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The rate of formation of O3(g) is 2.0 × 10^–7mol·L–1·s–1forthe following...

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lak-86 | Student, Undergraduate | (Level 1) Salutatorian

Posted August 18, 2013 at 9:49 AM via web

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The rate of formation of O3(g) is 2.0 × 10^–7mol·L–1·s–1for
the following reaction
`3O_2 (g)rarr 2O_3 (g)`

What is the rate of disappearance of O2(g).

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mvcdc | Student, Undergraduate | (Level 1) Associate Educator

Posted August 18, 2013 at 10:10 AM (Answer #1)

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Using the chemical equation, we see that for every 3 moles of oxygen gas, 2 moles of ozone are produced. This means that the rate of disappearance of oxygen is faster than the rate of formation of ozone. 

The rate of formation of ozone is `(d[O_3])/(dt).`

The rate of disappearance of oxygen is `-(d[O_2])/(dt).`

The negative sign on the rate of disappearance of oxygen is only to make the rate positive since `d[O_2]` will always be negative as oxgen is consumed.

Now, we refer back to the chemical equation, 3 moles of oxygen is consumed to form 2 moles of ozone. Hence, twice the rate of disappearance of oxygen is thrice the rate of formation of ozone:

`1/3 (d[O_2])/(dt) = 1/2 (d[O_3])/(dt)`

(In general, the coefficient of the species in the equation will be the denominator of its rate).

Hence, `(d[O_2])/(dt) = 3/2 (d[O_3])/(dt)` .` `

We know that `(d[O_3])/(dt) = 2.0 times 10^(-7) (mol)/(Ls)` .

Therefore, `(d[O_2])/(dt) = (3 * 2.0 times 10^(-7))/2 = 3 times 10^(-7).`

The rate of disappearance of oxygen gas is 3 x 10^-7 moles per liter per second.

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