
The mole ratio of \[S{O_2}\] and \[{O_2}\]in contact process is:
A. \[{\text{1:2}}\]
B. \[{\text{2:1}}\]
C. \[{\text{3:1}}\]
D. \[{\text{1:3}}\]
Answer
581.7k+ views
Hint: Contact process is the major industrial process and used for making sulphuric acid by oxidizing sulphur dioxide in the presence of catalyst and absorbing the resulting sulphur trioxide in water.
Complete step by step answer:
\[{H_2}S{O_4}\]is one of the most important industrial chemicals prepared by the contact process.
It involves three steps:
(i) Burning of sulphur or sulphide ores in the air to generate\[S{O_2}\]. The \[S{O_2}\]is purified by removing impurities such as arsenic compounds.
(ii) Conversion of \[S{O_2}\]to \[S{O_3}\]by reaction with oxygen in the presence of \[{V_2}{O_5}\](catalyst).
\[2S{O_2}\left( g \right) + 1{O_2}\left( g \right)\xrightarrow{{{V_2}{O_5}}}2S{O_3}\left( g \right){\text{ }}\Delta {\text{H}} = - 196.6{\text{ KJ/mol}}\]
\[2\] mole of \[S{O_2}\]reacts with one mole of \[{O_2}\] to form \[2\] mole of \[S{O_3}\].
The mole ratio of \[S{O_2}\]and \[{O_2}\] in this process is \[2:1\].
Absorption of \[S{O_3}\] in \[{H_2}S{O_4}\] to give oleum \[\left( {{H_2}{S_2}{O_7}} \right)\].
Finally, dilution of oleum with water gives \[{H_2}S{O_4}\]of desired concentration. \[96 - 98\% \]\[{H_2}S{O_4}\] is obtained. The mixture of sulphur dioxide and oxygen going into the reaction is in equal proportion by volume. Avogadro’s law says that equal volume of all gas contains an equal number of moles under the same temperature and pressure. This means that the gases are going into the reaction in the ratio of \[1\] mole of sulphur dioxide and \[1\] mole of oxygen. But an excess of oxygen relative to the proportions
Hence the correct option is (B).
Note:
According to Le – chatelier principle, increasing the concentration of oxygen in the mixture causes the position of equilibrium to shift toward right. Since the oxygen comes from air, this is a very cheap way of increasing the conversion of sulphur dioxide into sulphur trioxide. The contact process depends on the following factors like temperature, pressure, catalyst. Composition of equilibrium mixture and rate of reaction.
Complete step by step answer:
\[{H_2}S{O_4}\]is one of the most important industrial chemicals prepared by the contact process.
It involves three steps:
(i) Burning of sulphur or sulphide ores in the air to generate\[S{O_2}\]. The \[S{O_2}\]is purified by removing impurities such as arsenic compounds.
(ii) Conversion of \[S{O_2}\]to \[S{O_3}\]by reaction with oxygen in the presence of \[{V_2}{O_5}\](catalyst).
\[2S{O_2}\left( g \right) + 1{O_2}\left( g \right)\xrightarrow{{{V_2}{O_5}}}2S{O_3}\left( g \right){\text{ }}\Delta {\text{H}} = - 196.6{\text{ KJ/mol}}\]
\[2\] mole of \[S{O_2}\]reacts with one mole of \[{O_2}\] to form \[2\] mole of \[S{O_3}\].
The mole ratio of \[S{O_2}\]and \[{O_2}\] in this process is \[2:1\].
Absorption of \[S{O_3}\] in \[{H_2}S{O_4}\] to give oleum \[\left( {{H_2}{S_2}{O_7}} \right)\].
Finally, dilution of oleum with water gives \[{H_2}S{O_4}\]of desired concentration. \[96 - 98\% \]\[{H_2}S{O_4}\] is obtained. The mixture of sulphur dioxide and oxygen going into the reaction is in equal proportion by volume. Avogadro’s law says that equal volume of all gas contains an equal number of moles under the same temperature and pressure. This means that the gases are going into the reaction in the ratio of \[1\] mole of sulphur dioxide and \[1\] mole of oxygen. But an excess of oxygen relative to the proportions
Hence the correct option is (B).
Note:
According to Le – chatelier principle, increasing the concentration of oxygen in the mixture causes the position of equilibrium to shift toward right. Since the oxygen comes from air, this is a very cheap way of increasing the conversion of sulphur dioxide into sulphur trioxide. The contact process depends on the following factors like temperature, pressure, catalyst. Composition of equilibrium mixture and rate of reaction.
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