Question

The key step in the manufacture of sulfuric acid is the oxidation of sulfur dioxide in the presence of a catalyst, such as \(\mathrm{V}_{2} \mathrm{O}_{5}\) At \(727^{\circ} \mathrm{C}, 0.010 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) is injected into an empty \(2.00-\mathrm{L}\) container \(\left(K_{p}=3.18\right)\) (a) What is the equilibrium pressure of \(\mathrm{O}_{2}\) that is needed to maintain a \(1 / 1\) mole ratio of \(\mathrm{SO}_{3}\) to \(\mathrm{SO}_{2} ?\) (b) What is the equilibrium pressure of \(\mathrm{O}_{2}\) needed to maintain a \(95 / 5\) mole ratio of \(\mathrm{SO}_{3}\) to \(\mathrm{SO}_{2} ?\)

Short answer

(a) The equilibrium pressure of O_2 is approximately 0.314 atm. (b) The equilibrium pressure of O_2 is approximately 113.77 atm.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the oxidation of sulfur dioxide to sulfur trioxide is:\[2 \text{SO}_2 (g) + \text{O}_2 (g) \rightarrow 2 \text{SO}_3 (g)\]

Establish the equilibrium expression for the reaction

Using the balanced equation, set up the equilibrium expression for the reaction:\[K_p = \frac{P_{SO_3}^2}{(P_{SO_2}^2) \cdot (P_{O_2})}\]Given \(K_p = 3.18\).

Calculate the initial pressures of the gases

The initial pressure of \(\text{SO}_2\) injected into the container can be calculated using the ideal gas law:\[P_{SO_2,\text{initial}} = \frac{nRT}{V}\]Given: \[ n = 0.010 \text{ mol}, \text{ } T = 727 \text{ } \degree \text{C} = 1000 \text{ K}, \text{ } V = 2.00 \text{ L}, \text{ } R = 0.0821 \text{ L·atm/mol·K} \]Therefore, \[ P_{SO_2,\text{initial}} = \frac{0.010 \times 0.0821 \times 1000}{2.00} = 0.4105 \text{ atm}\]

Find the equilibrium pressures for part (a)

To maintain a 1:1 mole ratio of \(\text{SO}_3\) to \(\text{SO}_2\) at equilibrium, they must have the same partial pressure, denoted as \(P_{SO_3} = P_{SO_2} \ P_{SO_2} = 0.4105 \text{ atm}\).Using the equilibrium expression, solve for \(P_{O_2}\):\[K_p = 3.18 = \frac{P_{SO_3}^2}{(P_{SO_2}^2) \cdot (P_{O_2})} = \frac{(P_{SO_2})^2}{(P_{SO_2})^2 \cdot (P_{O_2})} = \frac{P_{SO_2}^2}{P_{SO_2}^2 \times P_{O_2}} = \frac{P_{SO_2}^2}{P_{SO_2}^2 \times P_{O_2}}\P_{O_2} = 1 / 3.18 \approx 0.314 \text{ atm}\]

Find the equilibrium pressures for part (b)

To maintain a 19:1 mole ratio of \(\text{SO}_3\) to \(\text{SO}_2\) at equilibrium, let’s assume \(P_{SO_3} = 0.95x\) and \(P_{SO_2} = 0.05x\). Then solve for \(x\):\[K_p = 3.18 = \frac{(0.95x)^2}{(0.05x)^2 \cdot P_{O_2}} = \frac{0.9025x^2}{0.0025x^2 \cdot P_{O_2}}\]Simplify to find \(P_{O_2}\):\[3.18 = \frac{0.9025}{0.0025 \cdot P_{O_2}} \Rightarrow P_{O_2} = \frac{0.9025}{3.18 \times 0.0025} = \approx 113.77 \text{ atm}\]

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse chemical reactions are equal.
In the context of sulfuric acid production, it involves the reaction:

2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
Once equilibrium is reached, the concentrations of reactants and products remain constant over time.
The equilibrium constant, denoted as Kp for gases, helps us understand the relationship between these concentrations or partial pressures.

The equilibrium expression for our reaction is:

\[K_p = \frac{P_{SO_3}^2}{(P_{SO_2}^2) \cdot (P_{O_2})} \]
This equation shows us how the partial pressures of the gases relate to one another at equilibrium.

Ideal Gas Law

The ideal gas law is crucial for calculating the initial pressures of gases in a reaction.
It is expressed as:

\[ PV = nRT \]
where P is pressure, V is volume, n is moles of gas, R is the gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin.
in our example, SO₂ is injected into a container, we first use this law to determine the pressure:
\[ P_{SO_2,\text{initial}} = \frac{nRT}{V} \]
With 0.010 mol of SO₂, at 1000 K, in a 2.00 L container, we find:
\[ P_{SO_2,\text{initial}} = \frac{0.010 \times 0.0821 \times 1000}{2.00} = 0.4105 \text{ atm} \]

Catalysis

Catalysis speeds up the chemical reaction without being consumed in the process.
In the production of sulfuric acid, vanadium(V) oxide (V₂O₅) is used as a catalyst.
It helps convert sulfur dioxide (SO₂) to sulfur trioxide (SO₃) quickly.
Catalysts are essential in industrial processes, making reactions more efficient.

They work by providing an alternative reaction pathway with a lower activation energy.
This means more molecules have enough energy to react when they collide.
In summary, catalysts like V₂O₅ enable the large-scale production of sulfuric acid by making the conversion of SO₂ to SO₃ faster and more energy-efficient.

Partial Pressures

Partial pressure is the pressure exerted by an individual gas in a mixture of gases.

In our problem, we deal with the partial pressures of SO₂, O₂, and SO₃.
At equilibrium, the total pressure is the sum of the partial pressures of all gases present.
Partial pressures help us understand how each gas contributes to the overall system.

For instance, we calculated the initial SO₂ pressure using the ideal gas law, and then used the equilibrium expression to determine O₂ partial pressure.
When the ratio of SO₃ to SO₂ is 1:1, we found:
\[P_{O_2} = \frac{P_{SO_2}^2}{K_p \times P_{SO_3}^2} = 0.314 \text{ atm} \]

This helps us ensure that the reaction proceeds with the desired balance of gases.

Mole Ratio Calculations

Mole ratio calculations are vital for understanding the relationships between reactants and products in chemical reactions.
We use the balanced chemical equation to determine these ratios.
In our problem, the ratio of SO₃ to SO₂ tells us how much O₂ is needed:

• For a 1:1 ratio (SO₃:SO₂), we maintained equal partial pressures
• For a 95:5 ratio (SO₃:SO₂), we used:
  \[K_p = 3.18 = \frac{(0.95x)^2}{(0.05x)^2 \times P_{O_2}} = \frac{0.9025x^2}{0.0025x^2 \times P_{O_2}} \]
  and solved for the oxygen partial pressure:
  \[P_{O_2} = \frac{0.9025}{3.18 \times 0.0025} \approx 113.77 \text{ atm} \]
  Understanding mole ratios helps ensure the reaction yields the desired product ratios efficiently.