Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a 2.50-L flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes? (Hint: First write a balanced equation for the decomposition.)

First, let's write the balanced equation for the decomposition of dinitrogen pentoxide gas: $$ \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 2 \mathrm{NO}_{2} + \frac{1}{2} \mathrm{O}_{2} $$ By examining the chemical equation, we can determine this relationship: For every mole of N2O5 that decomposes, 1/2 mole of O2 is formed.

We know the initial pressure and volume of N2O5 in the flask. Using the ideal gas law (PV=nRT), we can calculate the initial moles of N2O5, where: P = Pressure = \(756 \,\mathrm{mmHg} = 1 \,\mathrm{atm}\) (since 1 atm = 760 mmHg) V = Volume = \(2.50 \,\mathrm{L}\) R = Gas constant = \(0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\) T = Temperature = \(25^{\circ} \mathrm{C} = 298 \,\mathrm{K}\) Plug in the values: $$ n = \frac{PV}{RT} = \frac{(1 \, \mathrm{atm})(2.50 \, \mathrm{L})}{(0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}})(298 \, \mathrm{K})} $$ After calculating the expression, we get that the initial moles of N2O5 are approximately \(n_{0} = 0.101\, \mathrm{mol}\).

Since the reaction is first-order with respect to N2O5, we can write the rate law as follows: $$ \text{Rate} = k[\mathrm{N}_{2}\mathrm{O}_{5}] $$ Where k is the rate constant \(0.247 \, \mathrm{h}^{-1}\). The Integrated Rate Law for a first-order reaction is: $$ \ln{\frac{[\mathrm{N}_{2}\mathrm{O}_{5}]_{0}}{[\mathrm{N}_{2}\mathrm{O}_{5}]}} = kt $$ Where \([\mathrm{N}_{2}\mathrm{O}_{5}]_{0}\) is the initial concentration, \([\mathrm{N}_{2}\mathrm{O}_{5}]\) is the concentration at time t, and t is the time in hours.

In 135 minutes, the reaction will be running for: $$ \mathrm{t} = \frac{135 \, \mathrm{min}}{60 \, \frac{\mathrm{min}}{\mathrm{h}}} = 2.25 \, \mathrm{h} $$ Now, substitute the known values into the Integrated Rate Law: $$ \ln{\frac{[\mathrm{N}_{2}\mathrm{O}_{5}]_{0}}{[\mathrm{N}_{2}\mathrm{O}_{5}]}} = (0.247 \, \mathrm{h}^{-1})(2.25 \, \mathrm{h}) $$ We can then calculate \([\mathrm{N}_{2}\mathrm{O}_{5}]\), the concentration of N2O5 at time t and then calculate the moles of N2O5 left using the formula moles = concentration * volume. After that, find the change in moles: $$ \Delta n_{\mathrm{N}_{2}\mathrm{O}_{5}} = n_{0}-n_{\mathrm{N}_{2}\mathrm{O}_{5}} $$ Now, use the stoichiometric relationship derived from the balanced chemical equation: $$ \text{moles of O}_{2} \, \text{formed} = \frac{1}{2} \Delta n_{\mathrm{N}_{2}\mathrm{O}_{5}} $$ Calculating the final expression will give us the moles of oxygen formed after 135 minutes.