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?
1. Given: - $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ - $\mathrm{E}_{\mathrm{f}} = 180 \mathrm{~kJ} \mathrm{~mol}^{-1}$ - $\mathrm{E}_{\mathrm{b}} = 200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2. Calculate the enthalpy change ($\Delta \mathrm{H}$): \[ \Delta \mathrm{H} = \mathrm{E}_{\mathrm{f}} - \mathrm{E}_{\mathrm{b}} = 180 \mathrm{~kJ} \mathrm{~mol}^{-1} - 200 \mathrm{~kJ} \mathrm{~mol}^{-1} = -20 \mathrm{~kJ} \mathrm{~mol}^{-1} \]
3. Effect of catalyst: - Catalyst lowers the activation energy but does not change the Gibbs free energy change ($\Delta \mathrm{G}$) or the enthalpy change ($\Delta \mathrm{H}$) of the reaction.
Therefore, the correct answer is (1) Catalyst does not alter the Gibbs energy change of a reaction.
The rate of a reaction:
A + B −→ product
is given below as a function of different initial concentrations of A and B.
Experiment | \([A]\) (mol L\(^{-1}\)) | \([B]\) (mol L\(^{-1}\)) | Initial Rate (mol L\(^{-1}\) min\(^{-1}\)) |
---|---|---|---|
1 | 0.01 | 0.01 | \(5 \times 10^{-3}\) |
2 | 0.02 | 0.01 | \(1 \times 10^{-2}\) |
3 | 0.01 | 0.02 | \(5 \times 10^{-3}\) |
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: