1.

A gas phase reaction has energy of activation 200 "kJ mol"^(-1).If the frequency factor of the reaction is 1.6xx10^(13) s^(-1). Calculate the rate constant at 600 K . (e^(-40.09)=3.8xx10^(-18)) .

Answer»

SOLUTION :`k=Ae^((-(E_a)/(RT)))`
`k=1.6xx10^(13)s^(-1)_e^(-((200xx103"J mol"^(-1))/(8.314"JK "^(-1)mol^(-1)xx600K)))`
`k = 1.6xx10^(13)s^(-1)e^(-(40.1))`
`k=1.6xx10^(13)s^(-1)xx3.8xx10^(-18)`
`k=6.21xx10^(-5)s^(-1)`


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