1.

Show by using rate law, how much rate of the reaction, 2"NO"+"O"_(2)to2"NO"_(2), will change if the volume of the reaction vessel is reduced to one third of its initial value ?

Answer»


Solution :INITIALLY, rate `=K["NO"]^(2)["O"_(2)]=k" a"^(2)b("say").`
Whenvolume is reduced to one-third, the concentration of each reactant will BECOME THREE times.
New rate `=k(3" a")^(2)(3" b")=27" k a"^(2)b.`


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