Question

Explain how energy is conserved between this reaction and the and surrounding environment C3H8+4O2->3CO2+4H2O+ heat (-2218.6)

Answer 1

Answer:

The heat of formation = Heat of formation of the products - Heat of formation of the reactants

= -2323 + 104 = -2219 ≈ -2218.6 kJ/mol.

Explanation:

The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one system to another, but the total remains the same

The heat of formation of C₃H₈ is 3C + 4 H₂ → C₃H₈ $\Delta H^{\circ}_f$  -104 kJ/mol

The heat of formation of O₂ is O₂ (g) → O₂ (g)  $\Delta H^{\circ}_f$  0 kJ/mol

The heat of formation of H₂O is H₂(g) + 1/2 O₂→ H₂O (g)  $\Delta H^{\circ}_f$  -286kJ/mol

The heat of formation of CO₂ is C (s) + O₂ (g) → CO₂ (g)  $\Delta H^{\circ}_f$  -393 kJ/mol

Therefore, in the given  reaction we have;

C₃H₈ + 4 O₂ → 3 CO₂ + 4 H₂O

The heat of formation = Heat of formation of the products - Heat of formation of the reactants

The heat of formation = 3 × (-393) + 4 × (-286) - (-104) = -2219 ≈ -2218.6 kJ/mol.