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The rate constant of a first order react...

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

A

Greater the activation energy, greater will be the temperature coefficient `((K_(T+10))/(K_(T)))`

B

Freququency factor of the reaction is `2.1xx10^(10)`

C

Half life increases with increase of temperature

D

Activation energy of reaction is 5.4 is K.Cal

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
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
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