The decomposition of ammonium dichromate [(NH4)2Cr2O7] is called the volcano demonstration for its fiery display. The decomposition reaction involves breaking down ammonium dichromate into nitrogen gas, water vapor, and solid chromium(III) oxide. From the data in Appendix 4 and given kJ/mol and J/K mol for (NH4)2Cr2O7, calculate _G for the volcano reaction and calculate for ammonium dichromate.

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!!!

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

12 is it do you think so