For the overall rate constant:
\[ K = \frac{K_1 \cdot K_2}{K_3} = \frac{A_1 \cdot A_2}{A_3} \cdot e^{\left(\frac{E_{a1} + E_{a2} - E_{a3}}{RT}\right)} \]
Therefore,
\[ K = \frac{A \cdot e^{-E_a/RT}}{A_3} = \frac{A_1 A_2}{A_3} \cdot e^{\left(\frac{E_{a1} + E_{a2} - E_{a3}}{RT}\right)} \]
Given:
\[ E_a = E_{a1} + E_{a2} - E_{a3} = 40 + 50 - 60 = 30 \, \text{kJ/mol} \]
So, the correct answer is: 30
Step 1: Write the rate constant expressions using the Arrhenius equation.
\[ K_1 = A_1 e^{-E_{a1}/RT}, \quad K_2 = A_2 e^{-E_{a2}/RT}, \quad K_3 = A_3 e^{-E_{a3}/RT} \]
\[ K = \frac{K_1 K_2}{K_3} = \frac{A_1 A_2}{A_3} \, e^{-(E_{a1} + E_{a2} - E_{a3})/RT} \]
From the exponent, \[ E_a = E_{a1} + E_{a2} - E_{a3} \]
\[ E_a = 40 + 50 - 60 = 30\,\text{kJ/mol} \]
\[ \boxed{E_a = 30\,\text{kJ/mol}} \]
| Time (Hours) | [A] (M) |
|---|---|
| 0 | 0.40 |
| 1 | 0.20 |
| 2 | 0.10 |
| 3 | 0.05 |
Reactant ‘A’ underwent a decomposition reaction. The concentration of ‘A’ was measured periodically and recorded in the table given below:
Based on the above data, predict the order of the reaction and write the expression for the rate law.