Question

Consider the decomposition equilibrium for dinitrogen pentoxide: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ At a certain temperature and a total pressure of \(1.00 \mathrm{~atm}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0\), will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to \(0.50 \%\) ? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of \(10.0 .\)

Short answer

After increasing the volume by a factor of 10, the mole percent of decomposed dinitrogen pentoxide at equilibrium will be approximately \(25.3\%\).

Step-by-step solution

Le Chatelier's Principle Prediction

Using Le Chatelier's principle, we can predict how the equilibrium system will shift in response to a change in the volume of the container. When the volume is increased, the system will shift in the direction of the side with more moles of gas to minimize the change in pressure. In our case, there are 4 moles of gas on the right side (4 moles of NO₂ and 1 mole of O₂) and only 2 moles on the left side (2 moles of N₂O₅). Therefore, when the volume is increased by a factor of 10, the equilibrium will shift to the right (towards the decomposition) to minimize the change in pressure. As a result, we can answer part (a) of the question: the mole percent of decomposed dinitrogen pentoxide will be greater than 0.50% after increasing the volume by the factor of 10.

Set Up an ICE Table

To calculate the new equilibrium concentrations, we will use an ICE (initial-change-equilibrium) table. The initial concentrations can be represented as: - [N₂O₅] = 1 - \(0.005\) - [NO₂] = 0 - [O₂] = 0 Since the volume is increased by a factor of 10, the initial concentrations will be divided by 10: - [N₂O₅] = (1 - \(0.005\)) / 10 = \(0.0995\) - [NO₂] = 0 - [O₂] = 0 Now, let's denote x as the change in concentrations: - [N₂O₅] will decrease by 2x due to the stoichiometry of the balanced reaction. - [NO₂] will increase by 4x as a result of the decomposition. - [O₂] will increase by x due to its stoichiometry in the balanced reaction. At equilibrium, the concentrations will be: - [N₂O₅] = \(0.0995 - 2x\) - [NO₂] = \(4x\) - [O₂] = \(x\)

Calculate the New Equilibrium Concentrations

Now we can write the reaction quotient, Qp, as follows: \[Qp = \frac{[NO_2]^4[O_2]}{[N_2O_5]^2}\] Since the initial total pressure is 1.00 atm, we can convert it to partial pressures of the reactants and products. Recall that \(Qp = Kp\) at equilibrium. Then, we have: \[Kp = \frac{(4x)^4(x)}{(0.0995 - 2x)^2} = 1.00\] We need to solve the above equation for x to calculate the new equilibrium concentrations. Solving this equation can be quite challenging algebraically. However, we can get a rough estimation of x by first assuming that the mole ratio of 2N₂O₅(g) → 4NO₂(g) + O₂(g) is small (since we expect x to be relatively small due to the 0.50% mole percent of decomposition). Then, we have: \[Kp = \frac{(4x)^4(x)}{(0.0995)^2} = 1.00\] Now, we can solve for x: \[x = \sqrt[5]{\frac{1.00 \times (0.0995)^2}{(4)^4}} \approx 0.0126\]

Calculate the Mole Percent of Decomposed N₂O₅

With the value of x, we can now determine the mole percent of decomposed dinitrogen pentoxide. The mole percent decomposed is given by: Mole percent decomposed = 100% × \(\frac{amt.~decomposed}{amt.~initial}\) Mole percent decomposed = 100% × \(\frac{2x}{0.0995}\) Mole percent decomposed = 100% × \(\frac{2 * 0.0126}{0.0995}\) ≈ \(25.3\%\) So, after increasing the volume by a factor of 10, the mole percent of decomposed dinitrogen pentoxide at equilibrium will be approximately \(25.3\%\).

Key concepts

chemical equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes a state where the concentrations of reactants and products remain constant over time. This happens because the rates of the forward and reverse reactions are equal. In our specific exercise involving dinitrogen pentoxide (\(\mathrm{N}_{2}\mathrm{O}_{5}\)), the equilibrium represents a balance between the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{5}\) into nitrogen dioxide (NO₂) and oxygen (O₂).

Key characteristics of chemical equilibrium include:

• The system is dynamic, not static—molecules are constantly reacting, but there is no net change in concentrations.
• It can be influenced by changes in conditions like pressure, temperature, and concentration.
• At equilibrium, the reaction quotient ('\(Q\)') equals the equilibrium constant ('\(K\)').

Understanding this balance helps predict how the system will respond to changes, such as a volume change in this exercise.

reaction quotient

The reaction quotient, denoted as '\(Q\)', helps predict the direction a reaction will proceed to reach equilibrium. It's calculated using the same expression as the equilibrium constant (\(K\)), but with the current, nonequilibrium concentrations of reactants and products.

In our exercise scenario, the decomposition of dinitrogen pentoxide into nitrogen dioxide and oxygen can be represented by \[Qp = \frac{[NO_2]^4[O_2]}{[N_2O_5]^2}\]. The value of \(Qp\) determines if the reaction is at equilibrium (\(Q = K\)), or if it needs to precede in the forward or reverse direction to reach equilibrium:

• If \(Q < K\), the reaction will shift to the right, forming more products.
• If \(Q > K\), the reaction will shift to the left, forming more reactants.
• If \(Q = K\), the system is already at equilibrium.

This useful tool, the reaction quotient, facilitates understanding of the reaction’s status and potential shifts due to external changes.

ICE table

An ICE table, which stands for Initial, Change, and Equilibrium, is a structured way to keep track of changes in concentration for chemical reactions. It's particularly useful in equilibrium problems, like the one we're considering with dinitrogen pentoxide.

Here's how an ICE table works:

• Initial: List initial concentrations of all species involved in the reaction. For instance, initially, \([N_2O_5] = 0.0995\) while \([NO_2]\) and \([O_2]\) start at zero after volume change reduces concentrations.
• Change: Define changes in concentration using variables, often assuming a change of '\(x\)'. Our exercise indicates changes like \(-2x\) for \([N_2O_5]\), as it decomposes.
• Equilibrium: Determine the concentrations at equilibrium by adding the initial and change values.

By organizing information this way, one can systematically determine equilibrium concentrations, critical for calculating \(Q\) or \(K\).

mole percent decomposition

Mole percent decomposition expresses the extent of a reaction in terms of the percentage of a particular reactant that has been converted to products. In our exercise, we are particularly looking at how much \(\mathrm{N}_{2}\mathrm{O}_{5}\) has decomposed into \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\).

This concept is simply the ratio of the amount of decomposed reactant to the original amount, multiplied by 100 to convert it into a percentage. The formula for mole percent decomposition is:

• Mole percent decomposed = \(100\% \times \frac{\text{amount decomposed}}{\text{initial amount}}\)

In the exercise, by solving the equilibrium expressions, the mole percent of decomposed \(\mathrm{N}_{2}\mathrm{O}_{5}\) changes upon a volume increase, climbing from 0.50% to approximately 25.3%. This change is influenced by applying Le Chatelier’s Principle, which in this case, prompts a shift towards more decomposition due to increased volume.