Problem 51 - Chemistry

Question

A 1.00-L flask was filled with 2.00 moles of gaseous \(\mathrm{SO}_{2}\) and 2.00 moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K,\) for this reaction.

Step-by-Step Solution

Answer

The equilibrium constant, \(K\), for the given reaction is 3.45 when 1.30 moles of NO are present at equilibrium.

Step 1: Write the balanced chemical equation

The balanced equation for the given reaction is: \[\mathrm{SO}_{2(g)} + \mathrm{NO}_{2(g)} \rightleftharpoons \mathrm{SO}_{3(g)} + \mathrm{NO}_{(g)}\]

Step 2: Calculate the initial moles of each substance

According to the problem, initially there are: - 2.00 moles of \(\mathrm{SO}_{2}\) - 2.00 moles of \(\mathrm{NO}_{2}\) - 0 moles of \(\mathrm{SO}_{3}\) (Since they were not present initially) - 0 moles of \(\mathrm{NO}\) (Since they were not present initially)

Step 3: Calculate the change in moles for each substance

At equilibrium, there are 1.30 moles of NO. So, the change in moles of each substance can be represented as follows: - \[\Delta\, \mathrm{SO}_{2} = -x\] - \[\Delta\, \mathrm{NO}_{2} = -x\] - \[\Delta\, \mathrm{SO}_{3} = x\] - \[\Delta\, \mathrm{NO} = x\] Where \(x\) is the moles of \(\mathrm{SO}_{3}\) and \(\mathrm{NO}\) formed and the moles of \(\mathrm{SO}_{2}\) and \(\mathrm{NO}_{2}\) that reacted.

Step 4: Calculate the equilibrium moles of each substance and their concentrations

At equilibrium, the moles of each substance are: - Moles of \(\mathrm{SO}_{2} = 2.00 - x\) - Moles of \(\mathrm{NO}_{2} = 2.00 - x\) - Moles of \(\mathrm{SO}_{3} = x\) - Moles of \(\mathrm{NO} = x\) The volume of the container is 1.00 L, so the equilibrium concentrations can be found by dividing the moles by the volume of the container: - \[[\mathrm{SO}_{2}] = \frac{2.00 - x}{1.00\mathrm{L}} = 2.00 - x\] - \[[\mathrm{NO}_{2}] = \frac{2.00 - x}{1.00\mathrm{L}} = 2.00 - x\] - \[[\mathrm{SO}_{3}] = \frac{x}{1.00\mathrm{L}} = x\] - \[[\mathrm{NO}] = \frac{x}{1.00\mathrm{L}} = x\]

Step 5: Write the expression for the equilibrium constant

The equilibrium constant, \(K,\) can be expressed as: \[K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]}\] Substitute the equilibrium concentrations of each substance into the equation for \(K\): \[K = \frac{(x)(x)}{(2.00 - x)(2.00 - x)}\]

Step 6: Calculate the value of \(x\)

Since we know that at equilibrium, there are 1.30 moles of NO present, we can use this information to find the value of \(x\): \[x = 1.30\]

Step 7: Calculate the value of the equilibrium constant \(K\)

Substitute the value of \(x\) back into the equation for \(K\): \[K = \frac{(1.30)(1.30)}{(2.00 - 1.30)(2.00 - 1.30)}\] \[K = \frac{1.69}{0.49}\] Calculate the value of \(K\): \[K = 3.45\] The equilibrium constant, \(K\), for this reaction is 3.45.

Key Concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the forward and backward reactions happen at the same rate. In simpler terms, it's like a seesaw that remains balanced. When a reaction reaches equilibrium, no observable change is seen in the concentrations of substances involved, even though the reactions are still occurring.

Key things to remember about chemical equilibrium:

• The equilibrium state is dynamic, meaning reactions still continue but cancel each other out.
• The concentrations of reactants and products do not need to be equal; they just remain constant.
• Equilibrium can be achieved regardless of the starting concentrations.

In the exercise, the reaction reaches equilibrium when 1.30 moles of NO is present. It indicates that the system has balanced the forward and reverse reactions for the given concentrations of SO\(_2\), NO\(_2\), SO\(_3\), and NO.

Decoding Reaction Stoichiometry

Reaction stoichiometry involves the calculation of reactants and products in chemical reactions. To understand stoichiometry, it's essential to start with a balanced chemical equation. This equation represents the conservation of mass, indicating that matter cannot be created or destroyed.

In this particular exercise, the balanced chemical equation is:
\[\mathrm{SO}_{2}(g) + \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) + \mathrm{NO}(g)\]

This equation shows reactants on the left and products on the right. Each molecule's coefficient reflects the quantity involved in the reaction. Here, each reactant reacts in a 1:1 ratio with another reactant to produce products in a 1:1 ratio. Hence, for every mole of \(\mathrm{SO}_{2}\) consumed, one mole of \(\mathrm{SO}_{3}\) is produced, and similarly for \(\mathrm{NO}_{2}\) and \(\mathrm{NO}\).

Understanding stoichiometry allows us to calculate changes during reactions, such as the initial and equilibrium moles needed to find the equilibrium constant \(K\). In our case, knowing that \(x\) moles of \(\mathrm{NO}\) formed helped us to identify the other reactants' changes.

Mastering Concentration Calculations

Concentration calculations are crucial for understanding chemical reactions, especially when determining the equilibrium constant. Concentration describes how much of a substance exists in a given volume and is often measured in moles per liter (Molarity, M).

For the given reaction, the concentration of each substance can be calculated using the formula:
\[[A] = \frac{\text{moles of } A}{\text{Volume in Liters}}\]

The exercise informs us of the volume as 1.00 L, simplifying our concentration calculations, as the concentration equals the number of moles when the volume is 1 L:

• SO\(_2\): \(2.00 - x\)
• NO\(_2\): \(2.00 - x\)
• SO\(_3\): \(x\)
• NO: \(x\)

With \(x = 1.30\) moles, these values directly translate to concentrations. These concentrations are then plugged into the equilibrium expression, \(K = \frac{[\mathrm{SO}_3][\mathrm{NO}]}{[\mathrm{SO}_2][\mathrm{NO}_2]}\), to calculate the equilibrium constant \(K\). In this reaction, solving gives us \(K = 3.45\), simplifying these complex interactions into a single, measurable value.