1.

The rate constant of a first order reaction is given by K=2.1xx10^(10)"exp"(-2700//T). It meas that

Answer»

Greater the activation ENERGY, greater will be the TEMPERATURE coefficient `((K_(T+10))/(K_(T)))`
Freququency factor of the reaction is `2.1xx10^(10)`
Half life increases with increase of temperature
Activation energy of reaction is 5.4 is K.Cal

Solution :a. `K=2.1xx10^(10)XXE^(((-2700)/T)),K=Axxe^(((-EA)/(RT))),logk=logA-((Ea)/(2.303Rt))`
b. `A=2.1xx10^(10)`
c. If `Tuarr(t_(1//2))darr` so `(t_(1/2)alpha1/T)`
d. `(-2700)/T=(-epsilona)/(RT),epsilona=2700xxR=(2700xx2)/1000=5.4` Kcal


Discussion

No Comment Found

Related InterviewSolutions