The Haber process is the principal method for fixing nitrogen (converting \(\mathrm{N}_{2}\) to nitrogen compounds). $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$$ Assume that the reactant gases are completely converted to \(\mathrm{NH}_{3}(\mathrm{g})\) and that the gases behave ideally. (a) What volume of \(\mathrm{NH}_{3}(\mathrm{g})\) can be produced from 152 \(\mathrm{L} \mathrm{N}_{2}(\mathrm{g})\) and \(313 \mathrm{L}\) of \(\mathrm{H}_{2}(\mathrm{g})\) if the gases are measured at \(315^{\circ} \mathrm{C}\) and 5.25 atm? (b) What volume of \(\mathrm{NH}_{3}(\mathrm{g}),\) measured at \(25^{\circ} \mathrm{C}\) and\(727 \mathrm{mmHg},\) can be produced from \(152 \mathrm{L} \mathrm{N}_{2}(\mathrm{g})\) and \(313 \mathrm{L} \mathrm{H}_{2}(\mathrm{g}),\) measured at \(315^{\circ} \mathrm{C}\) and \(5.25 \mathrm{atm} ?\)

In the given chemical reaction, \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\), each volume of \(\mathrm{N}_{2}\) reacts with three volumes of \(\mathrm{H}_{2}\) to give two volumes of \(\mathrm{NH}_{3}\). Hence, given that 152 L of \(\mathrm{N}_{2}\) and 313 L of \(\mathrm{H}_{2}\) are reacting together, the amount of \(\mathrm{NH}_{3}\) that can be produced per volume of \(\mathrm{N}_{2}\) is 2 volumes. So, theoretically, we can produce \(2 \times 152 = 304\) L of \(\mathrm{NH}_{3}\). But we also need to confirm that we have sufficient \(\mathrm{H}_{2}\) available for the reaction.

From stoichiometry, it is known that 1 volume of \(\mathrm{N}_{2}\) requires 3 volumes of \(\mathrm{H}_{2}\). Hence, to completely react 152 L of \(\mathrm{N}_{2}\),we require \(3 \times 152 = 456\) L of \(\mathrm{H}_{2}\). Given we have only 313 L of \(\mathrm{H}_{2}\), the \(\mathrm{H}_{2}\) is the limiting reactant in this case. Therefore, the reaction will stop once all the \(\mathrm{H}_{2}\) is consumed. To find out how much \(\mathrm{NH}_{3}\) can be produced, the stoichiometry of reaction tells us that 3 volumes of \(\mathrm{H}_{2}\) forms 2 volumes of \(\mathrm{NH}_{3}\). Hence the volume of \(\mathrm{NH}_{3}\) produced is \(2/3 \times 313 = 208.67\) L (approximately).

Now we have to calculate the volume of \(\mathrm{NH}_{3}\) under two different conditions. The ideal gas law, \(PV=nRT\), allows us to do this. However, as we are dealing with volumes, not moles of gas, we need to use a derived form of the ideal gas law, \(P1V1/T1=P2V2/T2\), where P represents pressure, V volume, and T temperature (in kelvins) and the subscripts 1 and 2 refer to the initial and final states, respectively. (a) For part (a), P1 = 5.25 atm, V1 = 208.67 L, T1 = 315°C = 588.15 K ( by adding 273.15 to the temperature in °C to convert to kelvins), P2 = 5.25 atm, and T2 = 588.15 K. Here, the temperature and pressure are constant, so the volume of \(\mathrm{NH}_{3}\) would be the same, i.e., V2 = 208.67 L. (b) For part (b), we need to calculate the volume at a different temperature and pressure. P1 = 5.25 atm, V1 = 208.67 L, T1 = 588.15 K, P2 = 727 mmHg = 0.96 atm ( Converting mmHg to atm by dividing by 760), and T2 = 25°C = 298.15 K. Inserting these values into the equation gives V2 = \(P1 \times V1 \times T2 /(P2 \times T1) = 5.25 \times208.67 \times 298.15 / (0.96 \times 588.15) \approx 615.31\) L. Hence, the volume of \(\mathrm{NH}_{3}\) at these conditions is about 615.31 L.