1.

For reaction AtoB the rate constant k_(1)=A_(1)e^(-Ea_(1)//RT) and for the reaction PtoQ the rate constant k_(2)=A_(2)e^(-Ea_(2)//RT). If A_(1)=10^(8),A_(2)=10^(10) and E_(a_(1))=600,E_(a_(2))=1200, then the temperature at which k_(1)=k_(2) is

Answer»

`600/RK`
`(300xx4.606)/RK`
`600/(4.606R)K`
`4.606/(600R)K`

SOLUTION :`K_(1)=K_(2),A_(1).e^(-Ea_(1)//RT_(1))=A_(2).e^(-Ea_(2)//RT_(2),(A_(1))/(A_(2))=(e^((-epsilon_(2))/(RT))/(e^(-EPSILON a_(1)//RT))):(A_(1))/(A_(2))=e((epsilonn_(a_(1))-epsilon_(a_(2)))/(RT))`
`2.303"log"(10^(8))/(10^(10))=(600-1200)/(RT),2.303xx(-2)=(-600)/(RT),T=600/(2.303xx2xxR)=(600/4.606R))` Kelvin


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