A reaction proceeds five times more at `60^(@)`C as it does at `30^(@)`C. Estimate energy of activation.
Text Solution
AI Generated Solution
To estimate the energy of activation for the reaction that proceeds five times more at 60°C than at 30°C, we can follow these steps:
### Step 1: Convert temperatures to Kelvin
We need to convert the given temperatures from Celsius to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
- For \( T_1 = 30°C \):
\[ T_1 = 30 + 273 = 303 \, K \]
...
Topper's Solved these Questions
Chemical Kinetics
MOTION|Exercise SOLVED EXAMPLE (OBJECTIVE)|10 Videos
Chemical Kinetics
MOTION|Exercise SOLVED EXAMPLE (SUBJECTIVE)|24 Videos
CHEMICAL EQUILIBRIUM
MOTION|Exercise Exercise - 4|20 Videos
CLASSROOM PROBLEMS
MOTION|Exercise Electrochemistry|22 Videos
Similar Questions
Explore conceptually related problems
The rate of a particular reaction doubles when temperature changes from 27^(@) C to 37^(@)C . Calculate the energy of activation of such a reaction.
The rate constant of a reaction increases by five times on increase in temperature from 27^@C to 52^@C . The value of activation energy in kJ "mol"^(-1) is ______ (Rounded-off to the nearest integer)
A first order reaction is 50% completed in 30 min at 27^(@)C and in 10 min at 47^(@)C . Calculate the reaction rate constants at 27^(@)C and the energy of activation of the reaction in kJ mol^(-1) .
A 1st order reaction is 50% complete in 30 minutes at 27^(@)C and in 10 min at 47^(@)C . Calculate (i) rate constant for the reaction at 27^(@)C and 47^(@)C (ii) energy of activation for the reaction.