Question

Sulfuryl chloride decomposes according to the equation, $$ \mathrm{SO}_{2} \mathrm{CI}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{CI}_{2}(\mathrm{~g}) $$ Given that at \(300^{\circ} \mathrm{K}\) and 1 atm pressure, the degree of dissociation is 121, find the equilibrium constant, \(K_{p}\). What will be the degree of dissociation when pressure \(=10 \mathrm{~atm} ?\)

Short answer

In conclusion, the equilibrium constant \(K_p\) for the given dissociation reaction is equal to the initial concentration of \(SO_2Cl_2\), and the degree of dissociation at 10 atm pressure is 0.

Step-by-step solution

Write the reaction and the expression for equilibrium constant

The given reaction is: \[SO_2Cl_2 (g) \rightleftarrows SO_2(g) + Cl_2 (g)\] The equilibrium constant, \(K_p\), can be written as: \[K_p = \frac{[SO_2][Cl_2]}{[SO_2Cl_2]}\] We will later use the values of degree of dissociation to find \(K_p\).

Set up initial and equilibrium concentrations

Let's assume the initial concentration of \(SO_2Cl_2\) is \(C\) moles per liter. Since the degree of dissociation is 1, all the \(SO_2Cl_2\) will dissociate into \(SO_2\) and \(Cl_2\), so their final concentrations will also be \(C\) moles per liter. At equilibrium, \[[SO_2] = C\] \[[Cl_2] = C\] \[[SO_2Cl_2] = 0\]

Find the equilibrium constant \(K_p\) with given data

Using the equilibrium constant expression, we have: \[K_p = \frac{[SO_2][Cl_2]}{[SO_2Cl_2]} = \frac{C \times C}{0}\] However, we cannot divide by zero, so we should assume that at equilibrium, a very small amount, \(\alpha\) of \(SO_2Cl_2\), is still present. So, the initial concentration of \(SO_2Cl_2\) is \((C-\alpha)\), which for the given conditions, still represents virtually complete dissociation, and the expression for \(K_p\) becomes: \[K_p = \frac{C \times C}{(C-\alpha)}\] Given that degree of dissociation is 1 and the pressure is 1 atm: \[\alpha = \frac{C}{1} = C\] \[K_p = \frac{C \times C}{C} = C\]

Find the degree of dissociation at 10 atm pressure

We can find the degree of dissociation at 10 atm pressure, using the same procedure as in step 3. Let's say the new degree of dissociation is \(\beta\), then: \[K_p = \frac{C \times C}{(C-\beta)}\] Since \(K_p\) remains the same at the new pressure and we have already found its value to be C: \[C = \frac{C \times C}{(C-\beta)}\] Solving for \(\beta\) (the degree of dissociation at 10 atm pressure): \[\beta = C - \frac{C^2}{C} = C - C = 0\] So, the degree of dissociation of the system when the pressure is increased to 10 atm is 0, meaning there is no dissociation occurring. In conclusion, the equilibrium constant \(K_p\) for the given dissociation reaction is equal to the initial concentration of \(SO_2Cl_2\), and the degree of dissociation at 10 atm pressure is 0.

Key concepts

Chemical Equilibrium

In chemistry, chemical equilibrium is the state where the concentrations of reactants and products remain constant over time because the rate of the forward reaction equals the rate of the reverse reaction. This does not mean that the reactants and products are equal in concentration, but that their ratios stay consistent, abiding by the equilibrium constant, denoted as \( K_{eq} \).

Equilibrium can be altered by changing the reaction conditions such as temperature, pressure, and concentration. However, the value of the equilibrium constant itself only changes with temperature. Understanding the concept of equilibrium is crucial for chemists, as most chemical reactions in nature reach a state of balance. Equilibrium constants are derived from the Law of Mass Action, which gives a relationship between the concentrations of reactants and products of a chemical reaction at equilibrium.

Degree of Dissociation

The degree of dissociation, represented by the symbol \( \alpha \), is a dimensionless number that describes the extent to which a compound separates into its constituents upon reacting, such as in dissociation or ionization. \( \alpha \) is defined as the fraction of a mole of reactant that has dissociated at equilibrium. It is calculated by dividing the number of moles of the substance that has dissociated by the initial number of moles present before dissociation.

The degree of dissociation is an important concept when discussing reactions involving gases or electrolytes in solution. For instance, in a system where a gas dissolves into its ions or molecules, \( \alpha \) can significantly influence the equilibrium concentrations and, as a result, the equilibrium constant \( K_p \) for the reaction. In the provided exercise, the degree of dissociation changes with pressure, showing a direct connection with Le Chatelier's principle.

Le Chatelier's Principle

Le Chatelier's principle is a staple of chemical equilibrium, offering insight into how a system at equilibrium responds to disturbances. The principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle helps predict the effect of a change in condition on a chemical reaction.

Common disturbances include changes in concentration, pressure, or temperature, each of which can shift the equilibrium position. For example, when pressure is increased on a system where gases are involved, the equilibrium shifts to the side with fewer moles of gas to reduce pressure. This knowledge is particularly useful in industrial processes where yield optimization is crucial.

In the context of our exercise, when the pressure is increased from 1 atm to 10 atm, Le Chatelier's principle would predict a shift in equilibrium to oppose the increase in pressure. Since dissociation leads to an increase in the number of gas molecules, a higher pressure will favor association, leading to a reduction in the degree of dissociation (\( \beta \)), aligning with the calculations shown in the existing solution.