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 \(10.0 .\)

Short answer

When the volume is increased by a factor of 10.0, the mole percent of N2O5 decomposed at equilibrium will be greater than 0.5% due to the shifting of the equilibrium towards the right, according to Le Chatelier's principle. However, the exact mole percent cannot be calculated without more information.

Step-by-step solution

Balanced equation and reaction quotient

The balanced equation for the decomposition of dinitrogen pentoxide (N2O5) is given as: 2 N2O5(g) ⇌ 4 NO2(g) + O2(g) The reaction quotient (Q) for this reaction can be written as: Q = \(\frac{[NO_2]^4\cdot[O_2]}{[N_2O_5]^2}\) **Step 2: Analyze the effect of increasing volume on the equilibrium position using Le Chatelier's principle**

Effect of volume change on equilibrium

According to Le Chatelier's principle, if an external stress is applied to a system at equilibrium, the system will adjust itself to minimize the effect of the stress and re-establish equilibrium. In this case, increasing the volume of the system causes a decrease in pressure. Since there are more moles of gas on the right-hand side of the balanced equation (5 moles) than on the left-hand side (2 moles), the equilibrium will shift towards the right to counteract the decrease in pressure. Therefore, when the volume is increased by a factor of 10.0, the mole percent of N2O5 decomposed at equilibrium will be greater than 0.5%. **Step 3: Calculate the new mole percent of decomposed N2O5 at equilibrium when the volume is increased by a factor of 10.0**

Calculation of the new mole percent of decomposed N2O5

Let x represent the initial moles of N2O5, then the initial moles of NO2 and O2 are, respectively, 0.005x and 0. The volume is increased by a factor of 10.0, and at the new equilibrium, the decomposed N2O5 will be greater than 0.5% (from step 2). Let the change in moles of N2O5, NO2, and O2 be as follows: N2O5 decreases by 2y NO2 increases by 4y O2 increases by y Substitute these values in the expression for Q: Q = \(\frac{[(0 + 4y)/10V]^4\cdot[(0 + y)/10V]}{[(x - 2y)/10V]^2}\) Since the system is at equilibrium, Q = K: K = \(\frac{[(4y)^4\cdot y]}{[(x - 2y)^2]}\cdot \frac{1}{(10V)^5}\) To solve for y, we need more information, which is not provided. However, we've still determined that, as a result of increasing the volume by a factor of 10.0, the equilibrium will shift to the right, causing the mole percent of decomposed N2O5 to be greater than 0.5%.

Key concepts

Le Chatelier's Principle

Understanding Le Chatelier's Principle is key to predicting how a system at equilibrium will respond to changes. This principle helps us determine how a system will counteract an applied stress in order to re-establish equilibrium. When a change, such as pressure or concentration, affects a system in equilibrium, the system will shift its position to balance out these changes.

In the case of dinitrogen pentoxide decomposition, when we increase the volume, it decreases the pressure. According to Le Chatelier’s Principle, the system will compensate by shifting the equilibrium position in the direction that has more gas molecules, thereby increasing pressure again. This happens because there are more gas moles on the right side of the equation (the products side), so the equilibrium shifts to the right to increase the number of gas molecules when pressure decreases. This understanding is crucial for predicting how changes in conditions affect chemical reactions at equilibrium.

Dinitrogen Pentoxide Decomposition

Dinitrogen pentoxide (\(\mathrm{N}_2\mathrm{O}_5\)) can decompose into nitrogen dioxide (\(\mathrm{NO}_2\)) and oxygen gas (\(\mathrm{O}_2\)). This decomposition is a reversible reaction illustrated by:\[2 \mathrm{N}_2\mathrm{O}_5(g) \rightleftharpoons 4 \mathrm{NO}_2(g) + \mathrm{O}_2(g)\]

Understanding this decomposition reaction is important because it highlights the behavior of systems in equilibrium. Here, you can see that the number of moles increases from 2 moles of reactants to 5 moles of products. At equilibrium, the system maintains a balance between the forward reaction (decomposition of \(\mathrm{N}_2\mathrm{O}_5\)) and the reverse reaction (reformation of \(\mathrm{N}_2\mathrm{O}_5\)). The goal is to maintain the reaction quotient equal to the equilibrium constant \(K_c\). Thus, any changes in pressure or volume can shift this equilibrium, altering the concentrations of \(\mathrm{NO}_2\) and \(\mathrm{O}_2\).

Reaction Quotient

The reaction quotient \(Q\) is a mathematical expression that helps us compare the current state of a chemical reaction with its equilibrium state. Through analyzing \(Q\), we can determine whether the system is at equilibrium, or if it will shift to the left or right to reach equilibrium.

For the decomposition of \(\mathrm{N}_2\mathrm{O}_5\), the quotient \(Q\) is expressed as:\[Q = \frac{[\mathrm{NO}_2]^4 \cdot [\mathrm{O}_2]}{[\mathrm{N}_2\mathrm{O}_5]^2}\]This equation provides insight into the concentrations of reactants and products at any given point. If \(Q < K\), the reaction will proceed in the forward direction to produce more products. Conversely, if \(Q > K\), the reaction will favor the reverse direction to form more reactants. Understanding \(Q\) is essential for predicting the direction of the reaction at any moment.

Effect of Pressure Change on Equilibrium

The effect of changing pressure on chemical equilibrium is closely related to the number of moles of gas molecules in the reaction. According to Le Chatelier’s Principle, any change in pressure can cause a shift in equilibrium towards the side with fewer or more gas moles depending on the change.

When pressure in this system is decreased by increasing the volume, the equilibrium tends to shift towards the side with more gas moles. In the decomposition of \(\mathrm{N}_2\mathrm{O}_5\), the reaction shifts to the right with 5 moles of product gases compared to 2 moles of reactant gases. This demonstrates that the system compensates for the lowered pressure by forming additional gaseous molecules until the new equilibrium is established.

• If volume is increased, pressure decreases and the system shifts towards more gas moles.
• If volume is decreased, pressure increases and the system shifts towards fewer gas moles.

Understanding these principles allows us to predict how changes in conditions affect chemical reactions and maintain control over the reaction environment.