1. ACTIVATION ENERGY
Energy
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2. Energy EA
Reactants DH
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3. Energy
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4. Energy
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5. Energy EA
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6. What happens here?
Energy EA
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7. Effect of catalyst Without catalyst EA DH Reactants Products
Energy EA
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8. Effect of catalyst With catalyst DH Reactants Products
Energy
With catalyst
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9. Effect of catalyst EA With catalyst DH Reactants Products
Energy EA
With catalyst
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10. Effect of catalyst Without catalyst EA DH Reactants Products
Energy EA
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11. Effect of catalyst EA With catalyst DH Reactants Products
Energy EA
With catalyst
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12. Effect of catalyst Without catalyst EA EA With catalyst DH Reactants
Energy EA EA
With catalyst
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13. k = kNaOH = kHCl = kNaCl = kH2O
ACTIVATION ENERGY
Arrhenius equation
The algebraic equation that relates – rA to the species concentrations is called the kinetic expression or as you know rate law.
1NaOH + 1HCl 1NaCl + 1H2O
k = kNaOH = kHCl = kNaCl = kH2O
It was the great Swedish chemist Arrhenius who first suggested that the temperature dependence of the specific reaction rate, k, could be correlated by an equation of the type:
Where: A = pre-exponential factor, Ea = activation energy, J/mol or cal/mol,
R = gas constant = J/mol.K = cal/mol.K, T = absolute temperature (K).

14. ACTIVATION ENERGY y = mx +b ln(x) = 2.303log(x) ln k = − Ea R 1 T
Arrhenius equation
After taking the natural logarithm of Equation:
ln(x) = 2.303log(x)
ln k = − Ea R 1 T
+ ln A
y = mx +b

15. Arrhenius Equation y = m x + b ln k = − Ea R 1 T + ln A
Fig Graphical determination of activation energy
Plot of ln k vs 1/T
Slope = −Ea/R

16. The Arrehenius equation can be used to relate
rate constants k1 and k2 at temperatures T1 and T2.
ln k1 = − Ea R 1 T1
+ ln A
combine to give:
ln k2 = − Ea R 1 T2
+ ln A

17. 1NaOH + 1HCl 1NaCl + 1H2O R = 1.9858775×10−3 kcal/mol.K T (ºC) k 20
0.021 30
0.052 40
0.171 50
0.306 60
0.319
T (K)
1/T
ln k
293
303
0.0033
313
323
333
R = ×10−3 kcal/mol.K

18. Example:
The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures:
From these data, calculate the activation energy for the reaction.
(b) What is the value of the rate constant at K?

19. Solution
We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k.

20. A graph of ln k versus 1/T results in a straight line, as shown in Figure 14.17:
slope = − Ea/R = − 1.9 x 104

21. We use the value for the molar gas constant R in units of J/mol-K
We use the value for the molar gas constant R in units of J/mol-K. We thus obtain

22. (b) To determine the rate constant, k1, at T1 = 430
(b) To determine the rate constant, k1, at T1 = K, we can use Equation with Ea = 160 kJ/mol, and one of the rate constants and temperatures from the given data, such as:
k2 = × 10–5 s–1 and T2 = K:
Thus,
Note that the units of k1 are the same as those of k2.