1.

Two reactions R_(1)" and "R_(2) have identical perexponential factors. Activation energy of R_(1) the rate constants for reactions R_(1) and R_(2) respectively at 300 K, then ln (k_(2)//k_(1)) is equal to (R=8.314" J mole"^(-1)K^(-1))

Answer»

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Solution :`k_1=Ae^(-E_(a_(1))//RT), K_2=Ae^(-E_(a_(2))//RT)`
`:.""k_2/k_1=E^(1/(RT)(E_(a_(1))-E_(a_(2)))`
or`In k_2/k_1=(E_(a_(1))-E_(a_(2)))/(RT)=(10xx10^3)/(8.314xx300)=4`


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