Answer:
On the Moh's scale of hardness, aluminum oxide is positioned just below to diamond due to which it is considered as one of the hardest known compounds. This also shows that the compound exhibit an enormous amount of lattice energy, as to transform the oxide into its constituent ions, the energy is required to overcome.
Based on the chemical formula of the compound, that is, Al2O3, it is shown that the ions of Al3+ and O2- are kept close due to the activity of the strong electrostatic ionic bonds. The electrostatic forces and the ionic bonding between the ions are extremely robust due to the presence of the ions high charge density. Therefore, to dissociate the bonds, an enormous amount of energy is needed, and at the same time, a high amount of lattice energy is present.
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
It would be solid
hope this helps
