Q.The enthalpy of reaction for the reaction :
2H2(g) + O2(g) → 2H2O(l) is ΔrH° = -572 kJ mol^-1.
What will be standard enthalpy of formation of H2O(l)?

The key idea is that the standard enthalpy of formation ($\Delta_fH^\circ$) refers to the enthalpy change when one mole of a compound is formed from its elements in their standard states.

1. The given reaction, $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$, has an enthalpy change of $\Delta_rH^\circ = -572 \text{ kJ mol}^{-1}$. This value corresponds to the formation of 2 moles of $H_2O(l)$.
2. The standard formation reaction for $H_2O(l)$ is $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$, which produces 1 mole of $H_2O(l)$. …

When we talk about the standard enthalpy of formation ($\Delta_f H^\circ$) of a compound, we are referring to the enthalpy change that occurs when one mole of that compound is formed from its constituent elements, with all substances in their standard states (usually $298.15 \text{ K}$ and $1 \text{ bar}$ pressure).

The key here is "one mole" and "from its constituent elements in their standard states".

Let's break down the problem:

1. Understand the given reaction and its enthalpy. The reaction provided is:

The enthalpy of reaction for this specific process is given as $\Delta_r H^\circ = -572 \text{ kJ mol}^{-1}$. This value tells us that $572 \text{ kJ}$ of heat are released when 2 moles of gaseous hydrogen react with 1 mole of gaseous oxygen to produce 2 moles of liquid water.

2. Relate the given reaction to the definition of standard enthalpy of formation.

For the standard enthalpy of formation of $\text{H}_2\text{O}(l)$, we need to consider the reaction where one mole of $\text{H}_2\text{O}(l)$ is formed from its elements in their standard states.

The elements are hydrogen ($\text{H}_2$) and oxygen ($\text{O}_2$). Both are gases at standard conditions, which are their standard states.

So, the formation reaction for one mole of $\text{H}_2\text{O}(l)$ is:

Notice that the given reaction ($2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$) is exactly double the formation reaction for one mole of water. It produces 2 moles of $\text{H}_2\text{O}(l)$.

3. Calculate the standard enthalpy of formation for 1 mole of $\text{H}_2\text{O}(l)$.

Since the given $\Delta_r H^\circ = -572 \text{ kJ mol}^{-1}$ corresponds to the formation of 2 moles of $\text{H}_2\text{O}(l)$ from its elements, the enthalpy of formation for a single mole of $\text{H}_2\text{O}(l)$ will be half of this value.

> [!TIP]
> You can also use the general formula for enthalpy of reaction:
> > [!FORMULA] …