how is sulphuric acid manufactured by contact process
Answers
Answered by
3
hope this will help u......
Attachments:
Answered by
2
The process can be divided into five stages:
Combining of sulfur and oxygen (O2) to form sulfur dioxide
Purifying the sulfur dioxide in a purification unit
Adding an excess of oxygen to sulfur dioxide in the presence of the catalyst vanadium pentoxide at 450 °C and 1-2 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.
Purification of the air and sulfur dioxide (SO2) is necessary to avoid catalyst poisoning (i.e. removing catalytic activities). The gas is then washed with water and dried with sulfuric acid.
To conserve energy, the mixture is heated by exhaust gases from the catalytic converter by heat exchangers.
Sulfur dioxide and dioxygen then react as follows:
2 SO2(g) + O2(g) ⇌ 2 SO3(g) : ΔH = -197 kJ·mol−1
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 °C), medium pressures (1-2 atm), and vanadium(V) oxide (V2O5) 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 SO2 into SO3 by V5+:
2SO2 + 4V5+ + 2O2− → 2SO3 + 4V4+
Oxidation of V4+ back into V5+ by dioxygen (catalyst regeneration):
4V4+ + O2 → 4V5+ + 2O2−
Hot sulfur trioxide passes through the heat exchanger and is dissolved in concentrated H2SO4 in the absorption tower to form oleum:
H2SO4 (l) + SO3 (g) → H2S2O7 (l)
Note that directly dissolving SO3 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 H2SO4.
H2S2O7 (l) + H2O (l) → 2 H2SO4 (l)
Combining of sulfur and oxygen (O2) to form sulfur dioxide
Purifying the sulfur dioxide in a purification unit
Adding an excess of oxygen to sulfur dioxide in the presence of the catalyst vanadium pentoxide at 450 °C and 1-2 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.
Purification of the air and sulfur dioxide (SO2) is necessary to avoid catalyst poisoning (i.e. removing catalytic activities). The gas is then washed with water and dried with sulfuric acid.
To conserve energy, the mixture is heated by exhaust gases from the catalytic converter by heat exchangers.
Sulfur dioxide and dioxygen then react as follows:
2 SO2(g) + O2(g) ⇌ 2 SO3(g) : ΔH = -197 kJ·mol−1
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 °C), medium pressures (1-2 atm), and vanadium(V) oxide (V2O5) 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 SO2 into SO3 by V5+:
2SO2 + 4V5+ + 2O2− → 2SO3 + 4V4+
Oxidation of V4+ back into V5+ by dioxygen (catalyst regeneration):
4V4+ + O2 → 4V5+ + 2O2−
Hot sulfur trioxide passes through the heat exchanger and is dissolved in concentrated H2SO4 in the absorption tower to form oleum:
H2SO4 (l) + SO3 (g) → H2S2O7 (l)
Note that directly dissolving SO3 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 H2SO4.
H2S2O7 (l) + H2O (l) → 2 H2SO4 (l)
Similar questions