Animated Solution for Chemistry - Chemical Thermodynamics: For the equilibrium, $2H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$, the value of $\Delta G^{\circ }$ at $298\text{ K}$ is approximately

Imagine a beaker filled with absolutely pure water. To the naked eye, it looks perfectly still and uniform. However, at the molecular level, it is a bustling metropolis of activity. Water molecules are constantly moving, vibrating, and colliding with one another. Occasionally, a collision occurs with just the right energy and orientation that a proton ($H^{+}$) is transferred from one water molecule to another.

This microscopic event creates two distinct ions: a hydronium ion ($H_3{O}^{+}$) and a hydroxide ion ($O{H}^{-}$). This process is known as the autoionization of water, and it is represented by the equilibrium equation:

Even though this happens continuously, the fraction of water molecules that actually undergo this transformation at any given moment is incredibly small. At standard room temperature ($298\text{ K}$), the equilibrium constant for this specific reaction is a fundamental constant of nature that you likely already know: the ionic product of water, $K_w$.

This tiny number tells us that the equilibrium lies heavily to the left, favoring intact water molecules over the separated ions.

Thermodynamics and Equilibrium

In chemistry, we have two powerful lenses through which we can view a reaction. Thermodynamics tells us about the energy changes and whether a process is fundamentally spontaneous. Chemical Equilibrium tells us the extent to which a reaction proceeds before it balances out.

These two worlds are beautifully connected by a single, elegant master equation:

Here, $\Delta G^{\circ }$ is the standard Gibbs free energy change, $R$ is the universal gas constant, $T$ is the absolute temperature in Kelvin, and $K$ is the equilibrium constant. This equation is the Rosetta Stone of physical chemistry. It allows us to translate the language of equilibrium concentrations directly into the language of energy.

To make our calculations a bit more straightforward, we often convert the natural logarithm ($\ln$) to a base-10 logarithm (${\log }_{10}$). We do this by multiplying by the conversion factor $2.303$:

Crunching the Numbers

Now that we have our master equation and our known values, it is time to execute the calculation. Let's carefully substitute the numbers into the formula.

We know that the universal gas constant $R$ must be in energy units, so we use $8.314{\text{ J K}}^{-1}{\text{mol}}^{-1}$. The temperature $T$ is given as $298\text{ K}$. And our equilibrium constant $K$ is $10^{-14}$.

The logarithm term is the easiest part to handle. The base-10 logarithm of $10^{-14}$ is simply $-14$.

Notice something crucial here: the negative sign from the formula and the negative sign from the logarithm will multiply together to give a positive result.

Let's multiply the constants first:

Now, multiply this by $14$:

Why Positive?

Our calculated value is $79881.8{\text{ J mol}}^{-1}$. However, standard thermodynamic values are typically reported in kilojoules. To convert, we simply divide by $1000$:

Rounding to the nearest integer, as suggested by the options, gives us $80{\text{ kJ mol}}^{-1}$.

But let's pause and think about the physical meaning of this number. Why is $\Delta G^{\circ }$ a large positive value?

A positive standard Gibbs free energy change indicates that under standard conditions (where all reactants and products are at $1\text{ M}$ concentration), the forward reaction is non-spontaneous. This aligns perfectly with our intuition! If you had a beaker where the concentration of $H_3{O}^{+}$ and $O{H}^{-}$ were both artificially forced to be $1\text{ M}$, they would rapidly react to form water. The equilibrium naturally prefers the reactants (water molecules), which is why $K$ is so small and $\Delta G^{\circ }$ is so large and positive.

A Thought Experiment

We have successfully solved the problem, but true mastery comes from asking "What if?"

What if the temperature of the water was increased to $350\text{ K}$? The breaking of bonds to form ions requires energy, meaning the autoionization of water is an endothermic process. According to Le Chatelier's principle, increasing the temperature of an endothermic reaction shifts the equilibrium forward.

Therefore, at $350\text{ K}$, the value of $K_w$ would be greater than $10^{-14}$. If $K$ increases, the term ${\log }_{10}K$ becomes less negative. Consequently, the value of $\Delta G^{\circ }$ would become less positive. This beautiful interplay between temperature, equilibrium, and energy is the heart of chemical thermodynamics!