Problem 101 - Chemistry

Question

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\\2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{aligned}$$What mass of \(\mathrm{NH}_{3}\) (in grams) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the above procedure, assuming an 80 percent yield in each step? ( 1 ton = \(2000 \mathrm{lb} ; 1 \mathrm{lb}=453.6 \mathrm{~g} .)\)

Step-by-Step Solution

Answer

478,806.25 g of \(NH_{3}\) is needed to produce 1 ton of \(HNO_{3}\) assuming an 80 percent yield in each step.

Step 1: Convert Tons to Grams

For convenience, convert the mass of \(HNO_{3}\) from tons to grams. One ton equals 2000 lb, and one pound equals 453.6 g, so 1 ton of \(HNO_{3}\) equals \(1 \times 2000 \times 453.6 = 907,200\) g.

Step 2: Apply Stoichiometry

From the third equation in the Ostwald process, 1 mol of \(HNO_{3}\) is produced from one mol of \(NO_{2}\). From the second equation, 1 mol of \(NO_{2}\) is produced from 1 mol of \(NO\). And from the first equation, 1 mol of \(NO\) comes from 1 mol of \(NH_{3}\). Therefore, 1 mol of \(NH_{3}\) is needed to produce 1 mol of \(HNO_{3}\). Thus, calculate the number of moles of \(HNO_{3}\) produced by dividing the mass of \(HNO_{3}\) by its molar mass (63.01g/mol): \(907,200 \, g / 63.01 \, g/mol = 14,400 \, mol\). Hence, idealistically, 14,400 mol of \(NH_{3}\) is required to produce 1 ton of \(HNO_{3}\).

Step 3: Consider Efficiency

According to the question, each step in the Ostwald process has an 80% yield. So, the actual production of \(HNO_{3}\) will require more \(NH_{3}\) than the stoichiometric ratios imply. It is not enough to consider one 80% efficiency; since there are three steps in the process, the overall efficiency is \(0.8 \times 0.8 \times 0.8 = 0.512\) or 51.2%. Thus, the actual moles of \(NH_{3}\) needed is \(14,400 \, mol / 0.512 = 28,125 \, mol\).

Step 4: Convert to Mass

Finally, convert moles of \(NH_{3}\) to grams by multiplying by the molar mass of \(NH_{3}\) (17.03 g/mol): \(28,125 \, mol \times 17.03 \, g/mol = 478,806.25 \, g\).

Key Concepts

Industrial Production of Nitric Acid

Nitric acid (HNO3) is a highly corrosive and toxic strong acid with a wide range of industrial applications, from fertilizers to explosives. The industrial synthesis of nitric acid is predominantly done through the Ostwald process, a method that revolves around the oxidation of ammonia (NH3).

The Ostwald process can be summarized in three main reactions. First, ammonia is oxidized to nitric oxide (NO), then to nitrogen dioxide (NO2), and finally, nitrogen dioxide is combined with water to produce nitric acid. Each step involves careful control of temperature and pressure to ensure optimal yield. The efficiency of this process is vital to the cost-effectiveness of producing nitric acid on an industrial scale.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction, based on the coefficients from the balanced chemical equations.

For example, the Ostwald process outlines specific mole-to-mole conversions between reactants and products. Using this information, stoichiometry provides the calculations necessary to determine how much ammonia is needed to produce a specific amount of nitric acid. These calculations are crucial when scaling up reactions for industrial production, to ensure that all materials are used efficiently, thereby reducing waste and increasing yield.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, indicating the ratio between the actual yield and the theoretical yield (as predicted by stoichiometry) multiplied by 100%. It is an essential concept in both laboratory and industrial chemistry as it helps in evaluating the effectiveness of a reaction.

In industrial settings like the Ostwald process, factors such as incomplete reactions, side reactions, and loss of product can lead to a lower-than-expected percent yield. As in the given problem, if every step of the Ostwald process is 80% efficient, the overall percent yield decreases significantly with each subsequent reaction step. Understanding and calculating percent yield allows chemists to optimize reactions, improve production methods, and gauge the economic feasibility of the process.