Home
Class 12
CHEMISTRY
For a first order reaction A rarr P, the...

For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, are

A

`1.0 xx 10^(6) s^(-2) and 9.2 " kj mol"^(-1)`

B

`6.0 s^(-1) and 16.6 " kJ mol"^(-1)`

C

`1.0 xx 10^(6) s^(-1) and 16.6 " kJ mol"^(-1)`

D

`1.0 xx 10^(6) s^(-1) and 38.3 " kJ mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Properties of ionic solid
Promotional Banner

Similar Questions

Explore conceptually related problems

For a first order reaction ArarrP , the temperature (T) dependent rate constant (K) was found to follow the equation log k=-(2000)(1)/(T)+6.0 . The pre- exponential factor A and the activation energy E_(a) , respectively, are :

The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every 10^(@)C rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10^(@)C change in temperature. "Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C) Arrhenius gave an equation which describes aret constant k as a function of temperature k = Ae^(-E_(a)//RT) where k is the rate constant, A is the frequency factor or pre-exponential factor, E_(a) is the activation energy, T is the temperature in kelvin, R is the universal gas constant. Equation when expressed in logarithmic form becomes log k = log A - (E_(a))/(2.303 RT) Activation energies of two reaction are E_(a) and E_(a)' with E_(a) gt E'_(a) . If the temperature of the reacting systems is increased form T_(1) to T_(2) ( k' is rate constant at higher temperature).

The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every 10^(@)C rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10^(@)C change in temperature. "Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C) Arrhenius gave an equation which describes aret constant k as a function of temperature k = Ae^(-E_(a)//RT) where k is the rate constant, A is the frequency factor or pre-exponential factor, E_(a) is the activation energy, T is the temperature in kelvin, R is the universal gas constant. Equation when expressed in logarithmic form becomes log k = log A - (E_(a))/(2.303 RT) For a reaction E_(a) = 0 and k = 3.2 xx 10^(8)s^(-1) at 325 K . The value of k at 335 K would be

The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every 10^(@)C rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10^(@)C change in temperature. "Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C) Arrhenius gave an equation which describes aret constant k as a function of temperature k = Ae^(-E_(a)//RT) where k is the rate constant, A is the frequency factor or pre-exponential factor, E_(a) is the activation energy, T is the temperature in kelvin, R is the universal gas constant. Equation when expressed in logarithmic form becomes log k = log A - (E_(a))/(2.303 RT) For which of the following reactions k_(310)//k_(300) would be maximum?

The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every 10^(@)C rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10^(@)C change in temperature. "Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C) Arrhenius gave an equation which describes aret constant k as a function of temperature k = Ae^(-E_(a)//RT) where k is the rate constant, A is the frequency factor or pre-exponential factor, E_(a) is the activation energy, T is the temperature in kelvin, R is the universal gas constant. Equation when expressed in logarithmic form becomes log k = log A - (E_(a))/(2.303 RT) For the given reactions, following data is given {:(PrarrQ,,,,k_(1) =10^(15)exp((-2000)/(T))),(CrarrD,,,,k_(2) = 10^(14)exp((-1000)/(T))):} Temperature at which k_(1) = k_(2) is

The rate constant for the decompoistion of a certain reaction is described by the equation: log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T) Pre-exponential factor for this reaction is

The rate constant for a first order decomposition reaction is given by log K=10 - 1000/T .Then,what will be activation energy in kcal/mol?