Respuesta :
Answer:
Rate of formation of SO₃ [tex][\frac{d[SO_{3}] }{dt}][/tex] = 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒ [tex]-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}[/tex] -----------------------------(1)
Given that the rate of disappearance of oxygen = [tex]-\frac{d[O_{2} ]}{dt}[/tex] = 3.64 x 10⁻³ M/s
So the rate of formation of SO₃ [tex][\frac{d[SO_{3}] }{dt}][/tex] = ?
from equation (1) we can write
[tex]\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ][/tex]
⇒ [tex]\frac{d[SO_{3}] }{dt}[/tex] = 2 x 3.64 x 10⁻³ M/s
⇒ [tex][\frac{d[SO_{3}] }{dt}][/tex] = 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃ [tex][\frac{d[SO_{3}] }{dt}][/tex] = 7.28 x 10⁻³ M/s
The rate of formation of [tex]\rm SO_3[/tex] has been [tex]\rm 7.28\;\times\;10^-^3\;M/s[/tex]. Thus, option D is correct.
The rate of a chemical equation has been given as the disappearance of reactant or the formation of product in unit time.
The given balanced chemical reaction has been:
[tex]\rm 2\;SO_2\;+\;O_2\;\rightarrow\;2\;SO_3[/tex]
Computation for rate of formation of [tex]\rm SO_3[/tex]
The rate of reaction has been given as:
[tex]\rm Rate=\dfrac{1}{2} \dfrac{dSO_2}{dt} \;=\;\dfrac{dO_2}{dt}\;=\;\dfrac{1}{2} \dfrac{dSO_3}{dt}[/tex]
The given rate of disappearance of oxygen has been [tex]\rm 3.64\;\times\;10^-^3\;M/s[/tex]. The rate of formation of [tex]\rm SO_3[/tex] has been given as:
[tex]\rm O_2\;disappearance=\dfrac{1}{2}\;SO_3\;formation\\\\ 3.64\;\times\;10^-^3\;M/s=\dfrac{1}{2}\;SO_3\;formation\\\\SO_3\;formation=7.28\;\times\;10^-^3\;M/s[/tex]
The rate of formation of [tex]\rm SO_3[/tex] has been [tex]\rm 7.28\;\times\;10^-^3\;M/s[/tex]. Thus, option D is correct.
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