Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

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