Question

A sample of \(\mathrm{S}_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{~K}\) at an initial pressure of \(1.00 \mathrm{~atm}\), where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$ \mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{~S}_{2}(g) $$ At equilibrium, the partial pressure of \(S_{8}\) is \(0.25\) atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{~K}\).

Short answer

The equilibrium constant K_p for the given reaction at 1325 K is 324.

Step-by-step solution

Write down the balanced equation and equilibrium constant expression

We are given the balanced equation: \( \mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{~S}_{2}(g) \) The equilibrium constant expression for this reaction will be: \[ K_{p} = \frac{(\mathrm{[S_{2}]})^4}{\mathrm{[S_{8}]}} \]

Set up the I.C.E. table

To set up the I.C.E. table, we'll write the initial partial pressure of S₈ and S₂, the pressure change during the reaction, and the equilibrium partial pressures. I: Initial (atm): S₈: 1.00 S₂: 0 C: Change (atm): S₈: -x S₂: +4x E: Equilibrium (atm): S₈: 1-x S₂: 4x We are also given that the equilibrium pressure of S₈ is 0.25 atm. This means that: \(1-x = 0.25\)

Find the change in pressures and the equilibrium pressures

To find the change in pressures, we'll solve for x: \(1-0.25 = x\) \(x = 0.75\) Now, we can find the equilibrium pressure of S₂: S₂: 4x = 4(0.75) = 3.0 atm Hence, the equilibria partial pressures are: S₈: 0.25 atm, S₂: 3.0 atm

Calculate K_p using the equilibrium pressures

Now that we have the equilibrium pressures of S₈ and S₂, we can plug these values into the equilibrium constant expression: \[ K_{p} = \frac{(3.0 \thinspace \mathrm{atm})^4}{(0.25 \thinspace \mathrm{atm})} \] Calculate the value of K_p: \[ K_{p} = \frac{81.0 \thinspace \mathrm{atm^3}}{0.25 \thinspace \mathrm{atm}} = 324 \] Finally, the equilibrium constant K_p for this reaction at 1325 K is 324.

Key concepts

Pressure

Pressure is an important concept in maintaining chemical equilibrium in a closed system, like the one in our example. When we talk about pressure, we're referring to the force that gas molecules exert on the walls of their container. In chemical reactions involving gases, pressure can affect how the reaction proceeds.

In the given reaction, the initial pressure of the sulfur compound S₈ is 1.00 atm. This pressure represents the energy the gas molecules exert within the container. As the reaction progresses, the pressure changes due to the decomposition of S₈ to S₂. Monitoring the changes in pressure helps us understand how the reaction moves towards equilibrium.

Key points about pressure in chemical reactions:

• Pressure helps in determining how far a reaction has progressed.
• It affects the rate at which products are formed, especially in gaseous reactions.
• Understanding pressure changes is crucial when calculating equilibrium constants like Kₚ.

Overall, controlling and measuring pressure is vital to fully comprehend the behavior of gases during reactions.

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions happen at the same rate. In our reaction, the conversion of S₈ to S₂ and vice versa happens simultaneously. Over time, these reactions balance out, reaching a point where the concentrations of reactions and products remain constant.

At equilibrium, the system achieves a stable state, even though the reactions are still ongoing. This stability results in observable steady concentrations of all reacting species. For the given reaction, at equilibrium, the partial pressure of S₈ maintains at 0.25 atm, and that of S₂ at 3.0 atm. This equilibrium condition allows us to calculate the equilibrium constant Kₚ.

Key points about chemical equilibrium:

• Reversible reactions achieve a state where the rate of the forward reaction equals the reverse.
• At equilibrium, the ratio of the products to reactants stays the same.
• Equilibrium constants like Kₚ provide insights into the system's behavior under these conditions.

Understanding equilibrium gives us a glimpse of how reactions behave over time and under different conditions.

Decomposition Reaction

A decomposition reaction involves breaking down a compound into simpler substances. Here, S₈, a sulfur compound, decomposes into smaller molecules of S₂ gas. This reaction is a classic example of decomposition, where a single substance separates into distinct products.

Decomposition reactions are essential, as they help break down complex compounds, releasing energy or forming simpler products. In this specific case, S₈ undergoes decomposition when subjected to high temperature, resulting in the formation of S₂. The direction and extent of the decomposition can be predicted using the equilibrium constant, Kₚ, which gives a numerical value indicating favorability.

Key points about decomposition reactions:

• They typically require energy, such as heat, to break down compounds.
• Often result in the production of more than one substance.
• Rate and extent of decomposition can be affected by variables like temperature and pressure.

The understanding of decomposition reactions is crucial in many industrial and natural processes, helping us manage and predict the behavior and outcomes of chemical reactions.