Problem 26 - Chemistry: Principles and Reactions

Question

A sealed flask has \(0.541\) atm of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\). The following equilibrium is established. $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the partial pressure of oxygen is measured to be \(0.216\) atm. Calculate \(K\) for the decomposition of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\).

Step-by-Step Solution

Answer

Question: Calculate the equilibrium constant (K) for the decomposition of SO_3 at 1000 K, given the initial partial pressure of SO_3 as 0.541 atm, and the equilibrium partial pressure of O_2 as 0.216 atm. Answer: The equilibrium constant (K) for the decomposition of SO_3 at 1000 K is approximately 0.418.

Step 1: Write the equilibrium constant expression

For the given chemical reaction, the equilibrium constant expression (K) can be written in terms of the partial pressures of the reactants and products: $$K_p = \frac{(\mathrm{SO}_2)^2 \cdot (\mathrm{O}_2)}{(\mathrm{SO}_3)^2}$$

Step 2: Set up the ICE table

Use an ICE (Initial, Change, Equilibrium) table to represent the initial partial pressures, the changes during the reaction, and the equilibrium partial pressures of each compound involved in the reaction. Note that we are not given the initial partial pressures of SO_2 and O_2, so we can assume that they are initially 0. | | SO_3 | SO_2 | O_2 | |---------------|------|------|------| | Initial | 0.541| 0 | 0 | | Change | -x | +x | +x | | Equilibrium |0.541-x|x | 0.216|

Step 3: Identify the relationship between x and given partial pressures

Since we know the equilibrium partial pressure of O_2 (0.216 atm), we can now relate it to the change (x): $$ x = 0.216 $$

Step 4: Calculate equilibrium partial pressures of SO_2 and SO_3

Using the value of x, we can find the equilibrium partial pressures of SO_2 and SO_3: For SO_2: $$P_{SO_2} = x = 0.216 \ \mathrm{atm}$$ For SO_3: $$P_{SO_3} = 0.541-x = 0.541-0.216 = 0.325 \ \mathrm{atm}$$

Step 5: Calculate the equilibrium constant (K)

Now that we have the equilibrium partial pressures of each compound, plug them into the equilibrium constant expression to solve for K: \begin{aligned} K_p &= \frac{(\mathrm{SO}_2)^2 \cdot (\mathrm{O}_2)}{(\mathrm{SO}_3)^2} \\ K_p &= \frac{(0.216)^2 \cdot (0.216)}{(0.325)^2} \\ K_p &= 0.418 \end{aligned} The equilibrium constant for the decomposition of SO_3 at 1000 K is approximately 0.418.

Key Concepts

Understanding Partial Pressure

In a gaseous mixture, each gas exerts a pressure as if it were the only gas in the container. This pressure is known as the **partial pressure**. It is a crucial concept in understanding chemical equilibrium in gases.
The total pressure of the gas mixture is the sum of the partial pressures of the individual gases. For example, in the reaction \(2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)\), each gas in the mixture contributes to the overall pressure based on its concentration.
The partial pressure helps us determine how each component in a reaction contributes to the total pressure and the system's overall equilibrium state. By measuring the partial pressure of the components, like \(\mathrm{O}_{2}\), we can also determine other unknown values in a reaction.
Furthermore, the relationship between partial pressures and equilibrium is described by the equilibrium constant \(K_p\), which uses partial pressures to express the concentrations of the gases involved.

The Dynamics of Chemical Equilibrium

**Chemical equilibrium** occurs when the rate of the forward reaction equals that of the backward reaction, leading to a stable concentration of reactants and products.
In the case of the reaction \(2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)\), an initial asymmetry exists due to the different initial concentrations. However, as time progresses, the system reaches an equilibrium point where the amountof \(\mathrm{SO}_{3}\) decomposing to produce \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) balances with the recombination of these products back into reactants.
At equilibrium, the concentrations become constant over time, allowing us to use these values in the equilibrium constant expression. The equilibrium constant \(K_p\) provides a snapshot of the ratio of products to reactants, indicating the position of equilibrium and showing whether products or reactants are favored.

Using an ICE Table

An **ICE table** is a valuable tool for organizing information about initial concentrations, changes in concentration, and equilibrium concentrations in a reaction. This table helps us apply stoichiometry rules to each component of the chemical equation.

In our example, the ICE table is set up for the decomposition of \(\mathrm{SO}_{3}\). The table initially starts by recording the initial partial pressure provided (0.541 atm for \(\mathrm{SO}_{3}\)), with \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) both starting at 0 atm.
Then comes the change phase. Here, each substance's change in concentration is expressed in terms of \(x\), which represents the change as the reaction progresses towards equilibrium.
Finally, we capture each component's equilibrium concentration, using the known partial pressure of \(\mathrm{O}_{2}\) to solve for \(x\).These values are then used to calculate the equilibrium constant \(K_p\) to determine the system's balance of products and reactants at equilibrium.