# Rate of formation of \[S{O_3}\] in the following reaction, \[2S{O_2} + {O_2} \to 2S{O_3}\]\[100kg.{\min ^{ - 1}}\]. Hence, rate of disappearance of \[S{O_2}\] will be:

A.\[100kg.{\min ^{ - 1}}\]

B.\[80kg.{\min ^{ - 1}}\]

C.\[64kg.{\min ^{ - 1}}\]

D.\[32kg.{\min ^{ - 1}}\]

Last updated date: 24th Mar 2023

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Answer

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**Hint:**We need to remember that any component of a reaction with change in time is known as rate of formation. And the rate constant of a reaction is mainly expressed in terms of concentration of reactant and the product. The initial rate of a reaction is taken when the time t is equal to zero. And in the case of product rate of disappearance, it is always positive for reactants and it is negative for products.

**Complete answer:**

We have to know that the rate of disappearance of sulphur dioxide is not equal to \[100kg.{\min ^{ - 1}}\]. Because, the rate of formation of reactants will not equal the rate of disappearance of product. Hence, the option (A) is incorrect.

Here, two mole of sulphur dioxide are reacted with one mole of oxygen and there is a formation of three mole of sulphur trioxide.

\[2S{O_2} + {O_2} \to 2S{O_3}\]

Given, the rate of formation of \[S{O_3}\] is equal to \[100kg.{\min ^{ - 1}}\]. Convert \[100kg.{\min ^{ - 1}}\] into mol/min by dividing the given weight with molecular weight of \[S{O_3}\].

\[Mole = \dfrac{{given\,weight}}{{molecular\,mass\,\,S{O_3}}}\]

Now we can substitute the known values we get,

\[Mole = \dfrac{{100}}{{80}}\]

On simplification we get,

\[Mole = 1.25mol/\min \]

By differentiating the term of sulphur in equation one, will get

\[ - \dfrac{1}{2}\dfrac{{d\left[ {S{O_2}} \right]}}{{dt}} = \dfrac{1}{2}\dfrac{{d\left[ {S{O_3}} \right]}}{{dt}}\]

Substitute the molecular weight and given sulphur trioxide in the above equation.

\[ - \dfrac{1}{2}\dfrac{x}{{64}} = \dfrac{{100}}{{80}}\]

By rearranging the above equation, will get the value of ‘x’

\[x = 80kg/\min \]

Thus, the rate of disappearance of sulphur dioxide, \[S{O_2}\] is equal to \[80kg/\min \]. Hence, the option (B) is correct.

The rate of disappearance is not equal to \[64kg.{\min ^{ - 1}}\] and it is equal to \[80kg/\min \]. Hence, the option (C) is incorrect.

The rate of disappearance is not equal to \[32kg.{\min ^{ - 1}}\] and it is equal to \[80kg/\min \]. Hence, the option (D) is incorrect.

**Note:**

We need to remember that the rate of disappearance of sulphur dioxide, \[S{O_2}\] is equal to \[80kg/\min \]. It is calculated by taking the differentiation of sulphur dioxide as well as sulphur trioxide. . Any component of a reaction with change in time is known as rate of formation. Here, two moles of sulphur dioxide are reacted with one mole of oxygen and there is a formation of three mole of sulphur trioxide.

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