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351
3.
lf k1 and k2 are the rate constants at temp T1 and T2 respectively, then
Ea
log10k1 = log10A – 2.303 RT
1
Ea
log10k2 = log10A – 2.303 RT
2
⎡
⎤
⎡
⎤
Ea
Ea
– ⎢ log 10 A –
⎥
2.303RT2 ⎦
2.303RT1 ⎥⎦
⎣
⎣
Ea
Ea
–
2.303 RT1
2.303 RT2
∴ log10k2 – log10K1 = ⎢ log 10 A –
k2
∴ log10 k =
1
=
Ea
⎡1
1⎤
–
⎢
2.303 R ⎣ T1 T2 ⎥⎦
k2
⎡ T2 – T1 ⎤
Ea
∴ log10 k =
⎢
⎥
2.303 R ⎣ T1 T2 ⎦
1
(e) Arrhenius equation and energy of activation :
1.
2.
3.
4.
Arrhenius equation is k = A.e – E a /RT
Where k = Rate constant at temp T.
A = frequency factor
Ea = energy of activation
As the activation energy Ea decreases, the
energy barrier decreases and more reactant
molecules can cross the activation energy
barrier.
The graph shows when Ea decreases, the
fraction of molecules with kinetic energy greater
than or equal to Ea increases.
As Ea decreases,
Ea
⇒ E a /RT
⇒ – E a /RT ⇒ e –E a /RT ⇒ k
⇒ rate
increases
increases
decreases
decreases
increases
increases
(f) Determination of activation energy (Ea) :
1.
Consider a first order reaction B ⎯⎯
→C
2.
3.
The rate constants for the above reaction is determined at various temperatures.
By Arrhenius equation
k = A × e – E a /RT
Chapter - 5 Chemical Kinetics
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