Problem 9 - General Chemistry: Principles and Modern Applications

Question

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

Step-by-Step Solution

Answer

The total pressure when the reaction has gone to completion is \(3.000 \times 10^{3} \mathrm{mmHg}\). The total pressure when the partial pressure of \(A(g)\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg}\) is \(1.600 \times 10^{3} \mathrm{mmHg}\).

Step 1: Understand the relationship between reactant and products

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g)\), for every molecule of \(A(g)\) that reacts, it produces two molecules of \(B(g)\) and one molecule of \(C(g)\). Therefore, for every decrease in the pressure of \(A(g)\), the total pressure increases by an equivalent of three times that decrease (This is due to the increase of two molecules of \(B(g)\) plus one molecule of \(C(g)\)).

Step 2: Calculating the total pressure when reaction has gone to completion

If the reaction is complete, all the \(A(g)\) will be converted into \(B(g)\) and \(C(g)\). Therefore the pressure of \(A(g)\) will be 0. Since there was a total reduction of \(1.000 \times 10^{3} \mathrm{mmHg}\) in \(A(g)\), the total pressure increases by three times this which gives a total pressure of \(1.000 \times 10^{3} \times 3 \mathrm{mmHg}\) or \(3.000 \times 10^{3} \mathrm{mmHg}\) after the reaction.

Step 3: Determining the total pressure when partial pressure of \(A(g)\) has dropped to \(8.00 \times 10^{2} mmHg\)

In this case, the decrease in the pressure of \(A(g)\) is \(1.000 \times 10^{3}\) \(mmHg\) - \(8.00 \times 10^{2}\) \(mmHg\) = \(2.00 \times 10^{2} mmHg\). Applying the 1:3 ratio as identified in Step 1, the total pressure will increase by three times this decrease. Therefore, the total pressure in this case is equal to the initial pressure \(1.000 \times 10^{3} mmHg\) plus the increase which is \(2.00 \times 10^{2} \times 3 mmHg\) or \(6.00 \times 10^{2} mmHg\). Adding these together gives a total pressure of \(1.600 \times 10^{3} mmHg\).

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. In this process, reactants are converted to products. For the reaction provided, \(A(g) \rightarrow 2 B(g) + C(g)\), one molecule of \(A\) breaks down to form two molecules of \(B\) and one molecule of \(C\).

This specific reaction highlights a common principle: the number and type of product molecules can significantly differ from the reactants. Understanding this helps in predicting the conditions and outcomes in a closed system. Here, for every one molecule of \(A\) consumed, three new gas molecules emerge, increasing the gas molecules.

• This fundamental understanding above is crucial: reactants are transformed into products with specific stoichiometric relationships.
• In gas reactions, this transformation impacts gas pressure in a closed system, often increasing or decreasing total pressure depending on the moles of gas produced.

Consistency in recognizing these transformations, like increase in gas pressure when more products are formed, helps understand reaction behavior and effects.

Partial Pressure

Partial pressure refers to the pressure that one component of a gas mixture would exert if it alone occupied the entire volume. In the context of our reaction, as \(A(g)\) is used up and converted to \(B(g)\) and \(C(g)\), its partial pressure decreases, while the partial pressures of \(B\) and \(C\) increase.

Each gas in a mixture contributes to the total pressure proportionally to its partial pressure. The changes in partial pressure can be guided by Dalton's Law of Partial Pressures which states:

• The total pressure of a gas mixture is the sum of the partial pressures of each individual gas.
• In a closed system, as one gas's partial pressure decreases (like \(A(g)\)), the others' increase based on the change in moles of gases produced.

Understanding partial pressure changes gives insight into gas behavior and progress of reactions in a constant volume system. You'll see that as \(A(g)\) decreases, the total pressure increases due to the production of more gas molecules as shown in our example.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. It lets us calculate how much of a reactant is needed to produce a desired amount of product, or in this case, the change in pressure as a result of the reaction. In our example, \(A(g) \rightarrow 2B(g) + C(g)\), stoichiometry helps calculate the changes in pressure.

For every mole of \(A(g)\) reacting, three moles of products are created. Understanding this ratio (1:3) is invaluable for determining outcomes like total pressure:

• Initial changes in pressure of \(A(g)\) directly correlate to increased total pressure because 3 mole of gas emerges for every mole of \(A\) used up.
• When \(A(g)\) is fully utilized, knowing the reaction stoichiometry helps predict final pressure and state.
• Using stoichiometry allows for detailed predictions about quantity changes in reaction stages.

Grasping stoichiometric calculations helps interpret how physical quantities like pressure evolve as a reaction progresses, as it did in both completion and partial stages of our exercise's reaction.