91Ó°ÊÓ

The quantity of sulfuric acid used globally places it among the most plentiful of all commodity chemicals. In the modern chemical industry, synthesis of most sulfuric acid utilizes elemental sulfur as a feedstock. However, an alternative and historically important source of sulfuric acid was the conversion of an ore containing iron pyrites (FeS_) to sulfur oxides by roasting (burning) the ore with air. The following reactions occurred in an oven: $$\begin{array}{c} 2 \mathrm{FeS}_{2}(\mathrm{s})+\frac{11}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+4 \mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{3}(\mathrm{g}) \end{array}$$ The gases leaving the oven were fed to a catalytic converter in which most of the remaining \(\mathrm{SO}_{2}\) produced was oxidized to \(\mathrm{SO}_{3}\). Finally, the gas leaving the converter was sent to an absorption column where the \(S O_{3}\) was taken up by water to produce sulfuric acid \(\left(H_{2} S O_{4}\right)\) (a) The ore fed to the oven was 90.0 wt\% \(\mathrm{FeS}_{2}\), and the remaining material may be considered inert. Dry air was fed to the oven in \(30.0 \%\) excess of the amount required to oxidize all of the sulfur in the ore to \(S O_{3}\). Eighty-five percent of the \(\mathrm{FeS}_{2}\) was oxidized, and \(60 \%\) of the \(\mathrm{SO}_{2}\) produced was oxidized to \(S O_{3}\). Leaving the roaster were (i) a gas stream containing \(S O_{2}, S O_{3}, O_{2},\) and \(N_{2}\) and (ii) a solid stream containing unconverted pyrites, ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right),\) and the inert material. Calculate the required feed rate of air in standard cubic meters per \(100 \mathrm{kg}\) of ore fed to the process. Also determine the molar composition and volume (SCM/100 kg ore) of the gas leaving the oven. (b) The gas leaving the oven entered the catalytic converter, which operated at 1.0 atm. Reaction (2) proceeded to equilibrium, at which point the component partial pressures are related by the expression $$K_{\mathrm{P}}(T)=\frac{p_{\mathrm{SO}_{3}}}{p_{\mathrm{SO}_{3}} p_{\mathrm{O}_{2}}^{0.5}}$$ The gases were first heated to \(600^{\circ} \mathrm{C}\) to accelerate the rate of reaction, and then cooled to \(400^{\circ} \mathrm{C}\) to enhance \(S O_{2}\) conversion. The equilibrium constant \(K_{\mathrm{P}}\) at these two temperatures is 9.53 atm \(^{0.5}\) and 397 atm \(^{0.5}\), respectively. Calculate the equilibrium fractional conversions of \(S O_{2}\) at these two temperatures. (c) Estimate the production rate of sulfuric acid in \(\mathrm{kg} / \mathrm{kg}\) ore if all of the \(\mathrm{SO}_{3}\) leaving the converter was transformed to sulfuric acid. What would this value be if all the sulfur in the ore had been converted?

Step by step solution

01

Calculating the total molar amounts of FeS2 and O2

In 100 kg.feed to the oven, the mass of FeS2 = 90 kg because the ore is 90.0 wt% FeS2. Since the molar mass of FeS2 is about 119.975 g/mol, there are \((90000 g) / (119.975 g/mol) = 750.46 mol\) of FeS2. Only 85% of this FeS2 reacts in the oven, so the moles of FeS2 reacted in the oven = \((750.46 mol) * 0.85 = 637.892 mol \) FeS2. \nBased on the first stoichiometric equation, it can be seen that 11/2 = 5.5 moles of O2 are required to react with 2 moles of FeS2. Therefore, the theoretical moles of O2 required is \((637.892 mol) * (5.5 mol O2 / 2 mol FeS2) = 1754.197 mol O2.\nHowever, the problem states that 30.0% excess O2 is added, so the actual moles of O2 fed to the oven = \((1754.197 mol O2) * (1 + 0.30) = 2278.45 mol O2.\nThe molar volume of an ideal gas at STP is 22.414 L/mol, so the required feed rate of air =\((2278.45 mol) * (22.414 m^3 / 1000 mol) = 51.042 m^3 (=51.042 SCM) air per 100 kg ore.

02

Calculating the moles of each gas leaving the oven

Based on the first stoichiometric equation, it can be seen that 4 moles of SO2 are formed per 2 moles of FeS2. Therefore, moles of SO2 formed in the oven = \((637.892 mol FeS2) * (4 mol SO2 / 2 mol FeS2) = 1275.784 mol SO2. \nHowever, only 60% of this SO2 is oxidized to SO3 in the oven. Therefore, moles of SO2 remaining after the oven = \((1275.784 mol SO2) * (1.00 - 0.60) = 510.314 mol SO2.\nFrom the second stoichiometric equation, it can be seen that 1 mol of SO3 formed per mol of SO2. Therefore, moles of SO3 formed in the oven = \((1275.784 mol SO2) * 0.60 = 765.470 mol SO3.\n\nFrom the excess O2 calculation, the moles of O2 remaining after the oven = \((2278.455 mol O2) - (765.470 mol SO3 + 510.314 mol SO2) = 1002.671 mol O2.\nThe problem says air is used, which is about 21% O2 and 79% N2 by volume. So if there was 2278.455 mol of O2, then there are \((2278.455 mol O2) * (79 / 21) = 8550.295 mol N2.\

03

Finding the molar composition and volume of the gas leaving the oven

Total moles of gases leaving the oven = moles of SO2 + moles of SO3 + moles of O2 + moles of N2 = 510.314 mol + 765.470 mol + 1002.671 mol + 8550.295 mol = 10828.750 mol.\nThe molar composition (or molar fraction) of each gas is simply the moles of that gas divided by the total moles. \nTherefore, the molar fraction, or mole %, of SO2 = (510.314 mol / 10883.75 mol) = 0.047 (or 4.7%), for SO3 = (765.470 mol / 10883.75 mol) = 0.070 (or 7.0%), for O2 = (1002.671 mol / 10883.75 mol) = 0.092 (or 9.2%), and for N2 = (8550.295 mol / 10883.75 mol) = 0.790 (or 79%).\nThe volume of the gases leaving the oven = total moles of gas * the molar volume of an ideal gas at STP = \((10883.75 mol) * (22.414 m^3 / 1000 mol) = 243.82 m^3 (SCM) per 100 kg ore.

04

Calculating the equilibrium fractional conversions of SO2

The equilibrium reaction \(SO2(g) + 0.5 O2(g) ↔ SO3(g)\) has an equilibrium constant expressible as \(KP(T) = (pSO3) / (pSO2 * sqrt(pO2))\). \nIncreasing the temperature increases the reaction rate, but it decreases the equilibrium conversion of SO2 to SO3 because this is an exothermic reaction (it releases heat) so it tends to shift left (towards the reactants) with increasing temperature. Therefore, the equilibrium fractional conversion is higher at the lower temperature, 400°C (= 673.15 K), with a higher KP value, than at the higher temperature, 600°C (= 873.15 K). \nLet X be the fractional conversion of SO2 at equilibrium. Then the pressures of the gases are: \npSO2 = (1 - X) * (the initial partial pressure of SO2); pSO3 = X * (the initial partial pressure of SO2); and pO2 = (1 - X/2) * (the initial partial pressure of O2). \nSolving the equations \(KP(Ti) = (pSO3i) / (pSO2i * sqrt(pO2i))\) for Ti and X gives the equilibrium fractional conversions of SO2 at these two temperatures. To solve for X, use the quadratic formula based on this equation.

05

Estimating the production rate of sulfuric acid

In 100 kg.feed to the oven, the mass of FeS2 = 90 kg because the ore is 90.0 wt% FeS2. Since the molar mass of FeS2 is about 119.975 g/mol, there are \((90000 g) / (119.975 g/mol) = 750.46 mol\) of FeS2. Only 85% of this FeS2 reacts in the oven, so the moles of FeS2 reacted in the oven = \((750.46 mol) * 0.85 = 637.892 mol \) FeS2. \nBased on the first stoichiometric equation, it can be seen that 4 moles of SO2 are formed per 2 moles of FeS2. Therefore, moles of SO2 formed in the oven = \((637.892 mol FeS2) * (4 mol SO2 / 2 mol FeS2) = 1275.784 mol SO2. \nHowever, only 60% of this SO2 is oxidized to SO3 in the oven. Therefore, moles of SO3 formed in the oven = \((1275.784 mol SO2) * 0.60 = 765.470 mol SO3. \nThe reaction \(SO3(g) + H2O(l) → H2SO4(aq)\) shows that 1 mol of SO3 makes 1 mol of H2SO4. Since the molar mass of H2SO4 is 98.079 g/mol, if all of the SO3 was transformed to H2SO4 in the absorption column, the production rate of H2SO4 = \((765.470 mol SO3) * (98.079 g H2SO4 / mol SO3) = 75049.63 g H2SO4 = 75.05 kg H2SO4 per 100 kg ore. \nIf all of the sulfur in the FeS2 (which is 53.3333%) had been converted to H2SO4 (which is 32.6654% sulfur), the theoretical maximum production rate of H2SO4 = \((90 kg FeS2) * (53.3333% S in FeS2 / 32.6654% S in H2SO4) = 145.68 kg H2SO4 per 100 kg ore.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships of the reactants and products in chemical reactions. It's founded on the law of conservation of mass and the concept of the mole, both of which state that matter is neither created nor destroyed in a chemical reaction and that chemicals react in whole number ratios of moles.

For a student grappling with stoichiometry, the key is to understand that a balanced chemical equation serves as a recipe for the reaction, defining precisely how many moles of each substance are involved. When approaching a problem, you first interpret this 'recipe' by identifying the mole ratio between the reactants and products. Once you've established these ratios, you then use them to calculate how much of each reactant is needed, or how much of each product will be produced. This is crucial when scaling up from a laboratory to an industrial process like sulfuric acid production.

Furthermore, it's important for students to keep in mind that stoichiometry isn't just about the numbers. It also tests one's understanding of the chemical nature of the substances involved, as certain compounds may exist in multiple forms or stages throughout the process, which can impact the stoichiometric calculations. For instance, knowing that sulfur can exist as elemental sulfur, sulfur dioxide (SO_2), or sulfur trioxide (SO_3) is vital when calculating the stoichiometry for the production of sulfuric acid.

Balancing Chemical Reactions

Balancing chemical reactions is a fundamental skill that you'll need to perform stoichiometric calculations. A balanced reaction is one where the number of atoms for each element is the same in both the reactants and products. This reflects the physical reality that in chemical reactions, atoms are neither created nor destroyed; they are simply rearranged.

To balance a chemical equation, you follow several steps:

• Write the unbalanced equation to show the reactants and products.
• Count the number of atoms of each element in the reactants and products.
• Adjust the coefficients – the numbers in front of formulas – to get the same number of atoms of each element on both sides of the equation.
• Repeat the process until you achieve balance for all elements involved.
• Check your work by verifying that the equation is balanced.

It's important to practice this skill to build competence and confidence. Balancing equations is not just academic exercise; it's essential for scaling chemical reactions from the lab bench to industrial production, like the synthesis of sulfuric acid from sulfur dioxide.

Sulfuric Acid Production

Sulfuric acid production is an example of a large-scale, industrial chemical process that requires thorough understanding of stoichiometry and chemical reaction balancing. This acid is one of the most widely used chemicals in the world and is the backbone of many manufacturing processes. The Contact Process is the current method of choice for sulfuric acid production, but historically, the roasting of iron pyrite (FeS2) was also used.

In the provided exercise, we read about the intricate steps involved in producing sulfuric acid from iron pyrite. The reaction sequence involves the roasting of FeS2 to form sulfur dioxide (SO2), followed by the catalytic oxidation of SO2 to sulfur trioxide (SO3), and finally the absorption of SO3 in water to produce sulfuric acid (H2SO4).

Each step in the process demands careful calculation of the reagent ratios and understanding of the conditions needed to optimize yield, such as temperature and pressure. The reaction conditions are manipulated to maximize the conversion of SO2 to SO3, a critical aspect since the efficiency of SO2 conversion directly impacts the yield of sulfuric acid. For students, it is pivotal to recognize that in real-world applications, side reactions, incomplete conversions, and practical yield versus theoretical yield are all factors that need to be considered when discussing chemical processes.