1.

For the reaction , A to B , k_1=10^8 e^(-6000/"8.34 T") and P to Q , k_2=10^10 e^(-8000/"8.34T") The temperature at which k_1=k_2 is

Answer»

386 K
221 K
26 K
52 K

Solution :`k_1=k_2`
`10^8 e^(-6000/"8.34 T")=10^10 e^(-8000/"8.34 T")`
`10^8/10^10 = e^((-2000)/(8.34 T))`
ln `10^(-2) = (-2000)/(8.34 T) RARR -ln 100=(-2000)/(8.34 T)`
`rArr` 2.303 log 100= `2000/(8.34T) rArr T= 2000/(2.303xx2xx8.34)`= 52K


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