Question

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, andit is also responsible for the acid rain phenomenon. The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment 2.00 moles of \(\mathrm{SO}_{2}\) and 2.00 moles of \(\mathrm{O}_{2}\) were initially present in a flask. What must the total pressure be at equilibrium to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

Short answer

By substitution and solving the \(K_p\) equation, you can find the total pressure of the system to reach an 80% yield of \(\mathrm{SO}_{3}\). Please note that the real coefficient numbers must be used when they are plugged in.

Step-by-step solution

Identify Relevant Information

The balanced equation is: \[2 ~\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 ~\mathrm{SO}_{3}(g)\] The equilibrium constant \(K_P\) is 0.13. This is the ratio of the concentrations of the products and reactants. The initial amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) present in the flask are 2.00 moles each. The goal is to find the total pressure that leads to an 80% yield of \(\mathrm{SO}_{3}\).

Calculate the Amount of \(\mathrm{SO}_{3}\) Formed

If the yield of \(\mathrm{SO}_{3}\) is 80%, then 0.8 * (initial moles of \(\mathrm{SO}_{2}\) or \(\mathrm{O}_{2}\)) moles of \(\mathrm{SO}_{3}\) are formed. Hence, 0.8 * 2.00 = 1.60 moles of \(\mathrm{SO}_{3}\) are formed.

Adjust the Moles for the Reached Equilibrium

At equilibrium, the number of moles of \(\mathrm{SO}_{2}\), \(\mathrm{O}_{2}\) and \(\mathrm{SO}_{3}\) in the reaction can be represented as (2.00 - x), (2.00 - x/2) and x, respectively, where x is the moles of \(\mathrm{SO}_{3}\) produced, i.e., 1.60 mole. This step considers the stoichiometry of the equilibrium reaction.

Formulate the Equations

From the given \(K_p\), \[ K_p = \frac{P_{\mathrm{SO}_{3}}^2} {P_{\mathrm{SO}_{2}}^2 \times P_{\mathrm{O}_{2}}} = 0.13 \] The pressures of \(\mathrm{SO}_{2}\), \(\mathrm{O}_{2}\) and \(\mathrm{SO}_{3}\) at equilibrium can be expressed as: \(P_{\mathrm{SO}_{2}}\), \(P_{\mathrm{O}_{2}}\) and \(P_{\mathrm{SO}_{3}}\) are proportional to their mole ratios.

Solve for Total Pressure

Substitute the above results (from Step 4) into the \(K_p\) equation, and solve for total pressure (\(P_{Total}\)). The total pressure is the sum of the individual pressures at equilibrium, i.e., \(P_{Total} = P_{\mathrm{SO}_{2}} + P_{\mathrm{O}_{2}} + P_{\mathrm{SO}_{3}}\).

Key concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant, often denoted as \(K\), is a fundamental concept. It describes the ratio of the concentrations of products to reactants in a reaction at equilibrium. For reactions in the gas phase, specifically, we use \(K_P\) to denote the equilibrium constant expressed in terms of partial pressures of the gases involved.

The equilibrium constant for the reaction \(2\,\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\,\mathrm{SO}_3(g)\) is given as 0.13 at around \(830^\circ\, \mathrm{C}\). This indicates how balanced the formation of sulfur trioxide (\(\mathrm{SO}_3\)) and its decomposition into sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen (\(\mathrm{O}_2\)) is under specific conditions.

### Understanding \(K_P\)- A small \(K_P\) (like 0.13) indicates that at equilibrium, the reactants are favored over the products.- It helps in determining how far the reaction will proceed in the direction of the formation of \(\mathrm{SO}_3\).

This constant assists chemists in predicting the amounts of each substance present at equilibrium under certain conditions and is vital for industrial processes like the manufacture of sulfuric acid.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's based on the conservation of mass and the concept that matter cannot be created or destroyed in chemical processes.

### Reaction EquationFor the reaction \[2\,\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\,\mathrm{SO}_3(g)\], the stoichiometric coefficients tell us the ratio in which reactants and products participate. Here, two moles of sulfur dioxide react with one mole of oxygen to produce two moles of sulfur trioxide.

### Yield CalculationTo find the yield, if you start with 2.00 moles of \(\mathrm{SO}_2\) and 2.00 moles of \(\mathrm{O}_2\), and aim for an 80% yield of \(\mathrm{SO}_3\), you have:

• 80% of 2.00 moles \(\mathrm{SO}_2\) = 1.60 moles of \(\mathrm{SO}_3\).

At equilibrium, this means there are adjustments in the quantities of all species involved, demonstrating stoichiometry in balancing the reaction using theoretical and practical yields.

Gas Phase Reactions

Gas phase reactions involve reactants and products all in the gaseous state. These reactions are often sensitive to changes in pressure and temperature, which can influence the position of equilibrium.

### Characteristics- Gases are compressible, so changes in total pressure can shift equilibrium.- The partial pressure of each gas contributes to the total pressure, and these are linked to the mole fraction of each component.

In the example of \(2\,\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\,\mathrm{SO}_3(g)\), solving for total pressure at equilibrium involves:

• Using the stoichiometry of the reaction to find the partial pressures based on changes in mole numbers.
• Affecting these values via initial mole amounts and reaction yields to determine equilibrium positions.

Understanding these aspects helps in predicting and controlling the outcomes of gas phase reactions, which is crucial in both laboratory and industrial chemical processes.