3O2(g)---> 2O3(g)
At a given instant, the reaction rate in terms of [O2] is 2.17x10^-5 mol/L*s.
What is it in terms of O3?
2.17 x 10^-5 x (2/3) = ??
To determine the reaction rate in terms of O3, we need to use the stoichiometric coefficients from the balanced equation.
The balanced equation:
3O2(g) -> 2O3(g)
From the equation, we can see that 3 moles of O2 produce 2 moles of O3. Therefore, the rate of the reaction in terms of O3 can be calculated by dividing the rate in terms of O2 by the stoichiometric coefficient of O2:
Rate(O3) = Rate(O2) / Stoichiometric coefficient of O2
Rate(O3) = 2.17x10^-5 mol/L*s / 3
Rate(O3) = 7.23x10^-6 mol/L*s
So, the reaction rate in terms of O3 is 7.23x10^-6 mol/L*s.
To determine the reaction rate in terms of O3, we must use the stoichiometry of the balanced chemical equation.
From the balanced equation: 3O2(g) ---> 2O3(g)
The stoichiometric ratio between O2 and O3 is 3:2. This means that for every 3 moles of O2 reacted, 2 moles of O3 are produced.
The rate of the reaction in terms of O3 can be calculated by multiplying the rate in terms of O2 by the stoichiometric ratio of O3 to O2.
Given the rate in terms of O2 is 2.17x10^-5 mol/L*s, we can calculate the rate in terms of O3 as follows:
Rate in terms of O3 = Rate in terms of O2 * (2 moles O3 / 3 moles O2)
Rate in terms of O3 = 2.17x10^-5 mol/L*s * (2/3)
Rate in terms of O3 ≈ 1.45x10^-5 mol/L*s
Therefore, the reaction rate in terms of O3 is approximately 1.45x10^-5 mol/L*s.