Question

Ammonia is produced by the millions of tons annually for use as a fertilizer. It is commonly made from \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) by the Haber process. Because the reaction reaches equilibrium before going completely to product, the stoichiometric amount of ammonia is not obtained. At a particular temperature and pressure, \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{2}\) reacts with \(20.0 \mathrm{~g}\) of \(\mathrm{N}_{2}\) to form ammonia. When equilibrium is reached, \(15.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) has formed. (a) Calculate the percent yield. (b) How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are present at equilibrium?

Short answer

Percent yield is 61.76%. Moles of \(\text{N}_2\) and \(\text{H}_2\) at equilibrium are 0.273 and 3.677 respectively.

Step-by-step solution

- Write the balanced equation for the Haber process

The balanced chemical equation for the Haber process is: \[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

- Calculate the moles of reactants

Using the molar masses: Molar mass of \(\text{N}_2\) = 28 g/mol Molar mass of \(\text{H}_2\) = 2 g/mol Moles of \(\text{N}_2\): \[ \frac{20.0 \text{g}}{28 \text{g/mol}} = 0.714 \text{ moles} \] Moles of \(\text{H}_2\): \[ \frac{10.0 \text{g}}{2 \text{g/mol}} = 5.0 \text{ moles} \]

- Determine the limiting reactant

According to the stoichiometry of the balanced equation: 1 mole of \(\text{N}_2\) requires 3 moles of \(\text{H}_2\). For \(0.714 \text{ moles of N}_2\), required \(\text{H}_2\) is: \[0.714 \text{ moles N}_2 \times 3 \text{ moles H}_2/1 \text{ mole N}_2 = 2.142 \text{ moles H}_2\] Since 5.0 moles of \(\text{H}_2\) are available, \(\text{N}_2\) is the limiting reactant.

- Calculate the theoretical yield of NH3

Since \(\text{N}_2\) is the limiting reactant: \[\text{0.714 moles of N}_2 \rightarrow 2 \times 0.714 = 1.428 \text{ moles of NH}_3 \] Molar mass of NH3 is 17 g/mol. Theoretical mass of NH3: \[\text{1.428 moles NH}_3 \times 17 \text{ g/mol} = 24.276 \text{ g} \]

- Calculate the percent yield

Actual yield of NH3 is given as 15.0 g. Percent yield: \[\frac{15.0 \text{ g}}{24.276 \text{ g}} \times 100 = 61.76\text{ %} \]

- Calculate moles of NH3 at equilibrium

Actual yield of NH3 in moles: \[ \frac{15.0 \text{ g}}{17 \text{ g/mol}} = 0.882 \text{ moles} \]

- Calculate moles of N2 at equilibrium

Initial moles of \(\text{N}_2\): 0.714Moles of \(\text{N}_2\) reacted: \[ \text{0.882 moles NH}_3 /2 = 0.441 \text{ moles} \] Remaining moles of \(\text{N}_2\): \[0.714 - 0.441 = 0.273 \text{ moles} \]

- Calculate moles of H2 at equilibrium

Initial moles of \(\text{H}_2\): 5.0 Moles of \(\text{H}_2\) reacted: \[ \text{0.441 moles N}_2 \times 3 = 1.323 \text{ moles} \] Remaining moles of \(\text{H}_2\): \[5.0 - 1.323 = 3.677 \text{ moles} \]

Key concepts

Percent Yield

Percent yield is a crucial concept in chemistry, particularly in industrial processes like the Haber process. It measures the efficiency of a chemical reaction by comparing the actual yield (the amount of product actually obtained) to the theoretical yield (the maximum possible amount predicted by stoichiometry).

To calculate percent yield, you use the formula:
\[\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \]

For instance, in the exercise given, the actual yield of ammonia (\text{NH}_3) is 15.0 grams and the theoretical yield is 24.276 grams. Thus, the percent yield is:
\[\frac{15.0 \text{g}}{24.276 \text{g}} \times 100 = 61.76\text{ %} \]

A percent yield less than 100% indicates that the reaction did not go to completion, which is common in reactions that reach chemical equilibrium.

Limiting Reactant

Identifying the limiting reactant is a key step in determining how much product can be formed in a chemical reaction. The limiting reactant is the substance that is completely consumed first, thereby limiting the amount of products formed.

In the Haber process, you have \(\text{N}_2\) and \(\text{H}_2\) reacting to form ammonia (\text{NH}_3). Using the balanced equation:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

\(\text{To find the limiting reactant:}\)
• Calculate the moles of each reactant. For \(\text{N}_2\): \[ \frac{20.0 \text{g}}{28 \text{g/mol}} = 0.714 \text{ moles} \]
• For \(\text{H}_2\): \[ \frac{10.0 \text{g}}{2 \text{g/mol}} = 5.0 \text{ moles} \]
• Compare the mole ratio. According to the equation, 1 mole of \(\text{N}_2\) requires 3 moles of \(\text{H}_2\). For 0.714 moles of \(\text{N}_2\), required \(\text{H}_2\) is: \[ 0.714 \text{ moles N}_2 \times 3 \text{ moles H}_2/1 \text{ mole N}_2 = 2.142 \text{ moles H}_2 \]

\(\text{Since 5.0 moles of \(\text{H}_2\) are available, \(\text{N}_2\) is the limiting reactant.} \).

Chemical Equilibrium

In chemical reactions like the Haber process, reaching a state of chemical equilibrium is common. This means that the rate of the forward reaction (forming products) equals the rate of the reverse reaction (breaking down products). Consequently, the concentrations of reactants and products remain constant over time.

This reaction can be represented as:
\[ \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \]

Unlike reactions that go to completion, those that reach equilibrium do not produce the stoichiometric amount of product. This is why calculating the percent yield is important; it gives insight into how far the reaction went before equilibrium was established.

• In the given exercise, the reaction reaches equilibrium, making it necessary to calculate the remaining moles of \(\text{N}_2\) and \(\text{H}_2\) at equilibrium: \[ \text{Remaining moles of \(\text{N}_2\)} = 0.714 - 0.441 = 0.273 \] \[ \text{Remaining moles of \(\text{H}_2\)} = 5.0 - 1.323 = 3.677 \]

Molar Mass

Understanding molar mass is fundamental to converting between the mass of a substance and the number of moles, which is essential for stoichiometric calculations. The molar mass is the mass of one mole of a substance (atoms, molecules, or formula units) and is usually expressed in g/mol.

For example, in the Haber process exercise, you need to determine the number of moles of \(\text{N}_2\) and \(\text{H}_2\) initially present.

• The molar mass of \(\text{N}_2\) is 28 g/mol. To find the moles, use: \[ \frac{20.0 \text{g}}{28 \text{g/mol}} = 0.714 \text{ moles} \]
• The molar mass of \(\text{H}_2\) is 2 g/mol: \[ \frac{10.0 \text{g}}{2 \text{g/mol}} = 5.0 \text{ moles} \]

Similarly, calculating the theoretical yield of ammonia (\text{NH}_3) involves knowing its molar mass: \[ \text{Molar mass of \(\text{NH}_3\)} = 17 \text{ g/mol} \]
Theoretical yield in moles of \(\text{NH}_3\): \[ \text{0.714 moles of \(\text{N}_2\)}\times2 = 1.428 \text{ moles of \(\text{NH}_3\)} \]
Theoretical mass of \(\text{NH}_3\): \[ \text{1.428 moles} \times 17 \text{ g/mol} = 24.276 \text{ g} \]