)\), the mole fraction highlights what section of the initial gas has transformed. This can be mathematically expressed as the fraction of dissociated \(P_4(g)\) compared to its initial quantity:

• Fraction dissociated \(= \frac{x}{P_{4_0}}\)

The result highlights how much of \(P_4(g)\) has been converted into \(P_2(g)\) in terms of moles, providing insights into the efficiency and extent of the reaction. In our scenario, this fraction is calculated as 0.156, depicting that 15.6% of \(P_4(g)\) dissociated at equilibrium.