Salviya AntonySenior Executive - Content
Activation Energy: The minimum amount of extra energy needed by a reacting molecule to get converted into a product is called the activation energy. Activation energy is a concept in chemistry that refers to the minimum amount of energy that must be provided to a chemical reaction for it to occur. In other words, it is the energy barrier that must be overcome to initiate a reaction. We denote activation energy by Ea.
Every chemical reaction involves the breaking and formation of chemical bonds. The reactant molecules must acquire a certain amount of energy to reach an intermediate state called the transition state, from which the reaction can proceed to form the products. The activation energy represents the energy difference between the reactants and the transition state. We can also be define activation energy as the minimum amount of energy needed to activate or energise molecules or atoms so that they can undergo a chemical reaction or transformation. A reaction can occur when molecules collide with sufficient energy to overcome the activation energy barrier. The higher the activation energy, the less likely a reaction is to occur spontaneously. Catalysts are substances that can lower the activation energy of a reaction, making it easier for the reaction to proceed. In NCERT Class 12 Chemistry, activation energy comes in chapter Chemical Kinetics.
Activation Energy: SI Unit
Activation energy is measured in joules (J) and or kilo Joules per mole (kJ/mol) or kilocalories per mole (kcal/mol).
Activation Energy: Formula
Activation Energy, Ea is given by the formula
K = Ae-Ea/RT
K is the Rate Constant
A = Arrhenius Constant
Ea = Activation Energy
R = 8.34J/K/mol (Gas constant)
=8.314/1000 KJ/K/mol
= 2 cal/K/mol
= 0.0821 lit atm/K/mol
K = Ae-Ea/RT
Taking log on both sides
ln K = ln A – (Ea /RT)ln e
2.303 log K = 2.303 log A – Ea/RT
log K = log A – Ea /2.303RT
Factors Influencing Activation Energy
Activation energy depends on the effect of catalyst and nature of reactants.
Effect of Catalyst
Positive catalyst will give an alternate path in which the value of activation energy will be less. The negative catalyst gives an alternate path where activation energy will be more.
Nature of Reactants
For covalent reactant, the value of Ea will be high because energy is needed to break the older bonds. For ionic reactant, the value of activation energy will be less because there is an attraction between reacting species.
The concept of activation energy is often depicted graphically on an energy profile or reaction coordinate diagram, where the horizontal axis represents the progression of the reaction from reactants to products, and the vertical axis represents the energy changes. The activation energy is the energy difference between the reactants and the highest point on the energy profile, which corresponds to the transition state.
FAQs on Activation Energy
Q: The decomposition of hydrocarbon follows the equation k = (4.5 × 1011s–1) e-28000K/T. Calculate Ea
A:
The given equation is
k = (4.5 x 1011 s-1) e-28000K/T (i)
Comparing, Arrhenius equation k = Ae -E/RT (ii)
We get, Ea / RT = 28000K / T
⇒Ea = R x 28000K
= 8.314 J K-1mol-1 × 28000 K
= 232792 J mol–1 or 232.792 kJ mol–1
Q: The decomposition of A into product has value of k as 4.5× 103 s–1 at 10°C and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 × 104s–1?
A: From Arrhenius equation, we obtain
Also, k1 = 4.5 × 103 s -1
T1 = 273 + 10 = 283 K
k2 = 1.5 × 104 s -1
Ea = 60 kJ mol -1 = 6.0 × 104 J mol -1
Then,
→ 0.5229 = 3133.627 × (T2-283)/(283 × T2)
→ 0.0472T2 = T2-283 T2 = 297K or T2 = 240 C
Q: The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 × 1010s –1, calculate k at 318K and Ea.
A:
We know, time t = (2.303/k) × log([R]0/[R])
Where, k- rate constant [R]0-Initial concentration
[R]-Concentration at time ‘t’
At 298K, If 10% is completed, then 90% is remaining. t = (2.303/k) × log ([R]0/0.9[R]0)
t = (2.303/k) × log (1/0.9) t = 0.1054 / k
At temperature 308K, 25% is completed, 75% is remaining t’ = (2.303/k’) × log ([R]0/0.75[R]0)
t’ = (2.303/k’) × log (1/0.75) t’ = 2.2877 / k'
But, t = t’
0.1054 / k = 2.2877 / k' k' / k = 2.7296
From Arrhenius equation, we obtain log k2/k1 = (Ea / 2.303 R) × (T2 - T1) / T1T2
Substituting the values,
Ea = 76640.09 J mol-1 or 76.64 kJ mol-1 We know, log k = log A –Ea/RT
Log k = log(4 × 1010)-(76.64kJ mol-1/(8.314 × 318) Log k = -1.986
∴k = 1.034 x 10-2 s -1
Q: The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
A: Given, k2 = 4k1, T1 = 293K and T2 = 313K
We know, From Arrhenius equation, we obtain
On solving we get,
Ea = 58263.33 J mol-1 or 58.26 kJ mol-1
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