- At equilibrium, the Gibbs free energy change \( \Delta G \) is equal to zero. This is because, at equilibrium, the system has reached a state where there is no further net change in the composition.
- Therefore, \( \Delta G = 0 \) at equilibrium, indicating that the forward and reverse reactions occur at the same rate.
Conclusion:
The value of Gibbs free energy change at equilibrium is zero, so the correct answer is (D).
List-I (Sol) | List-II (Method of preparation) |
---|---|
A) \( \text{As}_2\text{S}_3 \) | I) Bredig's arc method |
B) \( \text{Au} \) | II) Oxidation |
C) \( \text{S} \) | III) Hydrolysis |
D) \( \text{Fe(OH)}_3 \) | IV) Double decomposition |
A closed-loop system has the characteristic equation given by: $ s^3 + k s^2 + (k+2) s + 3 = 0 $.
For the system to be stable, the value of $ k $ is: