In which thermochemical equations would the ΔH be considered a heat of solution? 2NaHCO3(s)→Na2CO3 (s) + H2O (g) + CO2 (g), ΔH = –129 kJ C6H6(s)→C6H6(l), ΔH=+9.87 kJ NH4NO3(s)→NH4+(aq)+NO3–(aq), ΔH=+25.7 kJ/mol 2C8H18(l)+25O2(g)→16CO2(g)+18H2O(l) , ΔH =–5,471 kJ/mol

NH4NO3 (s) --> NH4+ (aq) + NO3- (aq) H = +25.7 kJ/mol

Explanation:

Step 1: Data given

2NaHCO3(s) → Na2CO3 (s) + H2O (g) + CO2 (g) ΔH = –129 kJ

C6H6(s) → C6H6(l) ΔH=+9.87 kJ

NH4NO3(s )→ NH4+(aq) + NO3–(aq) ΔH=+25.7 kJ/mol

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) ΔH =–5,471 kJ/mol

Step 2:

2NaHCO3(s) → Na2CO3 (s) + H2O (g) + CO2 (g)  H = -129 kJ:

This is the decomposition reaction of NaHCO3.

NaHCO3 will break down and form Na2CO3, H2O and CO2

⇒ the heat here is heat of decomposition

C6H6 (s) --> C6H6 (l), H = +9.87 kJ  

In this reaction we'll not form a new product. Benzene will change phase. From its solid form to liquid benzene.

⇒ the heat here is heat of fusion

NH4NO3 (s) --> NH4+ (aq) + NO3- (aq) H = +25.7 kJ/mol  

When solid NH4NO3 is dossolved in water. it will dissociate in ammonium ion (NH4+) and nitrate ion (NO3-).

This reaction shows a solid dissolving in water and therefore the H represents the heat of solution.

⇒ The heat here is heat of solution

2C8H18 (l) + 25O2 (g) --> 16CO2 (g) + 18H2O (l), H = -5,471 kJ/mol: this is a combustion reaction

This is the combustion reaction of octane. CO2 and H2O will be produced.

⇒  The heat here is heat of combustion.

your answer would be “dry”