Question

4 moles each of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are allowed to react to form \(\mathrm{SO}_{3}\) in a closed vessel. At equilibrium, \(25 \%\) of \(\mathrm{O}_{2}\) is used up. The total number of moles of all the gases at equilibrium is a. \(2.0\) b. \(6.0\) c. \(7.0\) d. \(8.0\)

Short answer

c. 7.0

Step-by-step solution

Understand the Reaction

The chemical reaction is: \[ 2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2\mathrm{SO}_{3(g)} \]. This equation shows that 2 moles of \(\mathrm{SO}_2\) react with 1 mole of \(\mathrm{O}_2\) to form 2 moles of \(\mathrm{SO}_3\).

Determine Initial Moles

Initially, we have 4 moles of \(\mathrm{SO}_2\) and 4 moles of \(\mathrm{O}_2\).

Calculate Usage of \(\mathrm{O}_2\)

Since 25% of \(\mathrm{O}_2\) is used up, we will calculate the moles of \(\mathrm{O}_2\) reacted: \[ 4 \text{ moles of } \mathrm{O}_2 \times 0.25 = 1 \text{ mole reacted} \].

Determine Change in \(\mathrm{SO}_2\) and \(\mathrm{SO}_3\)

According to the stoichiometry of the reaction, 1 mole of \(\mathrm{O}_2\) will react with 2 moles of \(\mathrm{SO}_2\) to form 2 moles of \(\mathrm{SO}_3\).

Calculate Equilibrium Moles

At equilibrium, the moles of each substance are:\(\mathrm{SO}_2\): \(4 - 2 = 2 \text{ moles}\)\(\mathrm{O}_2\): \(4 - 1 = 3 \text{ moles}\)\(\mathrm{SO}_3\): \(0 + 2 = 2 \text{ moles}\)

Find Total Moles at Equilibrium

Add the equilibrium moles of all gases:\[ 2 \text{ moles of } \mathrm{SO}_2 + 3 \text{ moles of } \mathrm{O}_2 + 2 \text{ moles of } \mathrm{SO}_3 = 7 \text{ moles}\]

Key concepts

Stoichiometry in Chemical Reactions

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It's essentially the "math" behind chemical equations. In our example, the reaction involves \(2\mathrm{SO}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{SO}_{3}\).
This equation tells us the exact proportions in which the substances react. For every 2 moles of \mathrm{SO}_{2} that react with 1 mole of \mathrm{O}_{2}, 2 moles of \mathrm{SO}_{3} are produced.
The beauty of stoichiometry lies in its predictability; knowing the amount of one substance allows us to calculate the others.

In this exercise, the stoichiometric relationship helped us determine that 1 mole of \mathrm{O}_{2} reacting means 2 moles of \mathrm{SO}_{2} are consumed, which in turn produces 2 moles of \mathrm{SO}_{3}. These calculations ensure the reaction complies with the chemical law of conservation of mass.

Understanding Reaction Mechanisms

A reaction mechanism is the step-by-step description of how a chemical reaction occurs on the molecular level. It tells us the series of steps that transform reactants into products. This is crucial to differentiate simple reactants and products from what actually happens.
In this exercise, the mechanism shows that \(2\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) react directly to form \(2\mathrm{SO}_{3}\).

Mechanisms often involve the formation of short-lived intermediates or reaction intermediates that you don't see in the overall balanced equation. Thus, the mechanism can include several individual reactions that sum to give the overall reaction.
Knowing the mechanism gives us deep insights into the rate of reaction, which parts of the reaction are the slowest, and any potential for side reactions. This enables chemists to control conditions in practical applications for better yield or selectivity of the desired product.

Equilibrium Constant and Its Importance

The equilibrium constant provides insight into the composition of a reaction mixture at equilibrium, crucial for predicting the direction of the reaction. In a chemical equilibrium scenario, the forward and reverse reactions occur at the same rate, so concentrations remain constant.
The expression for the equilibrium constant (\(K_{eq}\)) is derived from the reaction:

• For \(2\mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2\mathrm{SO}_{3}\),
• \(K_{eq} = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2 [\mathrm{O}_2]}\),

based on the reactants and products' molar concentrations at equilibrium. A large \(K_{eq}\) value suggests that the equilibrium is far to the right, meaning more products are present.

In our example, knowing the equilibrium constant allows the prediction of the moles of various gases at equilibrium, thus providing validation for the calculations performed.

Gas Reactions and Volume Considerations

Gas reactions, like the one discussed here, behave under the unique rules of gas laws like the Ideal Gas Law. These reactions often occur in closed systems where pressure and volume may change.
In our context, a closed vessel implies that gas moles at equilibrium involve considerations of both volume and pressure as described by \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature.

Gas reactions are particularly sensitive to changes in conditions and may behave dramatically differently under varying temperatures or pressures. Understanding how moles of gases change in such a scenario allows for manipulation of reaction conditions.
This explains why, despite changes due to reaction advancement, the total number of moles was calculated to confirm the system's behavior accurately at equilibrium.