Question

If enthalpies of formation of C₂H₄(g), CO₂(g) and H₂O(l) at 25°C and 1 atm pressure are 52, – 394 and – 286 kJ/mol respectively, the change in ethalpy for combustion of C₂H₄ is equal to

विकल्प

• – 141.2 kJ/mol

• – 1412 kJ/mol

• + 14.2 kJ/mol

• + 1412 kJ/mol

Answer 1

– 1412kJ/mol

Explanation:

Enthalpy of formation of C₂H₄, CO₂ and H₂O are 52, – 394 and – 286 kJ/mol respectively. (Given)

The reaction is \[\ce{C2H4 + 3O2 -> 2CO2 + 2J2O}\]. change in enthalpy,

(ΔH) = \[\ce{ΔH_{products} – ΔH_{reactants}}\]

= 2 × (– 394) + 2 × (– 286) – (52 + 0) = – 1412 kJ/mol.