# Question #356d2

Jun 22, 2016

Here's my take on this.

#### Explanation:

The trick here is to realize that the enthlapy change for that reaction when $1$ mole of water is produced is actually the standard enthalpy of formation, $\Delta {H}_{f}^{\circ}$, for water.

The standard enthalpy of formation of a compound represents the enthalpy change that occurs when one mole of said compound is formed from its constituent elements in their stable form under standard conditions for pressure and temperature.

Standard conditions are defined as a pressure of $\text{101.3 kPa}$ and a temperature of $\text{298 K}$.

So, you know that you have

$2 \text{H"_ (2(g)) + "O"_ (2(g)) -> 2"H"_ 2"O"_ ((l)) + "energy}$

The thermochemical equation that describes the standard enthalpy change of formation is

$\text{H"_ (2(g)) + 1/2"O"_ (2(g)) -> "H"_ 2"O"_ ((l)) " } \Delta {H}_{f}^{\circ}$

Now, I assume that a table with some standard enthalpies of formation is available to you, because the problem doesn't provide you with any information that allows you to calculate $\Delta {H}_{f}^{\circ}$.

My guess would be that you have the value listed in a table that you can reference. The standard enthalpy of formation for water is

$\Delta {H}_{f}^{\circ} = - {\text{285.8 kJ mol}}^{- 1}$

https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation

The negative sign symbolizes the fact that heat is being given off.

So, when one mole of water is formed from its constituent elements in their stable form under standard conditions, $\text{285.8 kJ}$ of heat are being given off.

Therefore, the answer is (2) $\text{285.8 kJ}$.