Ncert Exercises & Solutions

The enthalpy of reaction for the reaction

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) is ∆_{r} H^{⊝} = - 572 kJ {mol}^{- 1}$

What will be standard enthalpy of formation of $H_{2} O (l)$ ?

Given that,

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l), ∆_{r} H^{⊝} = ?

This can be obtained by dividing the given authority by 2.

Therefore,

∆_{r} H^{⊝} (H_{2} O) = - \frac{572 kJ {mol}^{- 1}}{2} = - 286 kJ {mol}^{- 1}