Answer:
ΔG° for the reaction = -2345.81 kJ
ΔG°f for Ammonium dichromate = 619.91 kJ
Explanation:
The decomposition of (NH₄)₂Cr₂O₇ is represented as
(NH₄)₂Cr₂O₇ (s) → N₂ (g) + 4H₂O (g) + Cr₂O₃ (s)
From literature, the standard enthalpy of formation (ΔH°f) of the products are given as
For N₂, 0 kJ/mol
For H₂O, -242 kJ/mol
For Cr₂O₃, -1128 kJ/mol
For (NH₄)₂Cr₂O₇, -23 kJ/mol (given in the question)
Also, their standard entropies are given as
For N₂, 192 J/mol.K
For H₂O, 189 J/mol.K
For Cr₂O₃, 81 J/mol.K
For (NH₄)₂Cr₂O₇, 114 J/mol.K (given in the question)
Note that the enthalpy of the reaction is given as
ΔH° = ΔH°(products) - ΔH°(reactants)
ΔH°(products) = (1×0) + (4×-242) + (1×-1128) = -2096 kJ
ΔH°(reactants) = (1×-23) = -23 kJ
ΔH° = -2096 - (-23) = -2073 kJ
Note that the entropy of the reaction is given as
ΔS° = ΔS°(products) - ΔS°(reactants)
ΔS°(products) = (1×192) + (4×189) + (1×81) = 1029 kJ/K
ΔS°(reactants) = (1×114) = 114 kJ/K
ΔS° = 1029 - (114) = 915 J/K = 0.915 kJ/K
The Gibb's free energy is then calculated thus
ΔG° = ΔH° - TΔS°
where T = absolute temperature in Kelvin at which the reaction takes place = 298.15K
ΔG° = -2073 - (298.15×0.915) = -2345.81 kJ
To calculate the ΔG°f for Ammonium dichromate, we need the ΔG°f for the products.
For N₂, 0 kJ/mol
For H₂O, -228.6 kJ/mol
For Cr₂O₃, -811.5 kJ/mol
For (NH₄)₂Cr₂O₇, let the standard free energy be x
ΔG° = ΔG°(products) - ΔG°(reactants)
ΔG°(products) = (1×0) + (4×-228.6) + (1×-811.5) = -1,725.9 kJ
ΔG°(reactants) = (1×x) = x kJ
ΔG° = -1,725.9 - (x)
Recall that ΔG° = -2345.81 kJ
-2345.81 = -1,725.9 - x
x = -1,725.9 + 2345.81 = 619.91 kJ
Hope this Helps!!!
