Chapter 5: Stoichiometry Part 3 SABIS Grade 11 (Level M) Chemistry ~ SABIS ® Curriculum Chemistry Online Tuition

Chapter 5: Stoichiometry

5.3 The Manufacture of Sulfuric Acid

5.3.1 Cost

Many millions of tons of sulfuric acid, H2SO4, are produced every year. A manufacturing process called the “contact process” has widespread industrial importance and will be described in this chapter. This process is almost perfected and the cost of this useful chemical is only about $50.00 per ton!

Example

If H2SO4 costs $50.00 per ton, how many moles are obtained for a cent?

Given: Cost per ton ($50.00 per ton).

RTF: moles per ¢.

Solution

The units of the answer must be moles of H2SO4 per ¢. This problem is simply that of conversion of units. Since the answer ought to be in “moles per cent” we need to start with the cost as “quantity of acid per dollar”. In other words, instead of starting with “$50.00 per ton” we start with “ton per $50.00”. Next we keep multiplying by conversion factors until we end up with mol/¢.

$C o s t = \frac{1 t o n}{$ 50.00} \times \frac{$ 1}{¢ 100} \times \frac{1, 000 k g}{1 t o n} \times \frac{1, 000 g}{1 k g} \times \frac{1 m o l e}{98.0 g} = 2.0 m o l / ¢$

5.3.2 The Chemistry of Preparing Sulfuric Acid

The chemical reactions appear simple. They begin with pure sulfur (which occurs in natural deposits in the elemental state). First, sulfur is burned to give gaseous sulfur dioxide, SO2. Next, the SO2 is further oxidized, catalytically, to sulfur trioxide, SO3. Finally, the addition of water forms sulfuric acid. These reactions are written as:

S8(s) + 8O2(g) $⇌$ 8SO2(g)

8SO2(g) + 4O2(g) $⇌$ 8SO3(g)

8SO3(g) + 8H2O(l) $⇌$ 8H2SO4(l)
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Overall reaction: S8(s) +12O2(g) +8H2O(l) $⇌$ 8H2SO4(l)

Example

A shovelful of sulfur containing 1.00 kg S8 is to be converted to sulfuric acid. What mass of H2SO4 will be formed?

Given: 1.0 kg of sulfur
RTF: Mass of H2SO4 formed

Solution

Method (1):

Quantity of S8 = 1.00 kg of sulfur = $1 \times 10^{3} g \times \frac{1 m o l e}{8 \times 32 g} = \frac{10^{3}}{8 \times 32} m o l$

We can see that 1 mol of S8 produces 8 mol of H2SO4:

S8(s) + 12O2(g) + 8H2O(l) $⇌$ 8H2SO4(l)

1 mol S8 → 8 mol H2SO4

mol S8 → ?

The “reaction factor” is (8 mol H2SO4/ mol S8)

Moles of H2SO4 produced = $\frac{10^{3}}{8 \times 32} m o l S_{8}$ $x = \frac{8 m o l H_{2} S O_{4}}{m o l S_{8}} = \frac{10^{3}}{32} m o l H_{2} S O_{4}$

$M a s s = \frac{10^{3}}{32} m o l \times \frac{98.0 g}{1 m o l} = 3.06 \times 10^{3} g H_{2} S O_{4}$

Method (2):

A faster way to solve the problem is to forget about balancing equations, and to start as follows:

Every mole of H2SO4 contains 1 mole of sulfur atoms: 98.0 g of H2SO4 contain 32 g sulfur (S8 or any other form!). Therefore

$M a s s o f H_{2} S O_{4} p r o d u c e d = \frac{98 g H_{2} S O_{4}}{32 g S_{8}} \times 10^{3} g S_{8} = 3.06 \times 10^{3} g H_{2} S O_{4}$

5.3.3 Production of SO₂

The first step in the manufacture of H2SO4 is to burn sulfur to sulfur dioxide. Sulfur burns spontaneously in air, liberating heat.

S8(s) + 8O2(g) $⇌$ 8SO2(g) ΔH = –297 kJ/mole SO2

An important economic feature of the modern processes is the utilization of this heat in another step in which heat is absorbed.

Example

What mass of sulfur (to 3 significant figures) will burn to produce 100,000 liters of pure SO2 at 500°C and one atmosphere pressure?

Given: 1.00 ×105 dm3 SO2 at 500°C and 1.00 atm. RTF: Mass of S needed

Solution

The mass can be found from the equation

S8(s) + 8O2(g) $⇌$ 8SO2(g) ΔH = –297 kJ/mole SO2

1 mole 8 moles

To find the quantity of S8 we need to know how many moles of SO2 are given

Using PV = nRT, n = PV/RT

P = 1 atm, V = 105 dm3 , T = 500 + 273 = 773 K, R = 0.0821 dm3.atm/K.mol, n = ?

n = (1.00 × 105) / (0.0821 × 773) = 1.58 × 103 mol

Let us use the easy way to calculate.

Moles of S needed = moles of SO2 present × (1mol S/ 1 mol SO2) = 1.58 × 103 mol

Mass of S needed = 1.58 × 103 mol × (32g / 1 mol) = 50.8 × 103 g = 50.8 kg

Notice that whether the sulfur is found as S atoms or S8 molecules, the mass of sulfur will be always the same.

$M a s s o f H_{2} S O_{4} p r o d u c e d = \frac{98 g H_{2} S O_{4}}{32 g S_{8}} \times 10^{3} g S_{8} = 3.06 \times 10^{3} g H_{2} S O_{4}$

5.3.4 Production of SO₃

The second step in the manufacture of H2SO4 is to burn sulfur dioxide to sulfur trioxide. The process liberates heat:

SO2(g) + $\frac{1}{2}$ O2(g) $⇌$ SO3(g) ΔH = –98.2 kJ/mole SO3

The process uses a solid catalyst. Either finely divided platinum or vanadium pentoxide, V2O5, is effective. Because catalysis occurs where the gas contacts the surface of the catalyst, this process is called the contact process.

Example

What volume of air, at 500°C and one atmosphere pressure, is needed to react with the 1.00 x 105 dm3 of SO2 produced from 50.8 kilograms of sulfur?

Given: 1.00 × 105 dm3 SO2 at 500°C and 1.00 atm.

RTF: Volume of air needed.

Solution

Since the gases are at the same conditions (500°C and one atmosphere), Avogadro’s law tells us equal volumes contain equal number of moles. So the gases react in the ratio indicated below:

SO2(g) + 1/2 O2(g) $⇌$ SO3(g) ΔH = –23.5 kcal/mole SO3

1 mole 1/2 mol

1 dm3 0.5 dm3

Volume O2 needed = 1.00 × 105 dm3 SO2 × (0.500 mol O2 / mol SO2) = 5.00 × 104 dm3

Since air is only 20% oxygen:

Volume air needed = 5.00 × 104 dm3 O2 × (100 dm3 air / 20 dm3 O2) = 2.50 × 105 dm3

The last step in the preparation of commercial sulfuric acid is to allow the sulfur trioxide to react with steam*:

SO3(g) + H2O(g) ⇌ H2SO4(l)

*Actually, SO3 is first dissolved in H2SO4 to give “fuming H2SO4”, which is then added to water:

H2SO4.SO3(l) + H2O(g) ⇌ 2H2SO4(l)

This results in a concentrated sulfuric acid solution that contains 98% H2SO4 and 2% water. It is a viscous, colorless liquid. When it is mixed with water, so much heat is liberated that the operation must be carried out very cautiously. The sulfuric acid is slowly poured into the water, not the reverse. The density of this concentrated sulfuric acid solution (98 %) is 1.84 g/cm3 and its concentration is 18.4 M.

Example

Find the molar concentration of commercial sulfuric acid if it is 98% pure and has a density of 1.84g/ cm3.

Given: Density of 98% pure H2SO4 = 1.84g/ cm3.
RTF: its molar concentration

Solution

The density is really the concentration, but we need to express it in units of mol/ dm3 instead of g/ cm3:

$C o n c e n t r a t i o n = \frac{1.84 g s o l u t i o n}{1 c m^{3}} \times \frac{98 g H_{2} S O_{4}}{100 g s o l u t i o n} \times \frac{1 m o l}{98 g} \times \frac{10^{3} c m^{3}}{1 d m^{3}} = 18.4 M$