Question

2 moles of \(\mathrm{PCl}_{5}\) were heated in a closed vessel of a 2 litre capacity. \(\Lambda\) t cquilibrium \(40 \%\) of \(\mathrm{PCl}_{5}\) dissociated into \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\). The valuc of the equilibrium constant is (1) \(0.267\) (2) \(0.53\) (3) \(2.63\) (4) \(5.3\)

Short answer

The value of the equilibrium constant is 0.267.

Step-by-step solution

Determine Initial Moles of Each Component

Initially, there are 2 moles of \(\text{PCl}_{5}\). Therefore, initial moles of \(\text{PCl}_{5}\) = 2, \(\text{PCl}_{3}\) = 0, \(\text{Cl}_{2}\) = 0.

Determine the Change in Moles at Equilibrium

Given that 40% of \(\text{PCl}_{5}\) dissociates, we calculate the moles dissociated: \[ 0.40 \times 2 = 0.8 \text{ moles} \]. Thus, moles of \(\text{PCl}_{5}\) at equilibrium: \[ 2 - 0.8 = 1.2 \text{ moles} \]. Moles of \(\text{PCl}_{3}\) and \(\text{Cl}_{2}\) each formed: \[ 0.8 \text{ moles} \].

Calculate Concentrations at Equilibrium

The volume of the vessel is 2 liters. Thus, \[ [\text{PCl}_{5}] = \frac{1.2}{2} = 0.6 \text{ M} \] \[ [\text{PCl}_{3}] = \frac{0.8}{2} = 0.4 \text{ M} \] \[ [\text{Cl}_{2}] = \frac{0.8}{2} = 0.4 \text{ M} \]

Write the Expression for the Equilibrium Constant

The equilibrium constant (\text{K}_{c}) for the reaction is given by the expression: \[ K_c = \frac{[\text{PCl}_{3}][\text{Cl}_{2}]}{[\text{PCl}_{5}]} \]

Substitute the Equilibrium Concentrations into the Equilibrium Constant Expression

Using the calculated concentrations: \[ K_c = \frac{(0.4)(0.4)}{0.6} = \frac{0.16}{0.6} \]

Perform the Final Calculations

Calculate the value of \( K_c \): \[ K_c = \frac{0.16}{0.6} = 0.267 \]

Key concepts

chemical equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time. This does not mean the reactions stop; rather, they continue occurring at the same rate in both directions. Understanding this balance is crucial for solving equilibrium problems, such as calculating the equilibrium constant.

mole calculation

Mole calculation is fundamental in chemistry for understanding quantities of substances involved in reactions. In this exercise, we started with 2 moles of \(\mathrm{PCl}_{5}\) in a 2-liter vessel. To determine the moles dissociating into other products, we used the given dissociation percentage (40%). By multiplying the initial moles by the dissociation percentage, we calculated the moles changing during the reaction. For instance, \(2 \, \text{moles} \, \times \, 0.40 = 0.8 \, \text{moles}\).

dissociation reaction

A dissociation reaction involves a compound breaking down into smaller molecules or ions. In this specific problem, \(\mathrm{PCl}_{5}\) dissociates into \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\). The reaction can be represented as: \[ \mathrm{PCl}_{5} \leftrightarrow \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \] Given that 40% of \(\mathrm{PCl}_{5}\) dissociates, we were able to calculate the amounts of each substance at equilibrium. Understanding dissociation reactions helps to determine how substances interact and transform under different conditions.

concentration determination

Concentration determination is important for calculating the equilibrium constant and understanding the state of a reaction. Concentration (\text{M}) is found by dividing moles by volume. In this problem, we calculated the equilibrium concentrations for \(\mathrm{PCl}_{5}\), \(\mathrm{PCl}_{3}\), and \(\mathrm{Cl}_{2}\). The vessel's volume was given as 2 liters, simplifying these calculations. For example, the concentration of \(\mathrm{PCl}_{5}\) at equilibrium was \[ \frac{1.2\, \text{moles}}{2\, \text{liters}} = 0.6\, \text{M} \]. Knowing these concentrations allows us to use the equilibrium expression to find the equilibrium constant.