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
A(g) $ \rightarrow $ B(g) + C(g) is a first order reaction.
The reaction was started with reactant A only. Which of the following expression is correct for rate constant k ?
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?
Match List-I with List-II: List-I