i. Predict the sign (positive or negative) for the entropy change for the reaction.

ii. Explain why the entropy change for the above reaction is considerably more than for the reaction:

H₂(g) + I₂(g) → 2HI(g)

i. State what is oxidized and what is reduced during the thermal decomposition of ammonium dichromate(VI).

ii. Clearly the thermal decomposition of ammonium dichromate is very exothermic (ΔH^⦵ = −1794 kJ mol⁻¹).
What can you also deduce about the change in entropy?

i. The standard enthalpies of formation of CaCO₃(s), CaO(s) and CO₂(g) are – 1207 kJ mol^-1, – 635 kJ mol^-1 and – 394 kJ mol^-1 respectively.
Calculate the standard enthalpy change, ∆H^⦵, for the thermal decomposition of calcium carbonate.

Using Hess’s law

– 1207 + ∆H^⦵ = – 635 + (– 394)

∆H^⦵ = + 178 kJ mol^-1

40.0 J K^-1 mol^-1 and 214 J K^-1 mol^-1 respectively.
Calculate the change in entropy, ∆S^⦵, for the thermal decomposition of calcium carbonate.

∆S^⦵ = S^⦵products – S^⦵reactants

∆S^⦵ = (40.0 + 214) – 92.9 = + 161 J K^-1 mol^-1

iii. Calculate the change in standard Gibbs energy, ∆G^⦵, for the thermal decomposition of calcium carbonate at 298 K.

iv. State whether the thermal decomposition of calcium carbonate is spontaneous or non-spontaneous at 298K.

v. Explain why limestone cliffs are thermodynamically stable whereas limestone decomposes readily in a blast furnace.

i.  Use the information below to determine the value for the free enthalpy change for this reaction at 180 ^oC.

Using Hess’s law

∆H^⦵ = –104 – (20.4 + 0) = – 124.4 kJ mol^-1

∆S^⦵ = S^⦵products– S^⦵reactants

∆S^⦵ = 270 – (267 + 131) = – 128 J K^-1 mol^-1

∆G^⦵ = ∆H^⦵ – T∆S^⦵ = – 124.4 – [(180 + 273) x –128/1000] = – 66.4 kJ mol^-1

ii. At 180 ^oC this reaction is spontaneous.
Determine the temperature above which the reaction becomes non-spontaneous.