Home
Class 12
CHEMISTRY
The activation energy of a reaction is 9...

The activation energy of a reaction is `9.0 kcal//mol`.
The increase in the rate consatnt when its temperature is increased from `298 K` to `308 K` is

A

`10%`

B

`100%`

C

`50%`

D

`63%`

Text Solution

Verified by Experts

The correct Answer is:
d
`2.303 "log" K_(2)/K_(1)=E_(a)/R[(T_(2)-T_(1))/(T_(1)T_(2))]`
`:. 2.303 "log" K_(2)/K_(1)=9/(2xx10^(-3))[(10)/(298xx308)]`
`:. K_(2)/K_(1)=1.63, i.e., 63%` increse.
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 1|2 Videos

  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 3A|2 Videos

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 9 Advanced Numerical|12 Videos

Similar Questions

Explore conceptually related problems

The activation energy of a reaction is 9 kcal "mole "^(-1) . The increase in the rate cnstant when its temperature is raised from 295 to 300 approximately

Activation energy of reactions for which rate constant is doubled when the temperature is increased by 10^(@) (from 300 K to 310 K) is nearly………………… .

What is the activation energy (KJ/mol) for a reaction if its rate constant doubles when the temperature is raised from 300 K to 400 K ? (R="8.314 J mol"^(-1)"K"^(-1))

For two reactions (i) P rarr Product and (ii) Q rarr Product, the order of reaction (i) is 1 while that of reaction (ii) is 2 . At 347 K , the energy of activation of reaction (i) is 55 kJ mol^(-1) but whenever this reaction is carried out in the presence of catalyst at the same temperature, the energy of activation is 53 kJ mol^(-1) . Moreover, for reaction (ii), when the temperature is increased form 298 K to 308 K , the rate of reaction increases as many times as for reaction (i) in the presnece of catalyst. Calculate the rate constant of reaction (ii) at 318 K , if the pre-exponential factor for reaction (ii) is 3.56 xx 10^(9) mol^(-1) L s^(-1) . Note: Arrhenius equation does not depend on the order of reaction.