Question

Rate of formation of $S{O_3}$ ​ in the following reaction \[2S{O_2}\, + {O_2}\, \to \,\,2S{O_3}\]​ is \[100\;\,kg\,mi{n^{ - 1}}\]. Hence rate of disappearance of $S{O_2}$​ will be:
(A) \[100\;\,kg\,mi{n^{ - 1}}\]
(B) \[80\;\,kg\,mi{n^{ - 1}}\]
(C) \[64\,kg\,mi{n^{ - 1}}\]
(D) \[32\,kg\,mi{n^{ - 1}}\]

Answer 1

Hint: Rate of formation in chemical kinetics is defined as the change in concentration of the product with respect to time and the rate of disappearance is defined as the change in concentration of the reactant with respect to time. The concentration of the product increases as the reaction proceeds further so it is taken as positive while the concentration of reactant decreases so it is taken as negative.

Complete step by step answer:
For a reaction, $3X\, + \,2Y \to Z$. The rate of the formation of the reaction will be;
$rate\,\,of\,formation = \,\dfrac{{\Delta Z}}{{\Delta t}}$ and the rate of disappearance will be; $rate\,\,of\,disappearance\,of\,X = \,\dfrac{{\Delta X}}{{\Delta t}}$
Given reaction is; \[2S{O_2}\, + {O_2}\, \to \,\,2S{O_3}\], The rate of formation of $S{O_3}$ is $ = 100\;\,kg\,mi{n^{ - 1}}$
We have to convert the rate of formation into $moles/{\text{minute}}$ So,
Moles $ = \dfrac{{given\,weight\,}}{{molar\,weight}} = \dfrac{{100}}{{80}}$
Now, the rate of the reaction will be;
\[rate\;\, = \dfrac{1}{2}\dfrac{{[\Delta S{O_2}]}}{{\Delta t}}\, = \dfrac{{[\Delta {O_2}]}}{{\Delta t}}\, = \dfrac{1}{2}\dfrac{{[\Delta S{O_3}]}}{{\Delta t}}\]
Now we will calculate the rate of disappearance of $S{O_2}$​. It is given as;
The rate of formation of $S{O_3}$$ = $ the rate of disappearance of $S{O_2}$​
\[ + \dfrac{1}{2}\dfrac{{[\Delta S{O_3}]}}{{\Delta t}}\, = - \dfrac{1}{2}\dfrac{{[\Delta S{O_2}]}}{{\Delta t}}\]
\[\dfrac{1}{2}\dfrac{{weight\,of\,S{O_3}\,}}{{molar\,weight\,of\,S{O_3}\,}}\, = \dfrac{1}{2}\dfrac{{weight\,of\,S{O_2}}}{{molar\,weight\,of\,S{O_2}}}\]
Let the weight of $S{O_2}$ be $x$. Now we will solve the equation for $x$.
\[\dfrac{{100\,}}{{80\,}}\, = \dfrac{{x}}{{64}}\, = 1.25 = \dfrac{x}{{64}}\]
$x = \,64\,\, \times \,1.25 = 80$
The value of $x$ is \[80\;\,kg\,mi{n^{ - 1}}\]. Hence, the rate of disappearance of $S{O_2}$ will be \[80\;\,kg\,mi{n^{ - 1}}\].

Therefore, Option (B) is correct.
Note:
The rate of reaction is divided into two categories (a) average rate and (b) instantaneous rate. It depends on the amount of time period. If the time taken is finite, then it’s called average rate and if we talk about at a certain point of time it is called instantaneous rate. The rate depends upon temperature because as we increase the temperature the effective collisions between the particles will increase. It also depends on pressure, catalyst, and the stoichiometric coefficients of the reactants and the products.