Given Information:
Overall rate constant: \( K = \frac{k_1 k_2}{k_3} \)
Overall activation energy: \( E_a = 400 \, \text{kJ/mol} \)
Activation energies for each step:
\( E_{a1} = 300 \, \text{kJ/mol}, \, E_{a2} = 200 \, \text{kJ/mol}, \, E_{a3} = ? \)
Using the Arrhenius Equation:
The overall rate constant \( K \) and overall activation energy \( E_a \) can be determined by combining the individual rate constants and activation energies as follows:
\[ K = \frac{k_1 k_2}{k_3} \]
According to the Arrhenius equation, we can write: \[ \ln K = \ln \left(\frac{k_1 k_2}{k_3}\right) = \ln k_1 + \ln k_2 - \ln k_3 \] The corresponding activation energy \( E_a \) for \( K \) is: \[ E_a = E_{a1} + E_{a2} - E_{a3} \]
Substituting the Given Values:
\[ 400 = 300 + 200 - E_{a3} \]
Solving for \( E_{a3} \):
\[ E_{a3} = 500 - 400 = 100 \, \text{kJ/mol} \]
Conclusion:
The value of \( E_{a3} \) is \( 100 \, \text{kJ/mol} \).
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