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Question: Rate of formation \(S{{O}_{3}}\) according to the reaction \(2S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}\) ...

Rate of formation SO3S{{O}_{3}} according to the reaction 2SO2+O2→2SO32S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}} is 1.6×10−3kgmin⁡−11.6\times {{10}^{-3}}kg{{\min }^{-1}} . Hence rate at which SO2S{{O}_{2}} reacts is:
(A) 1.6×10−3kgmin⁡−11.6\times {{10}^{-3}}kg{{\min }^{-1}}
(B) 8.0×10−4kgmin⁡−18.0\times {{10}^{-4}}kg{{\min }^{-1}}
(C) 3.2×10−3kgmin⁡−13.2\times {{10}^{-3}}kg{{\min }^{-1}}
(D) 1.28×10−3kgmin⁡−11.28\times {{10}^{-3}}kg{{\min }^{-1}}

Explanation

Solution

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→aA+bBxX+yY\to aA+bB
The rate of reaction = rate of disappearance of reactant = rate of formation of products
r= −1xd[X]dt=−1yd[Y]dt=1ad[A]dt=1bd[B]dt-\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],&d[B]d[X],d[Y],d[A],\And d[B], is the change in the concentrations of reactants and products and dtdt will be changed in time.
Given reaction,
2SO2+O2→2SO32S{{O}_{2}}+{{O}_{2}}\to 2S{{O}_{3}}
The rate of reaction = rate of disappearance of SO2S{{O}_{2}} = rate of formation of SO3S{{O}_{3}}
= −12d[SO2]dt=+12d[SO3]dt-\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 SO3S{{O}_{3}} = 1.6×10−3kgmin⁡−11.6\times {{10}^{-3}}kg{{\min }^{-1}}
Molar mass of SO3S{{O}_{3}} = 80g/mol
d[SO3]dt=1.6×10−3kgmin⁡−1 ⇒d[SO3]dt=1.6×10−3molmin⁡−180=0.2×10−4mol/min⁡ \begin{aligned} & \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{aligned}
From equation (2),
d[SO2]dt=d[SO3]dt ⇒d[SO2]dt=0.2×10−4mol/min⁡ \begin{aligned} & \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{aligned}
Multiply the above rate of SO2S{{O}_{2}} with it molar mass = 64 g/mol, then
d[SO2]dt=0.2×10−4×64kgmin⁡−1=1.28×10−3kgmin⁡−1\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 SO2S{{O}_{2}} reacting is 1.28×10−3kgmin⁡−11.28\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.