77. In the early days of automobiles, illumination at night was provided by burning acetylene, C2H2. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC2:CaC2 (s) + 2H2O (l) ⟶Ca (OH) 2 (s) + C2H2 (g). Calculate the standard enthalpy of the reaction. TheΔHf°of CaC2is - 15.14 kcal/mol.

the standard enthalpy for the reaction is

ΔHreaction° = - 122.79kJ

Explanation:

Given the enthalpy of formation of Δf° of CaC₂ = - 15.14 kcal/mol

step1: convert 15.14kcal/mol to kJ/mol

1 kcal = 4.184 kJ

hence, - 15.14 kcal * 4.184 kJ = - 63.35 kJ/mol

Step 2: calculate the standard enthalpy of calcium carbide with water using equation from Hess's law;

ΔH°reaction = Σₙ * ΔH°product - Σₙ * ΔH°reactant

ΔH°reaction = [ 1 mol. Ca (OH) ₂ * - 985.2 kJ/mol. Ca (OH) ₂ + 1 mol. C₂H₂ * 227 kJ/mol. C₂H₂] - [1 mol. CaC₂ * - 63.36kJ/mol. CaC₂ + 2 mol. H₂O * - 285kJ/mol. H₂O]

ΔH°reaction = 1 (-985 kJ) + 1 (227.4 kJ) - 1 (-63.65 kJ) - 2 (-285.83)

= - 985.2 kJ + 227.4 kJ + 63.35 kJ + 571.66 kJ

ΔH°reaction = - 122.79 kJ

554.86kj

Explanation: Since 1 mole of CaC2=15.14kj yield 1mole of C2H2

The enthalpy change of H2O is 2*285=570

570+-15.14=554.86kj

Hence Hp is 554.86kj

He=Hp