Editing Contact process (section)

== Process ==

The process can be divided into four stages:

# Combining of [[sulfur]] and [[oxygen]] (O<sub>2</sub>) to form [[sulfur dioxide]], then purify the sulfur dioxide in a purification unit
# Adding an excess of oxygen to [[sulfur dioxide]] in the presence of the catalyst [[Vanadium(V) oxide|vanadium pentoxide]] at 450&nbsp;°C and 1-2&nbsp;atm
# The [[sulfur trioxide]] formed is added to [[sulfuric acid]] which gives rise to [[oleum]] (disulfuric acid)
# The oleum is then added to water to form sulfuric acid which is very concentrated. Since this process is an exothermic reaction, the reaction temperature should be as low as possible.

Purification of the air and sulfur dioxide (SO<sub>2</sub>) is necessary to avoid catalyst ''poisoning'' (i.e. removing catalytic activities). The gas is then washed with [[water (molecule)|water]] and dried with sulfuric acid.

To conserve energy, the mixture is heated by exhaust gases from the [[catalytic heater|catalytic converter]] by heat exchangers.

Sulfur dioxide and dioxygen then react as follows:

:2 SO<sub>2(g)</sub> + O<sub>2(g)</sub> ⇌ 2 SO<sub>3(g)</sub> : Δ''H'' = -197&nbsp;kJ·mol<sup>−1</sup>

According to the [[Le Chatelier's principle]], a lower temperature should be used to shift the chemical equilibrium towards the right, hence increasing the percentage yield. However too low of a temperature will lower the formation rate to an uneconomical level. Hence to increase the reaction rate, high temperatures (450&nbsp;°C), medium pressures (1-2&nbsp;[[atmosphere (unit)|atm]]), and [[vanadium(V) oxide]] (V<sub>2</sub>O<sub>5</sub>) are used to ensure an adequate (>95%) conversion. The catalyst only serves to increase the rate of reaction as it does not change the position of the [[thermodynamic equilibrium]]. The mechanism for the action of the catalyst comprises two steps:

# Oxidation of SO<sub>2</sub> into SO<sub>3</sub> by V<sup>5+</sup>:
#: 2SO<sub>2</sub> + 4V<sup>5+</sup> + 2O<sup>2−</sup> → 2SO<sub>3</sub> + 4V<sup>4+</sup>
# Oxidation of V<sup>4+</sup> back into V<sup>5+</sup> by dioxygen (catalyst regeneration):
#: 4V<sup>4+</sup> + O<sub>2</sub> → 4V<sup>5+</sup> + 2O<sup>2−</sup>

Hot sulfur trioxide passes through the heat exchanger and is dissolved in concentrated H<sub>2</sub>SO<sub>4</sub> in the absorption tower to form oleum.

: H<sub>2</sub>SO<sub>4</sub>  + SO<sub>3</sub>  → H<sub>2</sub>S<sub>2</sub>O<sub>7</sub> 

Note that directly dissolving SO<sub>3</sub> in water is impractical due to the highly exothermic nature of the reaction. Acidic vapor or mists are formed instead of a liquid.

Oleum is reacted with water to form concentrated H<sub>2</sub>SO<sub>4</sub>.

:H<sub>2</sub>S<sub>2</sub>O<sub>7</sub>  + H<sub>2</sub>O  → 2 H<sub>2</sub>SO<sub>4</sub>