Question

Calculate the activation parameters; Ea, H‡
, and S‡
at 100
o C from the data given for the
reaction below.
Temperature oC
k x 104
(s-1)
60.0
0.30
70.0
0.97
75.0
1.79
80.0
3.09
90.0
8.92
95.0
15.90

Answer 1

Using the Arrhenius equation, the activation energy $E_a$ can be calculated from the plot of $\text{ln}(k)$ in function of $1/T$ :

$\text{ln}(k) = \text{ln}(A) - \frac{E_a}{RT}$ ,

where $A$ is the frequency factor constant and $R$ is the gas constant, 8.314 J/(mol K).

The activation energy is calculated from the slope of the trend line:

$E_a = 13862\cdot8.314 = 115.25$ kJ/mol.

Now we can get the standard enthalpy of activation $\Delta H^‡$ at 100°C, using the following relation:

$\Delta H^‡ = E_a-RT$

$\Delta H^‡ = 115249-8.314\cdot(273.15+100) = 112.15$ kJ/mol.

However, in order to get the standard entropy of activation, we must use Eyring equation:

$\text{ln}(\frac{k}{T}) = \text{ln}(\frac{k_B}{h}) +\frac{\Delta S^‡}{R} - \frac{\Delta H^‡}{RT}$ .

Using its linearization, the slope will be $-\Delta H^‡/R$ , while the intercept will be a fuction of the entropy: $\Delta S^‡/R + \text{ln}(\frac{k_B}{h})$ , where $k_B$ is the Boltzmann constant 1.381 10^-23 J/K, and $h$ is the Plank constant, 6.626 10^-34 Js.

From these relations, one can get both the enthalpy $\Delta H^‡ = 13511\cdot8.314 = 112.33$ kJ/mol and the entropy of activation $\Delta S^‡ = (24.317-23.76)\cdot8.314 = 4.60$ (J/(mol K)). Note that the enthalpies obtained are very close: 112.33 and 112.15 kJ/mol.

Answer: $E_a = 115.25$ kJ/mol, $\Delta H^‡ =112.33$ (112.15) kJ/mol and $\Delta S^‡ = 4.60$ J/(mol K).