Home
Class 12
CHEMISTRY
The rate of the chemical reaction double...

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate `E_(a)`.

Text Solution

Verified by Experts

Here we are given that
When `T_(1)=298K, k_(1)=k` (say)
When `T_(2)=308K, k_(2)=2k`
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
Substituting these values in the equation, we get
`"log"(2k)/(k)=(E_(a))/(2.303xx8.314J K^(-1)"mol"^(-1))xx(308K-298K)/(298K xx 308K)`
or `log 2=(E_(a))/(2.303xx8.314) xx (10)/(298xx308)`
or `0.3010=(E_(a)xx10)/(2.303xx8.314xx298xx308)`
or `E_(a)=(0.3010xx2.303xx8.314xx298xx308)/(10)`
or `E_(a)=52897.8J mol^(-1)=53.6"kJ mol"^(-1)`.
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    U-LIKE SERIES|Exercise NCERT TEXTBOOK EXERCISES |29 Videos
  • CHEMICAL KINETICS

    U-LIKE SERIES|Exercise CASE BASED/SOURCE-BASED INTEGRATED QUESTIONS|15 Videos
  • CBSE EXAMINATION PAPER 2020 (CODE No.56/4/1)

    U-LIKE SERIES|Exercise SECTION -D|3 Videos
  • CHEMISTRY IN EVERYDAY LIFE

    U-LIKE SERIES|Exercise LONG ANSWER QUESTIONS - I|21 Videos

Similar Questions

Explore conceptually related problems

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea.

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 300K . Calculate E_(a) .

The rate constant of the chemical reaction doubled for an increase of 10 k in absolute temperature from 295 k. Calculate the (activation energy), E_(a).

The rate of the chemical reaction doubles for an increase of 10 K from 298 K. Calculate E_(a) .

The rate of a chemical reaction doubles for every 10^(@)C rise in temperature .If the temperature is increased by 60^(@)C the rate of reaction increases by about

The rate of a chemical reaction doubles for every 10^(@)C rise of temperature. If the temperature is raised by 50^(@)C , the rate of the reaction increases by about

The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A