Question

The rate of formation of \[S{O_3}\] in the following reaction \[2S{O_2} + {O_2} \to 2S{O_3}\] is \[{\text{100 g mi}}{{\text{n}}^{{\text{ - 1}}}}\] , Hence, the rate of disappearance of \[{O_2}\] is:
A. \[2{\text{ g }}{\min ^{ - 1}}\]
B. \[20{\text{ g }}{\min ^{ - 1}}\]
C. \[200{\text{ g }}{\min ^{ - 1}}\]
D. \[50{\text{ g }}{\min ^{ - 1}}\]

Answer 1

Hint: Rate of disappearance of reactant is equal to the rate of appearance of the product. The general representation for the reaction \[{\text{A}} + {\text{B}} \to {\text{C}}\] is given as follows:
\[\dfrac{{ - d\left[ {\text{A}} \right]}}{{dt}} = \dfrac{{ - d\left[ {\text{B}} \right]}}{{dt}} = \dfrac{{ + d\left[ {\text{C}} \right]}}{{dt}}\]
Here reactant A and B are denoted in negative because their concentration decreases with respect to time. And product C is positive because its concentration increases with respect to time.

Complete step by step answer:
In the following question we are provided with a change in weight (grams) of \[S{O_3}\] with respect to time. So we have to convert it into a change in the number of moles with respect to time by dividing it by molecular mass of \[S{O_3}\].
\[\dfrac{{d\left[ {S{O_3}} \right]}}{{dt}} = \dfrac{{100{\text{ g }}{{\min }^{ - 1}}}}{{80{\text{ g mo}}{{\text{l}}^{{\text{ - 1}}}}}}\]
By dividing we get:
\[\dfrac{{d\left[ {S{O_3}} \right]}}{{dt}} = 1.25{\text{ mol }}{\min ^{ - 1}}\] (equation 1)
Let us first write the given reaction
\[2S{O_2} + {O_2} \to 2S{O_3}\]
We know how to write the rate for the following reaction:
Rate of disappearance of \[S{O_2}\] = \[\dfrac{{ - 1}}{2}\dfrac{{d\left[ {S{O_2}} \right]}}{{dt}}\]
Here, the stoichiometric coefficient 2 of \[S{O_2}\] is divided with the rate of disappearance.
Rate of disappearance of \[{O_2}\] = \[\dfrac{{ - d\left[ {{O_2}} \right]}}{{dt}}\]
Here, the stoichiometric coefficient of \[{O_2}\] is 1.
Rate of the appearance of \[S{O_3}\] = \[ + \dfrac{1}{2}\dfrac{{d\left[ {S{O_3}} \right]}}{{dt}}\]
Here, the stoichiometric coefficient 2 of \[S{O_3}\] is divided with the rate of appearance.
Now, we know that the rate of disappearance of the reactant is equal to the rate of appearance of the product.
So, by combining all the three equation we get:
\[\dfrac{{ - 1}}{2}\dfrac{{d\left[ {S{O_2}} \right]}}{{dt}} = \dfrac{{ - d\left[ {{O_2}} \right]}}{{dt}} = + \dfrac{1}{2}\dfrac{{d\left[ {S{O_3}} \right]}}{{dt}}\]
Now we will only consider the term consisting of \[{O_2}\] and \[S{O_3}\] .
\[\dfrac{{ - d\left[ {{O_2}} \right]}}{{dt}} = + \dfrac{1}{2}\dfrac{{d\left[ {S{O_3}} \right]}}{{dt}}\]
By substituting the value from equation 1 we get:
\[\dfrac{{ - d\left[ {{O_2}} \right]}}{{dt}} = + \dfrac{1}{2}1.25{\text{ }}\left( {{\text{mol }}{{\min }^{ - 1}}} \right)\]
By solving the above equation we will get the rate of disappearance of \[{O_2}\] :
\[\dfrac{{ - d\left[ {{O_2}} \right]}}{{dt}} = {\text{ }} + 0.625\left( {{\text{mol }}{{\min }^{ - 1}}} \right)\]
Now to convert the value into grams we have to multiply it with the molecular mass of \[{O_2}\]:
\[\dfrac{{ - d\left[ {{O_2}} \right]}}{{dt}} = {\text{ }} + 0.625\left( {{\text{mol mi}}{{\text{n}}^{{\text{ - 1}}}}} \right) \times 32\left( {{\text{g mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)\]
So, we get the Rate of disappearance of \[{O_2}\] in grams
\[\dfrac{{ - d\left[ {{O_2}} \right]}}{{dt}} = {\text{ }}20{\text{ g }}{\min ^{ - 1}}\]

Therefore, we can conclude that the correct answer to this question is option B.

Note:
This type of question always focuses on the units associated with the rate of formation of product and rate of disappearance of reactant. Always convert grams per minute into moles per minute.