Question

When a carbonate solution is acidified, carbon dioxide is produced.

​2H+(aq) + CO32-(aq) → H2O(l) + CO2(g)

(a) Can we say anything about the sign of the entropy change without doing any calculations?

(b) Calculate the entropy change for this reaction.

please use the commonly found entropy values found online for each molecule.

Answer 1

(a)

Liquids have higher entropy than solids. Gases have higher entropy than liquids. So the an increasing the number of moles of gases implies a positive ∆S°.

Count the total number of moles of gas each side of the reaction. If products have more moles of gases then ∆S° is positive. If reactants have more moles of gases then ∆S° is negative.

Here products have more number of moles of gas. So the sign of entropy change is Positive.

(b)

∆S° = S° (products) - S° ( reactants)

∆S° = ( S° H2O(l) + S° CO2(g) ) - ( 2×S° H+(aq) + S°CO3^2-(aq))

∆S° ={ (69.91 + 213.74) - ( 2×0 + (-53.1) } J/mol.K

∆S° = (283.65 + 53.1 ) J/mol.K

∆S° = + 336.75 J/mol.K