1.

Show that time required for 99% completionis twice the time required for the completion of 90% of reactions for a first order reaction. The decomposition of hydrocarbon follow the equation K=(4.5xx10^(11)5^(-1))e^(-28000K//T)" Calculate Ea".

Answer»

Solution :`"For first ORDER reaction, t"=(2.303)/(R )LOG.(a_(0))/(a_(0))`
1)=(2)/(1)`
`rArr""t_(99%)=2xxt_(90%)`
`k=Ae^(-Ea//RT)`
`(Ea)/(RT)=(28000)/(T )`
`E_(a)=28000xxR`
`=28000xx8.314`
`E_(a)=232.79" KJ mol"^(-1)`


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