Sulfuric acid (H2 SO4), the most important industrial...

Sulfuric acid $\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)$, the most important industrial chemical in the world, is prepared by oxidizing sulfur to sulfur dioxide and then to sulfur trioxide. Although sulfur trioxide reacts with water to form sulfuric acid, it forms a mist of fine droplets of $\mathrm{H}_{2} \mathrm{SO}_{4}$ with water vapor that is hard to condense. Instead, sulfur trioxide is first dissolved in 98 percent sulfuric acid to form oleum $\left(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\right) .$ On treatment with water, concentrated sulfuric acid can be generated. Write equations for all the steps and draw Lewis structures of oleum based on the discussion in Example $9.11$.

Key Concepts

Contact Process

The Contact Process is the principal industrial method for producing sulfuric acid. It involves the sequential oxidation of elemental sulfur to sulfur dioxide and then to sulfur trioxide, which is subsequently absorbed into concentrated sulfuric acid to produce oleum. This process, renowned for its efficiency and scalability, underscores the integration of chemical reaction engineering with industrial production techniques.

Oxidation Reactions

Oxidation reactions are a fundamental type of redox process where a substance loses electrons. In the industrial synthesis of sulfuric acid, elemental sulfur is oxidized to sulfur dioxide, and sulfur dioxide is further oxidized to sulfur trioxide. Each step involves an increase in the oxidation state and requires careful control of reaction conditions such as temperature, pressure, and the presence of catalysts.

Oleum Formation

Oleum is produced by absorbing sulfur trioxide in concentrated sulfuric acid rather than water, because direct hydration of sulfur trioxide leads to the formation of a problematic mist of acid droplets. The resulting compound, oleum (H2S2O7), represents an intermediate that can be safely converted back to concentrated sulfuric acid upon subsequent dilution with water.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist. They are essential for visualizing the connectivity and electron distribution in molecules such as oleum. Constructing the Lewis structure for oleum involves accounting for all valence electrons, assigning formal charges where necessary, and depicting the resonance stabilization that can occur in such polyatomic ions.

Transcript

00:01 Hello, so today we'll be talking about how we make sulfuric acid.

00:06 So first what we do is we oxidize sulfur into sulfur dioxide.

00:11 So basically we're taking sulfur gas and we're reacting it with oxygen, oxidizing it.

00:22 And we're making sulfur dioxide gas.

00:28 And if we look, we'll see that this equation is balanced.

00:32 We've got one sulfur on both sides and two oxygens on both sides.

00:37 Next, we take the sulfur dioxide and we oxidize it again into sulfur trioxide.

00:44 So we're taking sulfur dioxide, reacting it again with oxygen, and we're making sulfur trioxide.

01:03 And what you'll notice is this equation is not balanced.

01:13 We have four oxygens on the left side and only three on the other.

01:18 So we have to find a way to balance this equation.

01:20 So first, how about we have two sulfur trioxide molecules? in order to have two sulphurs, we need to have two sulphurs on the left side.

01:34 So we'll have two sulfur dioxide molecules.

01:38 And then we see that we've got four, i mean six oxygen molecules.

01:44 On the left side and six on the right side so it's already balanced so we've balanced it now in the next step we dissolve sulfur troyoxide into water and we create sulfuric acid so sulfur dioxide is a gas and then we dissolve it into liquid water and that will form sulfuric acid and it's dissolved in the the water so it is aqueous and if we look at this we'll see we have two hydrogen on both sides one sulfur on both sides and four oxygens on both sides so it's already balanced now what we do is we dissolve more sulfur dioxide into the aqueous sulfuric acid to form oleum so we take sulfur sulfur peroyl peroxide, we react it with aquas, sulfuric acid, and we create oleum.

03:26 And this will also be aquis.

03:35 And so we see, if we look at both sides, we've got two hydrogens on both sides, two sulfurs on both sides, and seven oxygens on both sides.

03:47 So it is balanced.

03:51 And next, we take oleum and we dissolve it in water again to get sulfuric acid again.

04:03 So, and we'll see that this is not balanced.

04:36 We've got four hydrogens on one side, two hydrogens on the other.

04:43 So how about to counteract that? we added two to the side.

04:47 And now let's check if that will make it correct.

04:51 And we see we've got four hydrogens on both sides now.

04:55 And we've got two sulfurs on both sides.

04:59 And we've got eight oxygens on both sides.

05:01 So now it is balanced.

05:03 So we do equations for all of the steps and they're all balanced now.

05:10 So now let's try to draw allous structure for oleum.

05:18 So first let's count up the valence electron.

05:21 Hydrogen has one valence electron, sulfur has six and there are two of them, and oxygen has six and there are seven of them.

05:39 So seven times six is 42, 42 plus 12 is 54, plus another two is 56 electrons.

05:49 Now let's take a look at if everything follows that octet rule, how many bonds we will need.

06:00 So hydrogen has two, needs two valence electrons, and there are two of them...

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