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 \(10.0 .\)

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.

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] = \(x\)

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: \[x = \sqrt[5]{\frac{1.00 \times (0.0995)^2}{(4)^4}} \approx 0.0126\]

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 \(25.3\%\).