Consider the following reaction. 2C2H2(g)+5O2(g)⟶4CO2(g)+2H2O(g) 2C2H2(g)+5O2(g)⟶4CO2(g)+2H2O(g) Compound ΔH∘f (kJ/mol)ΔHf∘ (kJ/mol) C2H2(g)C2H2(g) 227.4227.4 O2(g)O2(g) 00 CO2(g)CO2(g) −393.5−393.5 H2O(g)H2O(g) −241.8−241.8 What is the ΔH∘ΔH∘ of the reaction?

ΔH° = -1255.8 kJ

Explanation:

Step 1: Data given

ΔHf∘ C2H2 = 227.4 kJ/mol

ΔHf∘ O2 = 0 kJ/mol

ΔHf∘ CO2 = -393.5 kJ/mol

ΔHf∘ H2O = -241.8 kJ/mol

Step 2: The balanced equation

2C2H2(g)+5O2(g) ⟶ 4CO2(g)+2H2O(g)

C2H2(g)+5/2O2(g) ⟶ 2CO2(g)+H2O(g)

Step 3: Calculate ΔH° of the reaction

ΔH° = (2*ΔHf∘ CO2 + ΔHf∘ H2O) - (ΔHf∘ C2H2)

ΔH° = (2* -393.5 kJ/mol + (-241.8) kJ/mol) - 227.4 kJ/mol

ΔH° = -787 - 241.8 - 227. kJ/mol

ΔH° = -1255.8 kJ