Problem 52 - Chemistry

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 \(\mathrm{S}_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at 1325 \(\mathrm{K}\) .

Step-by-Step Solution

Answer

The equilibrium constant Kp for the reaction \(\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)\) at 1325 K is \(324 \mathrm{atm^3}\).

Step 1: Write the reaction and its expression for Kp

We're given the reaction: \[\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)\] The expression for the reaction's Kp is: \[K_{p} = \frac{(P_{\mathrm{S_2}})^4}{(P_{\mathrm{S_8}})}\] where \(P_{\mathrm{S_2}}\) and \(P_{\mathrm{S_8}}\) are the equilibrium partial pressures of S₂ and S₈, respectively.

Step 2: Use initial and final pressures to find partial pressures of species at equilibrium.

Initially, the partial pressure of S₂ is 0 atm, since the container only has S₈. At equilibrium, the partial pressure of S₈ is 0.25 atm, which means the pressure change of S₈ during the reaction is: \[ \mathrm{∆P_{S_8}} = P_{\mathrm{Initial}} - P_{\mathrm{Equilibrium}} = 1.00 \mathrm{atm} - 0.25 \mathrm{atm} = 0.75 \mathrm{atm}\] Since 1 mole of S₈ produces 4 moles of S₂, we can find the pressure change of S₂ as a result of the reaction using the stoichiometry of the balanced equation: \[ \mathrm{∆P_{S_2}} = 4 \times \mathrm{∆P_{S_8}} = 4 \times 0.75 \mathrm{ atm} = 3.00 \mathrm{ atm}\] So, the equilibrium partial pressure of S₂ is: \[P_{\mathrm{S_2}} = \mathrm{Initial} + \mathrm{∆P_{S_2}} = 0 \mathrm{atm} + 3.00 \mathrm{atm} = 3.00 \mathrm{atm}\]

Step 3: Calculate Kp using equilibrium partial pressures

Now, we have the equilibrium partial pressures of both species. They are: \(P_{\mathrm{S_8}} = 0.25 \mathrm{atm}\) and \(P_{\mathrm{S_2}} = 3.00 \mathrm{atm}\) Substitute these values into the Kp expression: \[K_{p} = \frac{(P_{\mathrm{S_2}})^4}{(P_{\mathrm{S_8}})} = \frac{(3.00 \mathrm{atm})^4}{(0.25 \mathrm{atm})}\]

Step 4: Solve for Kp

Calculate Kp: \[K_{p} = \frac{(3.00 \mathrm{atm})^4}{(0.25 \mathrm{atm})} = \frac{81 \mathrm{atm^3}}{0.25 \mathrm{atm}} = 324 \mathrm{atm^3}\] So, the equilibrium constant Kp for this reaction at 1325 K is 324 atm³.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is the point at which the concentrations or pressures of the reactants and products remain constant over time. This is because the rate of the forward reaction equals the rate of the backward reaction.
In our problem, the decomposition of \[\mathrm{S}_8(g) \rightleftharpoons 4 \mathrm{S}_2(g)\] reaches equilibrium when the partial pressures of \(\mathrm{S}_8\) and \(\mathrm{S}_2\) no longer change.
Conditions like temperature and pressure can influence when equilibrium is reached and where the balance between products and reactants lies.

• Equilibrium does not mean equal concentrations, but rather that their ratios remain constant.
• The changes that occur do so on a microscopic level, with reactant and product molecules continuously interchanging.

Understanding equilibrium is crucial because it helps predict how the system behaves in response to changes in conditions.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It contributes to the total pressure of the gas mixture.
In the context of chemical equilibrium, we often work with partial pressures instead of concentration when dealing with gases.
For the reaction \[\mathrm{S}_8(g) \rightleftharpoons 4 \mathrm{S}_2(g)\] the problem gives the initial pressure of \(\mathrm{S}_8\) as 1.00 atm, and the equilibrium partial pressure as 0.25 atm.

• To track the shift towards equilibrium, the change in partial pressure is key.
• Partial pressures add up to the total pressure in the system.

Knowing partial pressures allows us to calculate the equilibrium constant \(K_p\) for the reaction.

Equilibrium Calculation

An equilibrium calculation involves determining the equilibrium constant \(K_p\).
This value quantitatively describes the ratio of product pressures to reactant pressures at equilibrium.
The formula for \(K_p\) is based on the balanced chemical equation:\[K_{p} = \frac{(P_{\mathrm{S_2}})^4}{(P_{\mathrm{S_8}})}\]

• Substitute the equilibrium partial pressures to find \(K_p\).
• Equilibrium constants are specific to the temperature and reaction.

In our problem, we find \(K_p\) using the pressures of 3.00 atm for \(\mathrm{S_2}\) and 0.25 atm for \(\mathrm{S_8}\), resulting in a \(K_p\) of 324 atm³. This detailed calculation gives insight into how strongly the reaction favors products over reactants.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. For the decomposition of \[\mathrm{S}_8(g) \rightarrow 4 \mathrm{S}_2(g)\], stoichiometry tells us that one mole of \(\mathrm{S}_8\) produces four moles of \(\mathrm{S}_2\).
This relationship allows us to determine changes in pressure or concentration when equilibrium is reached.

• Stoichiometry is vital for converting changes in reactant concentration to product concentration.
• Using stoichiometric ratios ensures the conservation of atoms and consistency within the reaction.

In our example, the initial pressure change due to \(\mathrm{S}_8\) decomposing determines that \(\Delta P_{\mathrm{S_2}} = 4 \times \Delta P_{\mathrm{S_8}}\), enabling us to find the equilibrium pressure of \(\mathrm{S}_2\). Mastery of stoichiometry is crucial for solving chemical equilibrium problems efficiently.