Question

Change in enthalpy for reaction 2H2O2(l) → 2H2O(l) + O2(g) If heat of formation of H2O2(l) and H2O(l) are – 188 & – 286 KJ/mol respectively : -

• A. -196 KJ/mol
• B. (+196 KJ/mol)
• C. (+198 KJ/mol)
• D. -198 KJ/mol

Answer 1

Answer: (A)

-196 KJ/mol

Answer 2

The correct option is(A): -196 KJ/mol. The change in enthalpy for the given reaction is calculated using the heats of formation of the reactants and products: ΔH = Σ(heats of formation of products) - Σ(heats of formation of reactants) ΔH = [2(-286 KJ/mol) + 0] - [2(-188 KJ/mol)] = -572 KJ/mol + 376 KJ/mol = -196 KJ/mol So, the change in enthalpy for the reaction 2H2O2(l) → 2H2O(l) + O2(g) is -196 KJ/mol.