Question

What is the change in enthalpy for the following reaction? 2H2O2(aq) 2H2O(l) + O2(g) Given: H2O: ∆H= -242 kJ H2O2: ∆H= -609 kJ

-286 kJ-572 kJ572 kJ286 kJ

Answer 1

The enthalpy change for the decomposition of hydrogen peroxide into water and oxygen is calculated using the enthalpies of products and reactants, resulting in a change in enthalpy of -196.8 kJ.

Step-by-step explanation:

To determine the change in enthalpy (ΔH) for the decomposition of hydrogen peroxide into water and oxygen, we need to consider the given enthalpies of the products and reactants and apply Hess's Law. For the reaction 2H2O2(aq) → 2H2O(l) + O2(g), the enthalpy changes for the products and reactants are as follows:

• H2O (l): ΔH = -286 kJ/mol
• H2O2 (aq): ΔH = -187.6 kJ/mol

Using Hess's Law, the overall change in enthalpy for the reaction can be calculated by the sum of the enthalpies of the products minus the sum of the enthalpies of the reactants:

ΔH = [2 * (-286 kJ)] - [2 * (-187.6 kJ)]

ΔH = (-572 kJ) - (-375.2 kJ)

ΔH = -572 kJ + 375.2 kJ

ΔH = -196.8 kJ

Therefore, the enthalpy change for the given reaction is -196.8 kJ.

Answer 2

ΔH = {2ΔH(H₂O) + ΔH(O₂)} - 2Δ(H₂O₂)

ΔH(H₂O) = -242 kJ
ΔH(O₂,g) = 0
Δ(H₂O₂) = -609 kJ

ΔH = 2*(-242) + 0 - 2*(-609) = 734 kJ