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Question: Rate of formation of \(S{O_3}\) ​ according to the \[2S{O_2}\, + {O_2}\, \to \,\,2S{O_3}\] ​is \[1.6...

Rate of formation of SO3S{O_3} ​ according to the 2SO2 +O2 →  2SO32S{O_2}\, + {O_2}\, \to \,\,2S{O_3} ​is 1.6×10−3 kgmin−11.6 \times {10^{ - 3}}\,kgmi{n^{ - 1}}. Hence rate at which SO2S{O_2} ​ reacts is:
(A) 1.6×10−3 kgmin−11.6 \times {10^{ - 3}}\,kgmi{n^{ - 1}}
(B) 8.0×10−4 kg min−18.0 \times {10^{ - 4}}\,kg\,mi{n^{ - 1}}
(C) 3.2×10−3 kg min−13.2 \times {10^{ - 3}}\,kg\,mi{n^{ - 1}}
(D) 1.28  ×10−3 kg min−11.28\,\, \times {10^{ - 3}}\,kg\,mi{n^{ - 1}}

Explanation

Solution

The 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.

Complete step by step answer:
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. For a reaction, 3X + 2Y→Z3X\, + \,2Y \to Z. The rate of the formation of the reaction will be;
rate  of formation= ΔZΔtrate\,\,of\,formation = \,\dfrac{{\Delta Z}}{{\Delta t}} and the rate of disappearance will be:rate  of disappearance of X= ΔXΔtrate\,\,of\,disappearance\,of\,X = \,\dfrac{{\Delta X}}{{\Delta t}}
Given reaction is 2SO2 +O2 →  2SO32S{O_2}\, + {O_2}\, \to \,\,2S{O_3} the rate of formation of SO3S{O_3} is 1.6×10−3 kgmin−11.6 \times {10^{ - 3}}\,kgmi{n^{ - 1}}.
Now, the rate of the reaction will be;
rate   =12[ΔSO2]Δt =[ΔO2]Δt =12[ΔSO3]Δtrate\;\, = \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}}
We have to convert the rate of formation into moles/minutemoles/{\text{minute}} So,
Moles=given weight molar weight=10080Moles = \dfrac{{given\,weight\,}}{{molar\,weight}} = \dfrac{{100}}{{80}}
Now we will calculate the rate of disappearance of ​ SO2S{O_2} . It is given as;
The rate of formation of SO3S{O_3} == the rate of disappearance of SO2S{O_2}​
+12[ΔSO3]Δt =−12[ΔSO2]Δt+ \dfrac{1}{2}\dfrac{{[\Delta S{O_3}]}}{{\Delta t}}\, = - \dfrac{1}{2}\dfrac{{[\Delta S{O_2}]}}{{\Delta t}}
−[ΔSO2]Δt =−22(1.6)(80)- \dfrac{{[\Delta S{O_2}]}}{{\Delta t}}\, = - \dfrac{2}{2}\dfrac{{(1.6)}}{{(80)}}
−[ΔSO2]Δt =1.680=0.2 moles/minute- \dfrac{{[\Delta S{O_2}]}}{{\Delta t}}\, = \dfrac{{1.6}}{{80}} = 0.2\,moles/{\text{minute}}
But the options are given in kg // min, so we have to convert for SO2S{O_2} . Now, as we know,
Moles=given weight molar weightMoles = \dfrac{{given\,weight\,}}{{molar\,weight}}
Mass=moles×molar massMass = moles \times molar\,mass
Mass=0.2×64×10−3=1.28×10−3 kg/min⁡Mass = 0.2 \times 64 \times {10^{ - 3}} = 1.28 \times {10^{ - 3}}\,kg/\min ∴\therefore The molar mass of SO2S{O_2} is 64×10−364 \times {10^{ - 3}}
Hence, the rate at which SO2S{O_2} react is 1.28×10−3kg min−11.28 \times {10^{ - 3}}kg\,{\text{mi}}{{\text{n}}^{ - 1}}. Therefore, option (D) is correct.

ADDITIONAL INFORMATION:
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.

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.