Problem 114 - Chemistry

Question

Consider the reaction $$\mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g)$$ 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 \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

Step-by-Step Solution

Answer

The equilibrium pressures of $\mathrm{P}_{4}(g)$ and $\mathrm{P}_{2}(g)$ are approximately 0.9084 atm and 0.1832 atm, respectively. The fraction of $\mathrm{P}_{4}(g)$ that has dissociated to reach equilibrium is approximately 0.0916 or 9.16%.

Step 1: Write down the Kp expression and the equilibrium pressure expressions

For the given reaction: $$\mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g)$$ The expression for the equilibrium constant Kp is given by: $$K_{\mathrm{p}}=\frac{[\mathrm{P}_{2}]^{2}}{[\mathrm{P}_{4}]}$$ Let the initial pressure of P4(g) be P, and since there is no P2(g) initially, the initial pressure of P2(g) is 0. At equilibrium, the pressure of P4(g) decreases by x atm, and the pressure of P2(g) increases by 2x atm. Therefore, the equilibrium pressures of P4(g) and P2(g) are (P-x) atm and 2x atm, respectively.

Step 2: Use the given total pressure and the equilibrium pressure expressions to create a system of equations

We are given that the total pressure at equilibrium is 1 atm. Therefore, we can write an equation relating the equilibrium pressures of P4(g) and P2(g): $$(P - x) + 2x = 1$$ Now, we can use the Kp expression and the equilibrium pressures to write another equation: $$K_{\mathrm{p}}=\frac{(2x)^{2}}{(P-x)}$$ We are given that Kp = 1.00 × 10⁻¹, so we can substitute that in the equation: $$1.00 \times 10^{-1}=\frac{(2x)^{2}}{(P-x)}$$ Now, we have a system of two equations with two unknowns (P and x).

Step 3: Solve the system of equations for P and x

First, we solve the equation (P-x) + 2x = 1 for P in terms of x: $$P = 1 - x$$ Now substitute this expression for P into the Kp equation: $$1.00 \times 10^{-1}=\frac{(2x)^{2}}{(1-x)}$$ Rearrange the equation to solve for x: $$(2x)^{2} = 1.00 \times 10^{-1} (1-x)$$ Solve this quadratic equation to obtain x: $$x\approx 0.0916$$ Now, we find the equilibrium pressures of P4(g) and P2(g) using the values of P and x: $$\mathrm{P}_{4} = P - x = 1 - 0.0916 \approx 0.9084 \:\text{atm}$$ $$\mathrm{P}_{2} = 2x \approx 0.1832 \:\text{atm}$$

Step 4: Calculate the fraction of P4(g) that has dissociated

Finally, we will find the fraction of P4(g) that has dissociated to reach equilibrium. The initial pressure of P4(g) is P, and the decrease in pressure is x. So, the fraction of dissociated P4(g) is the ratio of the decrease in pressure to the initial pressure: $$\frac{\text{Dissociated P}_4}{\text{Initial P}_4} = \frac{x}{P}$$ Substitute the values of x and P: $$\frac{0.0916}{1} = 0.0916$$ Hence, the fraction of P4(g) that has dissociated to reach equilibrium is approximately 0.0916, or 9.16%.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \(K_p\) when dealing with gaseous components and in terms of pressure, reflects the balance of concentrations of reactants and products in a chemical reaction at equilibrium.
For our reaction \(\mathrm{P}_4(g) \rightleftharpoons 2 \mathrm{P}_2(g)\), the \(K_p\) expression is \(K_p = \frac{[\mathrm{P}_2]^2}{[\mathrm{P}_4]}\). This specific setup embodies the relationship between the pressure of the products \(\mathrm{P}_2\) and the reactants \(\mathrm{P}_4\).

- **Key Aspects of \(K_p\)**: - **Temperature Dependence**: Equilibrium constants are specific to a given temperature, so \(K_p\) values change if temperature changes. - **Interpreting \(K_p\) Values**: - A large \(K_p\) signifies products are favored at equilibrium. - A small \(K_p\), like our \(1.00 \times 10^{-1}\), implies reactants are favored, indicating the extent of the reaction is low.Understanding \(K_p\) helps in predicting how much reactant will convert into product under equilibrium conditions, guiding calculations like those presented in steps 1 and 2 of the solution.

Partial Pressure

Partial pressure represents the pressure exerted by a single gas in a mixture. In an equilibrium setup, knowing the partial pressures helps determine the extent of a reaction.
In our reaction setup, initially, only \(\mathrm{P}_4(g)\) is present. As the reaction reaches equilibrium, some \(\mathrm{P}_4(g)\) dissociates into \(\mathrm{P}_2(g)\), creating distinct partial pressures for both gases.- **Calculating Partial Pressures**: - **Initial Condition**: Assume \(\mathrm{P}_4(g)\) is at its full pressure \(P\), while \(\mathrm{P}_2(g)\) starts at 0. - **Change at Equilibrium**: As the reaction proceeds, pressure decreases by \(x\) for \(\mathrm{P}_4(g)\) and increases by \(2x\) for \(\mathrm{P}_2(g)\), because two moles of \(\mathrm{P}_2\) form for every mole of \(\mathrm{P}_4\) that reacts.

Understanding partial pressures is crucial for using the \(K_p\) value effectively, as it requires these values to calculate the ratio of products to reactants at equilibrium.

Le Chatelier's Principle

Le Chatelier’s Principle helps predict how changes in conditions can shift the equilibrium of a reaction. When a system at equilibrium is exposed to a change such as pressure, temperature, or concentration, the equilibrium will shift to counteract this change.
In our context, if the total pressure of the system changes, or the temperature fluctuates from 1325 K, the balance between \(\mathrm{P}_4(g)\) and \(\mathrm{P}_2(g)\) will alter to restore equilibrium.- **Applications**: - **Pressure Adjustments**: Decreasing the volume of the container increases pressure, causing the system to shift toward fewer moles of gas (toward the formation of \(\mathrm{P}_4\) in this case). - **Temperature Changes**: - Endothermic reactions absorb heat. Increasing temperature shifts equilibrium towards products. - Exothermic reactions release heat. Increasing temperature shifts equilibrium toward reactants.

Thus, Le Chatelier's Principle is a valuable tool for controlling reactions in industrial settings, ensuring optimal efficiency and yield by adjusting environmental parameters.

Reaction Quotient

The Reaction Quotient, \(Q\), is similar to the equilibrium constant, but it represents current conditions rather than those at equilibrium.
This quotient is used to predict the direction in which a reaction will proceed to achieve equilibrium. For our chemical equation \(\mathrm{P}_4(g) \rightleftharpoons 2\mathrm{P}_2(g)\), \(Q\) is calculated in the same manner as \(K_p\), by plugging in current partial pressures.- **Uses of \(Q\) in Equilibrium**: - **When \(Q = K_p\)**: The system is at equilibrium. No net change occurs. - **When \(Q < K_p\)**: The reaction shifts to the right, producing more products to reach equilibrium. - **When \(Q > K_p\)**: The reaction shifts to the left, creating more reactants.Understanding \(Q\) is essential for adjusting reaction pathways in real-time, offering insights into how a system deviates from equilibrium and allowing for interventions to drive desired outcomes.