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Some webpages describe the contact process as this (I left out $\ce{SO2}$, $\ce{SO3}$ production):

$\ce{SO3 + H2SO4 -> H2S2O7}$

After this, it is controllable to add water to the oleum therefore making liquid sulfuric acid. However, some other sites describe this:

$\ce{SO3 + H2O -> H2SO4}$

I am confused about what really happens in the absorption tower. I've read about the wet acid, but it states that adding water causes gaseous $\ce{H2SO4}$ to form, which is condensed.

EDIT: In the absorption tower, trioxide sulfur enters as gas, and sulfuric acid comes down as liquid. As a result the liquid comes out as oleum (first eq.). The other contact process states different: the liquid entering is 98% sulfuric acid, and 2% water, this water reacts with the sulfur trioxide (this little water avoids the heat problem), giving strengthened sulfuric acid (let's say 99.5% sulfuric acid, and 0.5% water). Still confused.

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In principle, your sources are correct and not contradictory.

In order to avoid the formation of problematic sulfuric acid mist, sufur trioxide is not directly absorbed into water. Instead, it is absorbed in sulfuric acid of 98 %–99 % concentration, which approximately corresponds to an azeotrope where the partial pressures of sufur trioxide, sulfuric acid, and water are relatively low.

Sulfuric acid reacts with the absorbed sulfur trioxide to form disulfuric acid:

$$\ce{H2SO4 + SO3 <=> H2S2O7}$$

The disulfuric acid can be converted to sulfuric acid by reaction with added water:

$$\ce{H2S2O7 + H2O <=> 2 H2SO4}$$

Thus, the overall net reaction is as follows:

$$\begin{align} \ce{H2SO4 + SO3 \;&<=> H2S2O7}\\ \ce{H2S2O7 + H2O \;&<=> 2 H2SO4}\\ \hline\ce{SO3 + H2O \;&<=> H2SO4} \end{align}$$

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  • $\begingroup$ Thank you, but sources state either one of the equations as the contact process, which is what happens in the absorption tower. And adding the water occurs later, outside the tower. $\endgroup$ – Dave May 31 '15 at 21:38
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The answer to your question is thoroughly explained in Chemistry of the Elements (Second Edition), A. Earnshaw and Norman Greenwood, 1997, p. 708:

(...) The $\ce{SO3}$ gas cannot be absorbed directly in water because it would first come into contact with the water-vapour above the absorbant and so produce a stable mist of fine droplets of $\ce{H2SO4}$ which would then pass right through the absorber and out into the atmosphere. Instead, absorption is effected by 98% $\ce{H2SO4}$ in ceramic packed towers and sufficient water is added to the circulating acid to maintain the required concentration. (...)

Two pages following, the process is represented as:

$$\ce{SO3 + H2O\ ($\text{in 98% }$H2SO4) -> H2SO4}$$

In the Industrial Manufacture of Sulfuric Acid box (p. 708-710), the $\ce{H2S2O7}$ species is not mentioned. However, the discussion about concentrated Sulfuric Acid states:

It is clear that "pure" anhydrous sulfuric acid, far from being a single substance in the bulk liquid phase, comprises a dynamic equilibrium involving at least seven well-defined species. The concentration of the self-dissociation products in $\ce{H2SO4}$ at 25º (expressed in millimoles of solute per kg solvent) are:

$$ \begin{array}{lcr} \ce{HSO4-} & \ce{H3SO4+} & \ce{H3O+} & \ce{HS2O7-} & \ce{H2S2O7} & \ce{H2O} & \text{Total}\\ \hline 15.0 & 11.3 & 8.0 & 4.4 & 3.6 & 0.1 & 42.4 \end{array} $$

Considering the above, you can assume that the absorption of $\ce{SO3}$ in 98% $\ce{H2SO4}$ will lead to the formation of $\ce{H2S2O7}$ but also several other species. If you are really interested in industrial production details and properties of sulfuric acid, this reference is really valuable.

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  • $\begingroup$ Thanks. This is what I read how they add water to the sulfur trioxide in the absorption tower. Which is different to the other contact process equation, hence my confusion. $\endgroup$ – Dave May 31 '15 at 21:36

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