Respuesta :
Answer:
Rate in terms of formation of [tex]O_{3}[/tex] is [tex]1.45\times 10^{-5}mol/L.s[/tex]
Explanation:
- According to law of mass action for this reaction: Rate = [tex]-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [O_{3}]}{\Delta t}[/tex]
- [tex]-\frac{\Delta [O_{2}]}{\Delta t}[/tex] represents rate of disappearance of [tex]O_{2}[/tex] and [tex]\frac{\Delta [O_{3}]}{\Delta t}[/tex] represents rate of formation of [tex]O_{3}[/tex]
- Here, [tex]-\frac{\Delta [O_{2}]}{\Delta t}=2.17\times 10^{-5}mol/L.s[/tex]
- So, [tex]\frac{\Delta [O_{3}]}{\Delta t}=\frac{2}{3}\times -\frac{\Delta [O_{2}]}{\Delta t}=\frac{2}{3}\times (2.17\times 10^{-5}mol/L.s)=1.45\times 10^{-5}mol/L.s[/tex]
- Hence rate in terms of formation of [tex]O_{3}[/tex] is [tex]1.45\times 10^{-5}mol/L.s[/tex]
The rate of formation is the time taken by the reaction to yield the product by the chemical change in the reactants. The rate of the formation of the ozone is [tex]1.45 \times 10^{-5} \;\rm mol/Ls[/tex].
What is the law of mass action?
The law of mass action states that the rate of the reaction is proportional to the product of the reactant masses.
Rate according to the law of mass action:
[tex]\rm -\dfrac{1}{3}\dfrac {\Delta[O_{2}]}{\Delta t} = \rm \dfrac{1}{2}\dfrac{\Delta [O_{3}]}{\Delta t}[/tex]
Here,
- Rate of disappearance of oxygen [tex](\rm -\dfrac {\Delta[O_{2}]}{\Delta t} ) = 2.17 \times 10^{-5} \;\rm mol/Ls[/tex]
- Rate of formation of ozone =[tex]\rm \dfrac{\Delta [O_{3}]}{\Delta t}[/tex]
Substituting values in the above equation:
[tex]\begin{aligned}\rm \dfrac{\Delta [O_{3}]}{\Delta t} &= \dfrac{2}{3}\times - \rm \dfrac{\Delta [O_{2}]}{\Delta t}\\\\&= \dfrac{2}{3} \times (2.17 \times 10^{-5})\\\\&= 1.45 \times 10^{-5}\;\rm mol/Ls\end{aligned}[/tex]
Therefore, the rate of formation in the terms of ozone is [tex]1.45 \times 10^{-5} \;\rm mol/Ls.[/tex]
Learn more about the rate of formation here:
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