# Activation Energy Formula

Activation energy of a chemical reaction is defined as the least amount of energy necessary to initiate the reaction. It can be interpreted as the differential in energy content between molecules and atoms that causes it to be in an activation or transition-state configuration while the associated atoms and molecules stay in their initial configuration. It is known that to initiate a reaction, molecules must collide with other molecules and exchange kinetic energy or velocity. There will be no response if the collision does not occur or if the molecules do not have enough kinetic energy. This forms the basis for the concept of activation energy. Its standard unit of measurement is kilojoules per mole (kJ/mol).

**Formula**

E_{a}= 2.303 R (log k_{2}/k_{1}) [T_{1}T_{2}/ (T_{2}– T_{1})]where,

E

_{a}is the activation energy of the reaction,R is the ideal gas constant with the value of 8.3145 J/K mol,

k

_{1},k_{2}are the rates of reaction constant at initial and final temperature,T

_{1}is the initial temperature,T

_{2}is the final temperature.This is also known as the Arrhenius equation.

**Derivation**

Suppose a reaction takes place between two reactants P and Q to give a product R. The completion of a reaction is ensured and denoted by its rate constant.

We know the formula for rate constant is given by,

k = Ae

^{âˆ’E}_{a}^{RT}Taking log on both sides we get,

ln k = âˆ’E

_{a}RT + ln AFor initial temperature T

_{1}and rate of constant k_{1}, the equation is written as,ln k

_{1}= âˆ’E_{a}/RT_{1}+ ln A …….. (1)For final temperature T

_{2}and rate of constant k_{2}, the equation is written as,ln k

_{2}= âˆ’E_{a}/RT_{2}+ ln A …….. (2)Subtracting (2) from (1) we get,

ln k

_{2}– ln k_{1}= âˆ’E_{a}/RT_{2}+ ln A âˆ’ (âˆ’ E_{a}/RT_{1}+ ln A)ln (k

_{2}/k_{1}) = E_{a}R (1/T_{1}âˆ’ 1/T_{2})2.303 log (k

_{2}/k_{1}) = E_{a}R (1/T_{1}âˆ’ 1/T_{2})E

_{a}= 2.303 R (log k_{2}/k_{1}) [T_{1}T_{2}/ (T_{2}– T_{1})]This derives the formula for activation energy.

**Sample Problems**

**Problem 1. Calculate the activation energy of a reaction if its rate doubles when there is a temperature change from 200K to 400K. **

**Solution:**

We have,

k

_{2}= 2k_{1},T

_{1}= 200T

_{2}= 400Using the formula we get,

E

_{a}= 2.303 R (log k_{2}/k_{1}) [T_{1}T_{2}/ (T_{2}– T_{1})]= 2.303 (8.3145) (log 2) (80000/200)

= 461090.907/200

= 2305.45 KJ/mol

**Problem 2. Calculate the activation energy of a reaction if its rate triples when there is a temperature change from 100K to 300K.**

**Solution:**

We have,

k

_{2}= 3k_{1},T

_{1}= 100T

_{2}= 300Using the formula we get,

E

_{a}= 2.303 R (log k_{2}/k_{1}) [T_{1}T_{2}/ (T_{2}– T_{1})]= 2.303 (8.3145) (log 3) (30000/200)

= 274012.079/200

= 1370.060 KJ/mol

**Problem 3. Calculate the activation energy of a reaction if its rate quadruples when there is a temperature change from 150K to 400K.**

**Solution:**

We have,

k

_{2}= 4k_{1},T

_{1}= 150T

_{2}= 400Using the formula we get,

E

_{a}= 2.303 (8.3145) (log 4) (60000/250)= 691636.36/250

= 2766.55 KJ/mol

**Problem 4. Calculate the activation energy of a reaction if its rate becomes five times when there is a temperature change from 300K to 600K.**

**Solution:**

We have,

k

_{2}= 5k_{1},T

_{1}= 300T

_{2}= 600Using the formula we get,

E

_{a}= 2.303 (8.3145) (log 5) (180000/300)= 2405791.59/300

= 8019.30 KJ/mol

**Problem 5. Calculate the change in the rate of reaction if the activation energy of a reaction is 500 KJ/mol when there is a temperature change from 120K to 360K.**

**Solution:**

We have,

E

_{a}= 500T

_{1}= 120T

_{2}= 360Using the formula we get,

=> 3000 = 2.303 (log k

_{2}/k_{1}) [43200/240]=> log (k

_{2}/k_{1}) = (500/2.303) (240/43200)=> log (k

_{2}/k_{1}) = 7.23=> k

_{2}/k_{1}= 16=> k

_{2}= 16 k_{1}

**Problem 6. Calculate the change in the rate of reaction if the activation energy of a reaction is 200 KJ/mol when there is a temperature change from 50K to 100K.**

**Solution:**

We have,

E

_{a}= 200T

_{1}= 50T

_{2}= 100Using the formula we get,

=> 200 = 2.303 (log k

_{2}/k_{1}) [5000/50]=> log (k

_{2}/k_{1}) = (200/2.303) (50/5000)=> log (k

_{2}/k_{1}) = 0.86=> k

_{2}/k_{1}= 7.37=> k

_{2}= 7.37 k_{1}

**Problem 7. Calculate the change in the rate of reaction if the activation energy of a reaction is 450 KJ/mol when there is a temperature change from 100K to 200K.**

**Solution:**

We have,

E

_{a}= 450T

_{1}= 100T

_{2}= 200Using the formula we get,

=> 450 = 2.303 (log k

_{2}/k_{1}) [20000/100]=> log (k

_{2}/k_{1}) = (450/2.303) (100/20000)=> log (k

_{2}/k_{1}) = 0.97=> k

_{2}/k_{1}= 9.33=> k

_{2}= 9.33 k_{1}

## Please

Loginto comment...