The enthalpy change for the reaction H2O2(l)⟶H2O(l)+(1/2)O2(g) is −98.3 kJ mol−1 If the enthalpy of formation of H2O2(l) is −187.4 kJ mol−1the enthalpy of formation of H2O(l) would be

A
$- 285.8 kJ {mol}^{- 1}$
B
$285.8 kJ {mol}^{- 1}$
C
$- 384.1 kJ {mol}^{- 1}$
D
$384.1 kJ {mol}^{- 1}$

Solution:

The reaction to be considered is $H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l)$

Given data are

$\begin{array}{ll} H_{2} O_{2} (l) ⟶ H_{2} O (l) + (1 / 2) O_{2} (g) Δ H = - 98.3 kJ {mol}^{- 1} \\ H_{2} (g) + O_{2} (g) ⟶ H_{2} O_{2} (l) Δ H = - 187.4 kJ {mol}^{- 1} \end{array}$

Adding these two reactions gives the required reaction. Hence

$Δ H = - (98.3 + 187.4) kJ {mol}^{- 1} = - 285.7 kJ {mol}^{- 1}$

The enthalpy change for the reaction $H_{2} O_{2} (l) ⟶ H_{2} O (l) + (1 / 2) O_{2} (g)$ is $- 98.3 kJ {mol}^{- 1}$ If the enthalpy of formation of $H_{2} O_{2} (l)$ is $- 187.4 kJ {mol}^{- 1}$ the enthalpy of formation of $H_{2} O (l)$ would be

Solution:

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