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 \(10.0 .\)

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**

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**

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%.