The rate of disappearance of SO_2 in the reac | vedclass

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(D) The balanced chemical equation is $2SO_2 + O_2 	o 2SO_3$.According to the stoichiometry of the reaction,$2 	ext{ moles}$ of $SO_2$ produce $2

Vedclass · Accepted Answer

$1.60 	imes 10^{-3} \ g/sec$

Vedclass · Answer

(D) The balanced chemical equation is $2SO_2 + O_2 	o 2SO_3$.According to the stoichiometry of the reaction,$2 	ext{ moles}$ of $SO_2$ produce $2 	ext{ moles}$ of $SO_3$.Since the molar mass of $SO_2$ $(64 \ g/mol)$ is equal to the molar mass of $SO_3$ $(80 \ g/mol)$ is not the case,we must relate the mass rates.Rate of reaction $= -\frac{1}{2} \frac{d[SO_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$.In terms of mass rate: $-\frac{1}{2 	imes M_{SO_2}} \frac{d(mass_{SO_2})}{dt} = \frac{1}{2 	imes M_{SO_3}} \frac{d(mass_{SO_3})}{dt}$.Given $-\frac{d(mass_{SO_2})}{dt} = 1.28 	imes 10^{-3} \ g/sec$.$\frac{d(mass_{SO_3})}{dt} = \frac{M_{SO_3}}{M_{SO_2}} 	imes (	ext{Rate of disappearance of } SO_2) = \frac{80}{64} 	imes 1.28 	imes 10^{-3} \ g/sec$.$\frac{d(mass_{SO_3})}{dt} = 1.25 	imes 1.28 	imes 10^{-3} \ g/sec = 1.60 	imes 10^{-3} \ g/sec$.

The rate of disappearance of $SO_2$ in the reaction $2SO_2 + O_2 \to 2SO_3$ is $1.28 \times 10^{-3} \ g/sec$. What is the rate of formation of $SO_3$?

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