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The activation energy of the reaction: A...

The activation energy of the reaction: `A + B rarr` Products is `105.73 kJ mol^(-1)`. At `40^(@)C`, the Products are found at the rate of `0.133 mol L^(-1) min^(-1)`. What will be the rate of formation of Products at `80^(@)C`?

Text Solution

Verified by Experts

Let the rate law be defined as
At `" "T_(1) : r_(1) =k_(1)[A]^(x)[B]^(y)`
`At" "T_(2):r_(2) =k_(2)[A]^(x)[B]^(y)`
`rArr" "r_(2)=r_(1) ((k_(2))/(k_(1)))`
Using Arrhenius equation find k at `40^(@)C`.
`log_(10)""(k_(2))/(k_(1))=(E_(a))/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))`
`rArr log_(10)""(k_(2))/(k_(1))=(102.9xx10^(3))/(2.303xx8.31)((40)/(313xx353))`
`rArr log_(10)""(k_(2))/(k_(1))=1.95`
`rArr (k_(2))/(k_(1))=88.41`
`rArr r_(2)=0.133xx88.41=11.76` mol/L/min
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