3O2(g) ====> 2O3(g)

At a given instant, the reaction rate in terms of [O2] is 2.17x10^-5 mol/Lxs. What is it in terms of [O3]?

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3O2 ---> 2O3

We are given the above reaction as a balanced chemical equation. We are given the rate of the reaction in terms of O2 as 2.17 x 10^-5 mol/Ls. This means that the rate is 2.17 x 10^-5 mol O2/Ls, or 2.17 x 10^-5 moles per liter of O2 are consumed per second. In order to get the reaction rate in terms of O3 (the product), we must use the coefficients from the balanced equation. For every 3 moles of O2 that react, 2 moles of O3 are produced. Multiplying it out:

2.17 x 10^-5 mol O2/Ls * ([2 mol O3/Ls]/[3 mol O2/Ls]) = 1.45 x 10^-5 mol O3/Ls

**So the answer is 1.45 x 10^-5 mol/Ls in terms of O3.**

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