
Presence of traces of arsenious oxide ($A{{s}_{2}}{{O}_{3}}$ ) in the reacting gases $S{{O}_{2}}$ and ${{O}_{2}}$ in presence of platinised asbestos in contact process acts as:
(a) catalytic polymer
(b) catalytic poison
(c) dehydrating agent
(d) drying agent
Answer
588k+ views
Hint: A promoter is something which enhances the activity of a catalyst; while some compounds can reduce the catalytic activity of a catalyst. Presence of some impurities can decrease the activity of the catalyst.
Complete answer:
Contact process is an industrial method to produce sulphuric acid. Sulphur dioxide and oxygen is passed over a hot catalyst and unite to form sulphur trioxide which then combines with water to make sulphuric acid.
Sulphuric acid is produced in high concentrations by contact process. The process involves five stages.
- First step is combining sulphur and oxygen to form sulphur dioxide.
$2S{{O}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2S{{O}_{3(g)}}:\Delta H=-197kJmo{{l}^{-1}}$
- Sulphur dioxide is purified.
- Excess oxygen is added to sulphur dioxide in the presence of a catalyst at 450 degree Celsius and 1-2 atm.
$S{{O}_{2}}+{{O}_{2}}\to S{{O}_{3}}$
- From the above step sulphur trioxide is formed which is added to sulphuric acid to give oleum.
${{H}_{2}}S{{O}_{4}}(l)+S{{O}_{3(g)}}\to {{H}_{2}}{{S}_{2}}{{O}_{7}}(l)$
- To water, the oleum is added, we get sulphuric acid in high concentrations.
${{H}_{2}}{{S}_{2}}{{O}_{7}}(l)+{{H}_{2}}O(l)\to 2{{H}_{2}}S{{O}_{4}}(l)$
During the fourth step, the reaction is carried out in the presence of platinised asbestos which acts as catalyst.
If any arsenic compounds are present, it can reduce the catalytic activity of the platinised asbestos. Especially when traces of $A{{s}_{2}}{{O}_{3}}$ , it poisons the catalyst.
A promoter increases the activity of a catalyst while the substance which reduces the catalytic activity is called poison.
Thus, $A{{s}_{2}}{{O}_{3}}$ acts as a catalytic poison in contact process during the formation of sulphur trioxide and the correct option is (b) catalytic poison.
Note:
Nowadays vanadium pentoxide is used as a catalyst in the contact process. Its catalytic activity is not decreased due to the presence of arsenic impurities. So, it is more preferred.
Complete answer:
Contact process is an industrial method to produce sulphuric acid. Sulphur dioxide and oxygen is passed over a hot catalyst and unite to form sulphur trioxide which then combines with water to make sulphuric acid.
Sulphuric acid is produced in high concentrations by contact process. The process involves five stages.
- First step is combining sulphur and oxygen to form sulphur dioxide.
$2S{{O}_{2(g)}}+{{O}_{2(g)}}\rightleftharpoons 2S{{O}_{3(g)}}:\Delta H=-197kJmo{{l}^{-1}}$
- Sulphur dioxide is purified.
- Excess oxygen is added to sulphur dioxide in the presence of a catalyst at 450 degree Celsius and 1-2 atm.
$S{{O}_{2}}+{{O}_{2}}\to S{{O}_{3}}$
- From the above step sulphur trioxide is formed which is added to sulphuric acid to give oleum.
${{H}_{2}}S{{O}_{4}}(l)+S{{O}_{3(g)}}\to {{H}_{2}}{{S}_{2}}{{O}_{7}}(l)$
- To water, the oleum is added, we get sulphuric acid in high concentrations.
${{H}_{2}}{{S}_{2}}{{O}_{7}}(l)+{{H}_{2}}O(l)\to 2{{H}_{2}}S{{O}_{4}}(l)$
During the fourth step, the reaction is carried out in the presence of platinised asbestos which acts as catalyst.
If any arsenic compounds are present, it can reduce the catalytic activity of the platinised asbestos. Especially when traces of $A{{s}_{2}}{{O}_{3}}$ , it poisons the catalyst.
A promoter increases the activity of a catalyst while the substance which reduces the catalytic activity is called poison.
Thus, $A{{s}_{2}}{{O}_{3}}$ acts as a catalytic poison in contact process during the formation of sulphur trioxide and the correct option is (b) catalytic poison.
Note:
Nowadays vanadium pentoxide is used as a catalyst in the contact process. Its catalytic activity is not decreased due to the presence of arsenic impurities. So, it is more preferred.
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