Home
Class 12
CHEMISTRY
What is the activation energy for a reac...

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200 K to 400 K ? `(R = 8.314 JK^(-1)mol^(-1))`

A

`234.65 kJ mol^(-1) K^(-1)`

B

`434.65 kJ mol^(-1) K^(-1)`

C

`434.65 J mol^(-1) K^(-1)`

D

`334.65 J mol^(-1) K^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`T_(1) = 200K, k = k_(1) , T_(2) = 400K, k= k_(2) = 2k_(1)`
log `((k_(2))/(k_(1))) = (2.303 E_(a))/(R ) ((T_(2) - T_(1))/(T_(1) T_(2)))`
`log ((2k_(2))/(k_(1))) = (2.303 E_(a))/(8.314J K^(-1) mol^(-1)) ((400K - 200K)/(200 K xx 400 K))`
`E_(a) = (0.3010 xx 8.314J mol^(-1) xx 200 xx 400)/(2.303 xx 200)`
`E_(a) = 434.65 J mol^(-1)`
Promotional Banner

Topper's Solved these Questions

  • SAMPLE PAPER 10 (SOLVED)

    FULL MARKS|Exercise Part-II|9 Videos

  • SAMPLE PAPER 10 (SOLVED)

    FULL MARKS|Exercise Part-III|9 Videos

  • SAMPLE PAPER - 9 (SOLVED)

    FULL MARKS|Exercise PART-IV|10 Videos

  • SAMPLE PAPER 2 (SOLVED)

    FULL MARKS|Exercise PART IV|5 Videos

Similar Questions

Explore conceptually related problems

What would be the activation energy of a reaction when the temperature is increased from 27^(@)C to 37^(@)C ?

DeltaH and DeltaS for the reaction Ag_(2)O_((s))rarr2Ag_((s))+1/2O_(2_((g))) " are "30.56 kJ mol^(-1) and 66.0 Jk^(-1) mol^(-1) respectively. Calculate the temperature at which the free energy for this reaction will be zero. What will be the direction of reaction at this temperature and at temperature below this and why ? Given: DeltaH=30.56kJ mol^(-1)=30560 J mol^(-1) DeltaS=66.0JK^(-1)mol^(-1) DeltaG=0

The amount of heat energy required to raise the temperature of 1g of helium at NTP, from T_1K to T_2K is

Calculate the heat energy required to raise the temperature of 2 kg of water from 10^@C to 50^@C . Specific heat capacity of water is 4200 JKg^(-1) K^(-1)