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Question: Rate of formation of \[S{O_3}\] in the following reaction, \[2S{O_2} + {O_2} \to 2S{O_3}\]\[100kg.{\...

Rate of formation of SO3S{O_3} in the following reaction, 2S{O_2} + {O_2} \to 2S{O_3}$$$$100kg.{\min ^{ - 1}}. Hence, rate of disappearance of SO2S{O_2} will be:
A.100kg.min1100kg.{\min ^{ - 1}}
B.80kg.min180kg.{\min ^{ - 1}}
C.64kg.min164kg.{\min ^{ - 1}}
D.32kg.min132kg.{\min ^{ - 1}}

Explanation

Solution

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.min1100kg.{\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.
2SO2+O22SO32S{O_2} + {O_2} \to 2S{O_3}
Given, the rate of formation of SO3S{O_3} is equal to 100kg.min1100kg.{\min ^{ - 1}}. Convert 100kg.min1100kg.{\min ^{ - 1}} into mol/min by dividing the given weight with molecular weight of SO3S{O_3}.
Mole=givenweightmolecularmassSO3Mole = \dfrac{{given\,weight}}{{molecular\,mass\,\,S{O_3}}}
Now we can substitute the known values we get,
Mole=10080Mole = \dfrac{{100}}{{80}}
On simplification we get,
Mole=1.25mol/minMole = 1.25mol/\min
By differentiating the term of sulphur in equation one, will get
12d[SO2]dt=12d[SO3]dt- \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.
12x64=10080- \dfrac{1}{2}\dfrac{x}{{64}} = \dfrac{{100}}{{80}}
By rearranging the above equation, will get the value of ‘x’
x=80kg/minx = 80kg/\min
Thus, the rate of disappearance of sulphur dioxide, SO2S{O_2} is equal to 80kg/min80kg/\min . Hence, the option (B) is correct.
The rate of disappearance is not equal to 64kg.min164kg.{\min ^{ - 1}} and it is equal to 80kg/min80kg/\min . Hence, the option (C) is incorrect.
The rate of disappearance is not equal to 32kg.min132kg.{\min ^{ - 1}} and it is equal to 80kg/min80kg/\min . Hence, the option (D) is incorrect.

Note:
We need to remember that the rate of disappearance of sulphur dioxide, SO2S{O_2} is equal to 80kg/min80kg/\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.