Problem 116 - Chemistry

Question

The equilibrium constant \(\left(K_{P}\right)\) for the reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) is 2.93 at \(127^{\circ} \mathrm{C}\) Initially there were 2.00 moles of \(\mathrm{PCl}_{3}\) and 1.00 mole of \(\mathrm{Cl}_{2}\) present. Calculate the partial pressures of the gases at equilibrium if the total pressure is 2.00 atm.

Step-by-Step Solution

Answer

After calculation, we should get the equilibrium values of x and hence the moles at equilibrium. Assuming that the volume and the temperature remain constant, the partial pressure of each gas can be found by multiplying the total pressure by the mole fraction of each gas. The mole fraction is obtained using the equilibrium moles divided by the total moles.

Step 1: Establish the Initial, Change, and Equilibrium in Mols

Let's start by representing the number of moles at the start of the reaction and at equilibrium. Initially, we have 2.00 moles of \(\mathrm{PCl}_{3}\), 1.00 mole of \(\mathrm{Cl}_{2}\) and no \(\mathrm{PCl}_{5}\). Let the change in mols be x. Therefore, at equilibrium, the moles of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) would be \(2-x\) and \(1-x\) respectively while the moles of \(\mathrm{PCl}_{2}\) would be x.

Step 2: Use Kp Expression to Solve for x

The equation for \(K_{P}\) is given by \([P_{PCl_5}]/[P_{PCl_3}P_{Cl_2}]\). It is equal to 2.93. We know that partial pressure is given by \(P = nRT/V\) where \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is the temperature in Kelvin and \(V\) is the volume. Since the reaction takes place at constant temperature and volume, we can say that the partial pressure is directly proportional to the number of moles. Hence, we substitute the mole values in terms of x into the \(K_{P}\) equation and solve for x.

Step 3: Find the Partial Pressures Using the Mole Fractions and Total Pressure

Once x is found, substitute x back to find the mole values at equilibrium for \(\mathrm{PCl}_{3}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{PCl}_{5}\). The partial pressure of each gas can be found by multiplying the total pressure by the mole fraction of each gas. The mole fraction is given by the number of moles of the gas divided by the total number of moles. This will provide the equilibrium partial pressures of each gas.

Key Concepts

Le Chatelier's Principle

Le Chatelier's principle is a fundamental concept in chemistry that predicts how a system at equilibrium will respond to changes in concentration, temperature, or pressure. When a system at equilibrium is subjected to a change in one of these conditions, the system will shift in a direction that counteracts the change, striving to re-establish equilibrium.

Let's apply this principle to a chemical reaction, such as the one in the exercise involving \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g)\). If we were to increase the pressure of the system by decreasing the volume, Le Chatelier's principle suggests that the equilibrium would shift towards the side with fewer gas particles, in an attempt to reduce pressure. Conversely, if the temperature is increased, and the reaction is exothermic, the equilibrium would shift towards the reactants, as the system attempts to absorb the added heat and cool down. Understanding this principle helps in predicting the behavior of the reaction when subjected to stress, thus allowing for better control over the chemical process.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. It is a dynamic state, as individual molecules continue to react, but their overall concentrations remain constant. Equilibrium does not mean that the reactants and products are present in equal amounts, but rather that their ratios are constant.

In the context of the exercise, when \(\mathrm{PCl}_{3}(g)\) combined with \(\mathrm{Cl}_{2}(g)\) react to form \(\mathrm{PCl}_{5}(g)\), they do so until the equilibrium state is achieved. The equilibrium constant \(K_{P}\) quantifies this state, representing the ratio of the products to reactants at equilibrium, with each raised to the power of their stoichiometric coefficients. Here, a \(K_{P}\) of 2.93 indicates the particular ratio at which the system achieves equilibrium at \(127^\circ \mathrm{C}\). Calculating the changes in mole amounts and their corresponding partial pressures requires a clear understanding of equilibrium and how these values relate in the equilibrium constant expression.

Partial Pressure

Partial pressure is an essential concept when dealing with gases. It represents the pressure that a single gas component in a mixture of gases would exert if it occupied the entire volume alone at the same temperature. The partial pressure of a gas is directly proportional to its mole fraction in the gas mixture, which aligns with Dalton's Law of Partial Pressures.

To calculate the partial pressures at equilibrium as in our exercise, we first determine the mole fraction for each gas by dividing its number of moles by the total number of moles present. With the total pressure known to be 2.00 atm, we multiply the mole fraction of each gas by this total pressure to obtain their respective partial pressures. Understanding partial pressures is not only crucial for solving this type of exercise but also for mastering gas behavior and reactions under various pressure conditions in real-world applications.