Problem 108 - Chemistry An Atoms First Approach

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 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

Step-by-Step Solution

Answer

a. The mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium will be greater than 0.50\% when the volume increases by a factor of 10.0, as the equilibrium will shift towards the products side, which has more moles of gas. b. After solving the equation \(\frac{(0.01)^4 \cdot 0.005}{(0.995)^2} = \frac{(4x)^4 \cdot x}{(1-x)^2}\) for x, convert it back to mole percent by multiplying by 100. This will be the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium after increasing the volume by a factor of 10.0.

Step 1: Analyze the equilibrium reaction based on the given information

The given decomposition equilibrium for dinitrogen pentoxide is: \[2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2(g)}+\mathrm{O}_{2}(g)\] At a certain temperature and a total pressure of 1.00 atm, the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is 0.50\% decomposed (by moles) at equilibrium.

Step 2: Determine the effect of increasing the volume on the equilibrium system

When the volume of the system increases, the pressure decreases. According to Le Chatelier's principle, the equilibrium will shift in the direction that opposes the change. In this case, it will shift towards the side with more moles of gas since this will help increase the pressure. Since there are more moles of gas on the product side (4 moles of NO2 and 1 mole of O2) than on the reactant side (2 moles of N2O5), the equilibrium will shift towards the products when the volume is increased. a. Therefore, the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium will be greater than 0.50\% when the volume increases by a factor of 10.0.

Step 3: Set up the equilibrium constant expression

Because the percent decomposition is given, we can use partial pressures in the equilibrium constant expression: \[K_p=\frac{P_{NO_2}^4 P_{O_2}}{(P_{N_2O_5})^2}\]

Step 4: Calculate the new mole percent of decomposed N2O5

We start with the initial conditions: - N2O5: \(1.00\,atm - 0.005(1.00\,atm) = 0.995\,atm\) (as only 0.50% is decomposed at equilibrium) - NO2: \(0.005(1.00\,atm) \times 2 = 0.01\,atm\) (2 moles of NO2 are formed for every mole of N2O5 decomposed) - O2: \(0.005(1.00\,atm) = 0.005\,atm\) We now increase the volume by a factor of 10: - New total pressure will be \(0.1\,atm\) - N2O5: \(1 - x\) - NO2: \(4x\) - O2: \(x\) Apply the above (changing the pressure from atm to the fractional change x) in the equilibrium constant expression: \[K_p = \frac{(4x)^4 \cdot x}{(1-x)^2}\] Since the system's temperature does not change and the reaction is still in equilibrium, the value of \(K_p\) stays the same. Solve for x: \[K_{p_{initial}} = K_{p_{final}} \Longrightarrow \frac{(0.01)^4 \cdot 0.005}{(0.995)^2} = \frac{(4x)^4 \cdot x}{(1-x)^2}\] Now, solve this equation for x (using algebraic techniques or numerical methods) to calculate the new mole percent of decomposed N2O5. b. After calculating the value of x, convert it back to mole percent by multiplying by 100. This will be the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium after increasing the volume by a factor of 10.0.

Key Concepts

Le Chatelier's Principle

Le Chatelier's principle is a cornerstone concept in chemistry that predicts how a change in conditions can shift a chemical equilibrium. When a system at equilibrium is disturbed, it will adjust in a way that counteracts the change and re-establishes equilibrium. Common disturbances include changes in concentration, pressure, volume, or temperature.

For example, consider the equilibrium involving dinitrogen pentoxide (N2O5) and its decomposition products, nitrogen dioxide (NO2) and oxygen (O2). If we increase the volume of the container housing this gaseous equilibrium, the principle tells us that the reaction will favor the production of more gas molecules to occupy the new space, thus reducing the pressure. In this case, because the number of gas molecules increases as N2O5 decomposes, an increase in volume would lead to a higher degree of decomposition.

To visualize it, imagine standing in a crowded room (the original volume), and then walls suddenly move outward to make the room bigger (increase in volume). Intuitively, the people (gas molecules) will spread out to fill the additional space. Similarly, by increasing the volume of the reaction vessel, you encourage more N2O5 to decompose, producing a greater number of NO2 and O2 molecules to fill the expanded volume.

Equilibrium Constant Expression

The equilibrium constant expression (K) is a quantitative measure of the position of equilibrium. It is defined for a particular reaction at a given temperature and relates the concentrations (or partial pressures in gaseous systems) of the products to those of the reactants.

In the case of the decomposition of dinitrogen pentoxide (N2O5), the equilibrium constant expression can be written in terms of partial pressures (since all the species involved are gases) as follows:\[K_p = \frac{P_{NO_2}^4 P_{O_2}}{(P_{N2O5})^2}\]
The equilibrium constant remains unchanged as long as the temperature is constant. If the volume is changed (as in our example problem), the pressures of all gases change, but the ratio defined by the equilibrium constant expression does not—this is key to predicting the system's behavior.

In your homework problem, if the volume is increased by a factor of 10, the individual pressures of each gas decrease, but the ratio of their pressures as expressed in the equilibrium constant must stay the same since temperature is constant. This requirement dictates how much N2O5 will decompose to achieve the same constant value for Kp in the new conditions.

Decomposition Equilibrium

Decomposition equilibrium refers to a particular type of chemical equilibrium involving the breaking down of a compound into simpler substances. The equilibrium lies between the forward reaction, where the compound decomposes, and the reverse reaction, where the products recombine to form the original compound.

In our example, we are analyzing the decomposition equilibrium of dinitrogen pentoxide into nitrogen dioxide and oxygen:\[2 N2O5(g) \rightleftharpoons 4 NO2(g) + O2(g)\]
This reaction is dynamic, meaning even at equilibrium, both the forward and reverse reactions are still occurring; they just occur at the same rate, so there's no net change in the concentrations of reactants and products. The mole percent of N2O5 decomposed tells us the extent to which the reaction has proceeded towards the products side.

The exercise involved calculating a change in this percentage when the volume in which the reaction occurs is increased. According to Le Chatelier's principle, as discussed, the increased volume favors the forward reaction and increases the decomposition. The calculated mole percent reflects the new equilibrium position, representing the response of the decomposition equilibrium to the imposed change.