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, \[ E_a = E_{a1} + E_{a2} - E_{a3} \]
\[ E_a = 40 + 50 - 60 = 30\,\text{kJ/mol} \]
\[ \boxed{E_a = 30\,\text{kJ/mol}} \]
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
For $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ $\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following statement is correct?
