Question

The rate of formation of $S{{O}_{3}}$ in the following reaction $2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}$ is $100g{{\min }^{-1}}$ . Hence, the rate of disappearance of ${{O}_{2}}$​ is:
(A) $50g{{\min }^{-1}}$
(B) $100g{{\min }^{-1}}$
(C) $200g{{\min }^{-1}}$
(D) $20g{{\min }^{-1}}$

Answer 1

Hint: We know that rate of a reaction tells the speed at which the chemical reaction is taking place and during the reaction, concentration of the reactants decreases and the concentration of the product decreases which means that even they affect the rate of the reaction. Rate of formation means the speed at which the product is getting produced whereas rate of disappearance tells the speed at which reactants are getting disappeared.
Formula used:
-For a reaction: ${{n}_{1}}A\to {{n}_{2}}B$
Rate of reaction = $-\dfrac{1}{{{n}_{1}}}\dfrac{[dA]}{[dt]}=\dfrac{1}{{{n}_{2}}}\dfrac{[dB]}{[dt]}$

Complete answer:
Let us first understand the concept of rate of appearance and disappearance followed by calculating the rate of disappearance of ${{O}_{2}}$​ in the given reaction as follows:-
-Rate of formation: It describes the speed at which the product is getting produced and mathematically is shown as:-
Rate of formation =$\dfrac{\text{ }\!\!\Delta\!\!\text{ }\!\![\!\!\text{ products }\!\!]\!\!\text{ }}{\text{ }\!\!\Delta\!\!\text{ t}}$
-Rate of disappearance: It describes the speed at which reactants are getting disappeared and mathematically is shown as:-
Rate of disappearance =$-\dfrac{\text{ }\!\!\Delta\!\!\text{ }\!\![\!\!\text{ reactants }\!\!]\!\!\text{ }}{\text{ }\!\!\Delta\!\!\text{ t}}$
For any reaction${{n}_{1}}A\to {{n}_{2}}B$, the rate of reaction = $-\dfrac{1}{{{n}_{1}}}\dfrac{[dA]}{[dt]}=\dfrac{1}{{{n}_{2}}}\dfrac{[dB]}{[dt]}$
Therefore for$2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}$, the rate of reaction = $-\dfrac{1}{2}\dfrac{[S{{O}_{2}}]}{[dt]}=-\dfrac{[{{O}_{2}}]}{[dt]}=\dfrac{1}{2}\dfrac{[S{{O}_{3}}]}{[dt]}$
-Calculation of the rate of disappearance of ${{O}_{2}}$​as follows:-
Given that rate of formation of $S{{O}_{3}}$= $100g{{\min }^{-1}}$
The molar mass of $S{{O}_{3}}$= 80g/mol and the molar mass of ${{O}_{2}}$​= 32g/mol
The rate of formation of $S{{O}_{3}}$in terms of moles = $\dfrac{100g{{\min }^{-1}}}{80gmo{{l}^{-1}}}=1.25mol{{\min }^{-1}}$
So the rate of disappearance of ${{O}_{2}}$​= $-\dfrac{[{{O}_{2}}]}{[dt]}=\dfrac{1}{2}\dfrac{[S{{O}_{3}}]}{[dt]}$
On substituting the values, we get:-
$\begin{align}
  & \Rightarrow -\dfrac{[{{O}_{2}}]}{[dt]}=\dfrac{1}{2}\dfrac{[S{{O}_{3}}]}{[dt]} \\
 & \Rightarrow -\dfrac{[{{O}_{2}}]}{[dt]}=\dfrac{1}{2}(1.25mol{{\min }^{-1}}) \\
 & \Rightarrow -\dfrac{[{{O}_{2}}]}{[dt]}=0.625mol{{\min }^{-1}} \\
\end{align}$ So the rate of disappearance of ${{O}_{2}}$​in terms of grams= $0.625mol{{\min }^{-1}}\times 32gmo{{l}^{-1}}=20g{{\min }^{-1}}$

Hence the rate of disappearance of ${{O}_{2}}$​ is: (D) $20g{{\min }^{-1}}$

Note:
-Remember to balance the chemical reaction given provided in the question if it isn’t as the coefficients play a vital role in the calculation of reaction rate from the concentration.
-For a reaction: ${{n}_{1}}A\to {{n}_{2}}B$
Rate of reaction =$-\dfrac{1}{{{n}_{1}}}\dfrac{[dA]}{[dt]}=\dfrac{1}{{{n}_{2}}}\dfrac{[dB]}{[dt]}$. The negative sign shows that the concentration of reactant ‘A’ is decreasing and concentration of product ‘B’ is increasing.