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Hint: We know that the contact process is used for the formation of sulphuric acid. In this process there is an occurrence of three reactions, the intermediate product is sulphur dioxide. So, write down the balanced equations occurring in the contact process.
Complete step by step solution:
Now, we will discuss the contact process. As mentioned, it is used for the manufacturing of sulphuric acid.
Let us discuss the three steps. In the first step we have the manufacturing of sulphur dioxide.
We can write the chemical reaction as:
S (g) + O$_2$(g) $\rightarrow$ SO$_2$ (g)
In this reaction, sulphur, and oxygen combines; and leads to the formation of sulphur dioxide.
Now, we have the second step; in this step the reaction will move forward, i.e. we will use the sulphur dioxide, and it will react in the presence of oxygen gas.
So, the chemical reaction can be written as:
2SO$_2$ (g) + O$_2$(g) $\rightleftharpoons$ 2SO$_3$ (g)
In this reaction, vanadium oxide is used as a catalyst. We can see that the product formed is sulphur trioxide.
Talking about the third step we have the use of sulphur trioxide, and in this step the sulphur dioxide will react with the sulphuric acid. The chemical reaction is as follows:
SO$_3$ (g) + H$_2$SO$_4$ $\rightarrow$ H$_2$S$_2$O$_7$ (l)
In this reaction, oleum is formed.
Now, we have the second reaction in the third step, i.e. further the reaction will happen in the presence of water. The reaction is
H$_2$S$_2$O$_7$ (l) + H$_2$O(l) $\rightarrow$ 2H$_2$SO$_4$
In this step, there is formation of concentrated sulphuric acid.
In the last we can conclude that the concentrated sulphuric acid is formed from the sulphur dioxide as mentioned above.
Note: Don’t get confused about why there is formation of oleum first, then it reacts with the water to form sulphuric acid. The reason behind this is that sulphur trioxide is unable to dissolve in the water directly, it will form the fog.
Complete step by step solution:
Now, we will discuss the contact process. As mentioned, it is used for the manufacturing of sulphuric acid.
Let us discuss the three steps. In the first step we have the manufacturing of sulphur dioxide.
We can write the chemical reaction as:
S (g) + O$_2$(g) $\rightarrow$ SO$_2$ (g)
In this reaction, sulphur, and oxygen combines; and leads to the formation of sulphur dioxide.
Now, we have the second step; in this step the reaction will move forward, i.e. we will use the sulphur dioxide, and it will react in the presence of oxygen gas.
So, the chemical reaction can be written as:
2SO$_2$ (g) + O$_2$(g) $\rightleftharpoons$ 2SO$_3$ (g)
In this reaction, vanadium oxide is used as a catalyst. We can see that the product formed is sulphur trioxide.
Talking about the third step we have the use of sulphur trioxide, and in this step the sulphur dioxide will react with the sulphuric acid. The chemical reaction is as follows:
SO$_3$ (g) + H$_2$SO$_4$ $\rightarrow$ H$_2$S$_2$O$_7$ (l)
In this reaction, oleum is formed.
Now, we have the second reaction in the third step, i.e. further the reaction will happen in the presence of water. The reaction is
H$_2$S$_2$O$_7$ (l) + H$_2$O(l) $\rightarrow$ 2H$_2$SO$_4$
In this step, there is formation of concentrated sulphuric acid.
In the last we can conclude that the concentrated sulphuric acid is formed from the sulphur dioxide as mentioned above.
Note: Don’t get confused about why there is formation of oleum first, then it reacts with the water to form sulphuric acid. The reason behind this is that sulphur trioxide is unable to dissolve in the water directly, it will form the fog.
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