Question

Rate of formation $S{{O}_{3}}$ according to the reaction $2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}$ is $1.6\times {{10}^{-3}}kg{{\min }^{-1}}$ . Hence rate at which $S{{O}_{2}}$ reacts is:
(A) $1.6\times {{10}^{-3}}kg{{\min }^{-1}}$
(B) $8.0\times {{10}^{-4}}kg{{\min }^{-1}}$
(C) $3.2\times {{10}^{-3}}kg{{\min }^{-1}}$
(D) $1.28\times {{10}^{-3}}kg{{\min }^{-1}}$

Answer 1

Hint: As the reaction proceeds, the number of reactants decreases and the number of products increase in a chemical reaction. The number of reactants consumed or the number of products formed depends on the overall rate of the reaction. The rate of the formation or the rate of disappearance is equal to the overall rate of the reaction.

Complete step by step solution:
For a chemical reaction, the measure of the change in concentration of reactants or the change in concentration of products per unit time is known as the reaction rate.
Consider the stoichiometrically complicated reaction,
$xX+yY\to aA+bB$
The rate of reaction = rate of disappearance of reactant = rate of formation of products
r= $-\dfrac{1}{x}\dfrac{d[X]}{dt}=-\dfrac{1}{y}\dfrac{d[Y]}{dt}=\dfrac{1}{a}\dfrac{d[A]}{dt}=\dfrac{1}{b}\dfrac{d[B]}{dt}$  (Equation 1)
Where r = rate of overall reaction
$d[X],d[Y],d[A],\And d[B]$, is the change in the concentrations of reactants and products and $dt$ will be changed in time.
Given reaction,
$2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}$
The rate of reaction = rate of disappearance of $S{{O}_{2}}$ = rate of formation of $S{{O}_{3}}$
= $-\dfrac{1}{2}\dfrac{d[S{{O}_{2}}]}{dt}=+\dfrac{1}{2}\dfrac{d[S{{O}_{3}}]}{dt}$ (Equation 2)
According to the given reaction, rate of formation of $S{{O}_{3}}$ = $1.6\times {{10}^{-3}}kg{{\min }^{-1}}$
Molar mass of $S{{O}_{3}}$ = 80g/mol
$\begin{align}
& \dfrac{d[S{{O}_{3}}]}{dt}=1.6\times {{10}^{-3}}kg{{\min }^{-1}} \\
& \Rightarrow \dfrac{d[S{{O}_{3}}]}{dt}=\dfrac{1.6\times {{10}^{-3}}mol{{\min }^{-1}}}{80}=0.2\times {{10}^{-4}}mol/\min \\
\end{align}$
From equation (2),
$\begin{align}
& \dfrac{d[S{{O}_{2}}]}{dt}=\dfrac{d[S{{O}_{3}}]}{dt} \\
& \Rightarrow \dfrac{d[S{{O}_{2}}]}{dt}=0.2\times {{10}^{-4}}mol/\min \\
\end{align}$
Multiply the above rate of $S{{O}_{2}}$ with it molar mass = 64 g/mol, then
$\dfrac{d[S{{O}_{2}}]}{dt}=0.2\times {{10}^{-4}}\times 64kg{{\min }^{-1}}=1.28\times {{10}^{-3}}kg{{\min }^{-1}}$
Hence, the rate of $S{{O}_{2}}$ reacting is $1.28\times {{10}^{-3}}kg{{\min }^{-1}}$

So, the correct answer is option D.

Note: Based on the collision theory, reactant molecules collide with each other to form products. The number of colliding particles will increase, then the rate of reaction increases by increasing the concentration of reaction.