On the basis of the following thermochemical data
[\Delta_f G^\circ H^+(aq) = 0]
H2O(l) - H^+(aq) +

Q: Why is the enthalpy of formation of $H^+(aq)$ taken as zero?

It is an arbitrary reference point chosen by convention. Just as we define sea level as zero altitude to measure the heights of mountains, chemists define the standard enthalpy of formation of $H^+(aq)$ as zero to measure and compare the enthalpies of all other aqueous ions.

Q: Why are the enthalpies of formation of $H_2(g)$ and $O_2(g)$ zero?

By definition, the standard enthalpy of formation is the energy change when one mole of a substance is formed from its constituent elements in their most stable standard states. Since $H_2(g)$ and $O_2(g)$ are already in their most stable elemental forms at standard conditions, no energy is required to 'form' them.

Q: Could we solve this using Hess's Law by directly adding or subtracting the equations?

Absolutely! If you reverse the water formation equation, its $\Delta H$ becomes $+286.20 \text{ kJ}$. Adding this reversed equation to the ionization equation cancels out $H_2O(l)$ and gives the formation of $H^+(aq)$ and $OH^-(aq)$ directly from $H_2$ and $O_2$. The resulting $\Delta H$ will be $57.32 + 286.20 = 343.52 \text{ kJ}$. Since $H^+$ is zero, the remaining energy belongs to $OH^-$. Wait, reversing it means we are breaking water. The formation of $OH^-$ requires the elements to form the ion. The algebraic method used in the solution is much less prone to sign errors.

Q: What does the negative sign in the final answer indicate?

A negative enthalpy of formation indicates that the formation of the $OH^-(aq)$ ion from its constituent elements is an exothermic process; it releases energy and results in a more thermodynamically stable state compared to the free elements.

Q: Why did the question mention $\Delta_f G^\circ H^+(aq) = 0$ instead of $\Delta_f H^\circ$?

It is likely a slight typographical error in the original exam paper. While it is true that the standard Gibbs free energy of formation for $H^+(aq)$ is also zero by convention, the calculation here strictly requires the standard enthalpy of formation, $\Delta_f H^\circ(H^+, aq) = 0$, to be consistent with the given $\Delta H$ values.

Question

On the basis of the following thermochemical data
$[\Delta_f G^\circ H^+(aq) = 0]$
$H_2O(l) \longrightarrow H^+(aq) + OH^-(aq); \Delta H = 57.32 	ext{ kJ}$
$H_2(g) + \frac{1}{2}O_2(g) \longrightarrow H_2O(l); \Delta H = -286.20 	ext{ kJ}$
The value of enthalpy of formation of $OH^-$ ion at $25^\circ	ext{C}$ is

Options:

(A) $-22.88 	ext{ kJ}$

(B) $-228.88 	ext{ kJ}$

(C) $+228.88 	ext{ kJ}$

(D) $-343.52 	ext{ kJ}$

Sigma Percentile · Accepted Answer

### Decoding the Thermochemical Puzzle

Thermochemistry often feels like a grand accounting system where energy is the currency. In this problem, we are tasked with finding the standard enthalpy of formation of the hydroxide ion, $\Delta_f H(OH^-, aq)$, using two given thermochemical equations.

Let's lay out our given data:
1. The ionization of water: 
$$H_2O(l) \longrightarrow H^+(aq) + OH^-(aq); \Delta H_1 = 57.32 	ext{ kJ}$$
2. The formation of water from its elements: 
$$H_2(g) + \frac{1}{2}O_2(g) \longrightarrow H_2O(l); \Delta H_2 = -286.20 	ext{ kJ}$$

### The Master Equation: Enthalpy of Reaction

The cornerstone of solving such problems is the fundamental relationship derived from Hess's Law. The enthalpy change of any reaction is simply the difference between the total enthalpy of formation of the products and the total enthalpy of formation of the reactants:

$$\Delta_r H = \sum \Delta_f H(	ext{products}) - \sum \Delta_f H(	ext{reactants})$$

### Step 1: Finding the Enthalpy of Formation of Water

Let's apply our master equation to the second reaction, the formation of liquid water.

$$\Delta_r H_2 = \Delta_f H(H_2O, l) - \left[ \Delta_f H(H_2, g) + \frac{1}{2}\Delta_f H(O_2, g) ight]$$

Here is a crucial thermodynamic convention: **The standard enthalpy of formation of any element in its most stable, standard state is exactly zero.** Since hydrogen gas ($H_2$) and oxygen gas ($O_2$) are in their standard states, their formation enthalpies vanish from the equation.

$$-286.20 = \Delta_f H(H_2O, l) - 0 - 0$$

This directly gives us the enthalpy of formation of liquid water:

$$\Delta_f H(H_2O, l) = -286.20 	ext{ kJ}$$

### Step 2: Analyzing the Ionization of Water

Now, we turn our attention to the first reaction, where water dissociates into hydrogen and hydroxide ions. Applying the master equation again:

$$\Delta_r H_1 = \Delta_f H(H^+, aq) + \Delta_f H(OH^-, aq) - \Delta_f H(H_2O, l)$$

We encounter another vital convention here. In aqueous thermodynamics, the standard enthalpy of formation of the hydrogen ion, $H^+(aq)$, is arbitrarily set to zero. It acts as the baseline reference for all other aqueous ions, much like sea level is the reference for altitude.

Substituting our known values into the equation:

$$57.32 = 0 + \Delta_f H(OH^-, aq) - (-286.20)$$

### Final Calculation

All that remains is simple algebra. We isolate the term for the hydroxide ion:

$$57.32 = \Delta_f H(OH^-, aq) + 286.20$$

$$\Delta_f H(OH^-, aq) = 57.32 - 286.20$$

$$\Delta_f H(OH^-, aq) = -228.88 	ext{ kJ}$$

And there we have it! The standard enthalpy of formation of the hydroxide ion is **$-228.88 	ext{ kJ}$**, which corresponds perfectly to option (b).

Select Answer:

Visualized Solution

\text{Given Thermochemical Equations}

\text{Enthalpy of Reaction Formula}

\text{Enthalpy of Formation of } H_2O(l)

\text{Standard States Convention}

\text{Analyzing the Ionization Reaction}

\text{Substituting Known Values}

\text{Final Calculation}

The Sigma Insight: Enthalpy and Hess's Law

Decoding the Thermochemical Puzzle

The Master Equation

Step 1

Step 2

Final Calculation

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