How is the rate at which ozone disappears related to the rate at which oxygen appears in this reaction 2O3(g) -----> 3O2 (g) ?

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hey
Explanation: 

lets consider a general reaction:

aA + bB ----> cC + dD

Rate = - 1/a delta A / delta t = - 1/b deltaB / delta t = - 1/c delta C / delta t = - 1/d delta D / delta t

So,the speed of any event  is measured by the change that  occurs in any  interval of time.  The speed of a reaction or reaction rate is expressed as the change in concentration of a reactant or product over a certain amount of time.

Since concentration of the reactant decreases over time, final concentration minus the initial will give a negative value . And that is why we see a negative sign above in the rate equation of the reactants A and B .
look at the rate of appearance of a product. As a product appears, its concentration increases. The rate of appearance is a positive quantity.  
 
 rate of appearance of a product is equal to the negative of rate of disappearance of a reactant
 
In the reaction  :
    2O3(g) -----> 3O2 (g)
the rate of the disaperence of O3 is the -1 / 2  delta O3 / delta t
and the rate of the appearence of the O2 is 1/3 O2/ delta t and so,
 
-1/2 delta O3 / delta t = 1/3 O2/ delta t
 
 
 
Answer: 

 

-1/2 delta O3 / delta t = 1/3 O2/ delta t
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