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 decomposition of a compound A follows first-order kinetics. The concentration of A at time t = 0 is 1.0 mol L-1. After 60 minutes, it reduces to 0.25 mol L-1. What is the initial rate of the reaction at t = 0? (Take ln 2 = 0.693)
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).