Problem 3 - Chemistry in 30 days

Question

Phosphorous pentachloride dissociates as follows $$ \mathrm{PCl}_{5}(\mathrm{~g}) \text { ? } \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ If total pressure at equilibrium is \(P\) and the degree of dissociation of \(\mathrm{PCl}_{5}\) is \(x\), the partial pressure of \(\mathrm{PCl}_{3}\) will be (a) \(\left(\frac{x}{x+1}\right) P\) (b) \(\left(\frac{2 x}{1-x}\right) P\) (c) \(\left(\frac{x}{x-1}\right) P\) (d) \(\left(\frac{x}{1-x}\right) P\)

Step-by-Step Solution

Answer

The partial pressure of \( \mathrm{PCl}_3 \) is \( \left( \frac{x}{x+1} \right) P \), corresponding to option (a).

Step 1: Understand the Reaction

The given reaction is a dissociation reaction for phosphorous pentachloride: \[\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\] This means that \( \mathrm{PCl}_5 \) dissociates into \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \) with the same degree of dissociation.

Step 2: Set Initial Concentrations

Initially, assume we have 1 mole of \( \mathrm{PCl}_5 \) and no \( \mathrm{PCl}_3 \) or \( \mathrm{Cl}_2 \). Therefore: - Initial moles of \( \mathrm{PCl}_5 \) = 1- Initial moles of \( \mathrm{PCl}_3 \) = 0- Initial moles of \( \mathrm{Cl}_2 \) = 0

Step 3: Consider Changes at Equilibrium

At equilibrium, let \( x \) be the degree of dissociation of \( \mathrm{PCl}_5 \). Therefore, the changes will be:- Moles of \( \mathrm{PCl}_5 \) = \( 1 - x \)- Moles of \( \mathrm{PCl}_3 \) = \( x \)- Moles of \( \mathrm{Cl}_2 \) = \( x \)

Step 4: Establish Equilibrium Partial Pressures

The total moles at equilibrium will be \( 1 - x + x + x = 1 + x \). The total pressure is \( P \). The partial pressure of \( \mathrm{PCl}_3 \) is given by the mole fraction of \( \mathrm{PCl}_3 \) multiplied by the total pressure: \[\text{Partial pressure of } \mathrm{PCl}_3 = \left( \frac{x}{1+x} \right) P\]

Step 5: Compare to Given Options

We compare the derived expression for the partial pressure of \( \mathrm{PCl}_3 \), \( \left( \frac{x}{1+x} \right) P \), with the given options:- Option (a): \( \left( \frac{x}{x+1} \right) P \)- Option (b): \( \left( \frac{2x}{1-x} \right) P \)- Option (c): \( \left( \frac{x}{x-1} \right) P \)- Option (d): \( \left( \frac{x}{1-x} \right) P \)Clearly, option (a) \( \left( \frac{x}{x+1} \right) P \) matches the derived expression.

Key Concepts

Degree of Dissociation

When a chemical compound breaks down into its components, especially at equilibrium, we often talk about its "degree of dissociation." In this context, it refers to how much of the original molecule breaks down into smaller molecules or ions. For our reaction, phosphorous pentachloride, denoted as \(\mathrm{PCl}_5\), dissociates into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\). If we originally have 1 mole of \(\mathrm{PCl}_5\), the equation can help us determine how much of it converts into products at equilibrium.

- If \( x \) represents the degree of dissociation, this is the fraction of the total \(\mathrm{PCl}_5\) that has dissociated. If \( x = 0.2 \), for example, 20% of \(\mathrm{PCl}_5\) has dissociated into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\).
- At equilibrium, the amount of \(\mathrm{PCl}_5\) remaining is \(1-x\), while the amount of \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) formed from dissociation is equal to \(x\) moles each, assuming 100% conversion per mole dissociated.

Partial Pressure

In a gaseous mixture, each gas exerts its own pressure that contributes to the total pressure of the system. This is what we call "partial pressure." Every gas in a mixture will exert pressure proportionate to its mole fraction in the mixture. The partial pressure is important in chemical equilibria because it affects how the reaction shifts between reactants and products based on changes in pressure.

During our dissociation case of \(\mathrm{PCl}_5\), let’s suppose equilibrium has been reached. Here, the total pressure \(P\) is shared among the gases based on their quantity. To find the partial pressure, we use the mole fraction. The mole fraction of \(\mathrm{PCl}_3\) is \(\frac{x}{1+x}\), meaning the partial pressure of \(\mathrm{PCl}_3\) at equilibrium is \(\left( \frac{x}{1+x} \right)P\).
- The mole fraction is calculated by taking the moles of the gas of interest and dividing by the total moles in the mixture.- This formula helps determine how the gases exert individual pressures, which is crucial for predicting the behavior of reactions at equilibrium.

PCl5 Dissociation

Understanding the dissociation of \(\mathrm{PCl}_5\) is crucial for analyzing its behavior in chemical reactions. \(\mathrm{PCl}_5\) is a compound composed of phosphorus and chlorine, and it is well known to dissociate into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) when heated or under specific conditions such as pressure changes. This dissociation is a reversible process, aligning with the concept of chemical equilibrium.

- \(\mathrm{PCl}_5\) starts dissociating when it reaches a particular temperature or pressure, and eventually, equilibrium is established where the rate of dissociation equals the rate of recombination.- At this equilibrium state, a balance between \(\mathrm{PCl}_5\), \(\mathrm{PCl}_3\), and \(\mathrm{Cl}_2\) is achieved with a constant degree of dissociation \(x\).- Understanding the equilibrium condition and the partial pressures involved can help predict how altering pressure or temperature would shift the reaction, either favoring the formation of products or reforming the original reactant, \(\mathrm{PCl}_5\).