Question

If $\Delta H_{f}^{o}$ for $H_{2}O_{2}$ and $H_{2}O$ are -188 kJ/mol and -286 kJ/mol, what will be the enthalpy change of the reaction:

$2H_{2}O_{2}\left(l\right) \rightarrow 2H_{2}O\left(l\right)+O_{2}\left(\right.g\left.\right)$

• A. -196 kJ
• B. -494 kJ
• C. 146 kJ
• D. -98 kJ

Answer 1

Answer: (A)

Given reaction is $2H_{2}O_{2}\left(l\right) \rightarrow 2H_{2}O\left(l\right)+O_{2}\left(\right.g\left.\right)$

$\text{ΔH}_{\text{Reaction}}^{\text{o}} = \text{ΔH}_{\text{f}}^{\text{o}} \left(\text{Products}\right) - \text{ΔH}_{\text{f}}^{\text{o}} \left(\text{Reactants}\right)$


Enthalpy change $=2\times \left(- 286\right)-\left(2 \times - 188\right)$

$=-196 \, k J$