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?
To determine the reaction rate in terms of O3, we need to use the balanced equation and stoichiometry of the reaction.
The balanced equation for the given reaction is:
3O2(g) ---> 2O3(g)
From the equation, we can see that for every 3 moles of O2 consumed, 2 moles of O3 are produced.
The reaction rate in terms of O3 can be calculated using the formula:
Rate(O3) = (1/2) * Rate(O2)
Since the reaction rate in terms of [O2] is given as 2.17 × 10^-5 mol/L*s, we can substitute this value into the above formula to find the reaction rate in terms of O3:
Rate(O3) = (1/2) * (2.17 × 10^-5 mol/L*s)
Rate(O3) = 1.09 × 10^-5 mol/L*s
Therefore, the reaction rate in terms of O3 is approximately 1.09 × 10^-5 mol/L*s.
To determine the reaction rate in terms of O3, we need to use the stoichiometry of the reaction. The balanced equation tells us that for every 3 moles of O2 that react, we produce 2 moles of O3.
Given that the rate of the reaction in terms of [O2] is 2.17x10^-5 mol/L*s, we can use this information to find the rate of the reaction in terms of O3.
Since there is a 3:2 mole ratio between O2 and O3, we can set up a proportion:
(2.17x10^-5 mol/L*s) / (3 mol O2) = x / (2 mol O3)
Simplifying the proportion:
x = (2.17x10^-5 mol/L*s) * (2 mol O3) / (3 mol O2)
x = 1.44x10^-5 mol/L*s
Therefore, the reaction rate in terms of O3 is 1.44x10^-5 mol/L*s.