# Question #899a4

Apr 25, 2017

$\Delta {H}_{\text{combustion}}^{\circ} = - 242 \cdot k J \cdot m o {l}^{-} 1$.

#### Explanation:

$\Delta {H}^{\circ}$ values are ALWAYS quoted per mole of reaction as written.

And thus:

$2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 2 {H}_{2} O \left(g\right)$ $\Delta {H}^{\circ} = - 484 \cdot k J \cdot m o {l}^{-} 1$.

equivalently,

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow 2 {H}_{2} O \left(g\right)$ $\Delta {H}^{\circ} = - 242 \cdot k J \cdot m o {l}^{-} 1$.

Now, by definition, $\Delta {H}_{\text{combustion}}^{\circ}$ is the enthalpy change when ONE mole of substance undergoes combustion under standard conditions.

And thus for $\text{dihydrogen}$, $\Delta {H}_{\text{combustion}}^{\circ} = - 242 \cdot k J \cdot m o {l}^{-} 1$.

And if we combust a $1 \cdot g$ mass of ${H}_{2}$, i.e. a half molar quantity with respect to dihydrogen, the energy released will reflect the molar quantity, i.e. $\Delta {H}^{\circ} = - 141 \cdot k J$.