Question

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 \(5.25 \mathrm{atm} ?\)

Short answer

(a) The volume of \(\mathrm{NH}_{3}(\mathrm{g})\) produced is 208.67 L \n(b) The volume of \(\mathrm{NH}_{3}(\mathrm{g})\) produced is approximately 615.31 L

Step-by-step solution

Stoichiometric Calculations

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.

Confirming Limiting Reactant

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).

Calculation of \(NH_3\) volume at different condition for (a) and (b)

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.

Key concepts

Stoichiometry

In chemical reactions, stoichiometry quantifies the relationships between reactants and products. It involves using coefficients from the balanced chemical equation to determine the relative amounts of substances that participate in a reaction. This is crucial for understanding how much product can be formed from given reactants.

In the Haber process reaction:

• 1 mole of nitrogen gas (\(\mathrm{N}_2\)) reacts with 3 moles of hydrogen gas (\(\mathrm{H}_2\)) to produce 2 moles of ammonia (\(\mathrm{NH}_3\)).
• This means for every one volume of \(\mathrm{N}_2\), three volumes of \(\mathrm{H}_2\) are needed, resulting in two volumes of \(\mathrm{NH}_3\).

In this exercise, we start with 152 L of \(\mathrm{N}_2\). According to stoichiometry, if enough \(\mathrm{H}_2\) is available, up to 304 L of \(\mathrm{NH}_3\) can be produced (double the amount of nitrogen). However, it's important to consider actual quantities available to determine if the ideal stoichiometric conditions are met.

Ideal Gas Law

The Ideal Gas Law is a key equation for the relationships between pressure, volume, temperature, and moles of gas. Written as \(PV = nRT\), where:

• \(P\) is pressure,
• \(V\) is volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant, and
• \(T\) is the temperature in Kelvin.

In this problem, the derived formula \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) is used because it's simpler when dealing with gas volumes under different conditions without knowing the number of moles directly.

For example:

• For part (a), both pressure and temperature remain constant, making the volume of ammonia remain 208.67 L.
• For part (b), temperatures and pressures differ, so calculations shift volumes using given temperature and pressure to find the volume of ammonia as 615.31 L.

Limiting Reactant

A limiting reactant is the substance that is entirely consumed first in a reaction, determining the maximum amount of product that can be formed. It's like a bottleneck in the reaction process.

For the Haber process example:

• We have 152 L of \(\mathrm{N}_2\) and 313 L of \(\mathrm{H}_2\).
• To react all 152 L of \(\mathrm{N}_2\), we need 456 L of \(\mathrm{H}_2\), calculated as \(3 \times 152\).
• However, only 313 L of \(\mathrm{H}_2\) is available, making \(\mathrm{H}_2\) the limiting reactant.

Thus, the reaction is constrained by the amount of \(\mathrm{H}_2\). Consequently, calculations of produced \(\mathrm{NH}_3\) must be based on the available \(\mathrm{H}_2\) amounts, which yields 208.67 L of \(\mathrm{NH}_3\) produced.