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For the autoionization of water at \(25^{\circ} \mathrm{C}:\) $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{w}}\) is \(1.0 \times 10^{-14}\). What is \(\Delta G^{\circ}\) for the process?

Short Answer

Expert verified
\(\Delta G^{\circ} = 184 \, \text{kJ/mol}\) for water autoionization at 25°C.

Step by step solution

01

Write the Formula for ΔG°

The standard Gibbs free energy change (abla G^{\circ}) is related to the equilibrium constant (K) by the formula:\[ \Delta G^{\circ} = -RT \ln K\]where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin.
02

Convert Temperature to Kelvin

The given temperature is 25°C. Convert it to Kelvin by adding 273.15:\[ T = 25 + 273.15 = 298.15 \, \text{K}\]
03

Insert Values into the Formula

Now use the known values:- R = 8.314 J/mol·K- T = 298.15 K- K_w = 1.0 \times 10^{-14}Insert these into the formula:\[ \Delta G^{\circ} = -(8.314 \, \text{J/mol·K})(298.15 \, \text{K}) \ln (1.0 \times 10^{-14})\]
04

Calculate Natural Logarithm

Calculate the natural logarithm of the equilibrium constant:\[ \ln (1.0 \times 10^{-14}) = -32.236\]
05

Calculate ΔG°

Substitute the value from Step 4 into the calculation for \( \Delta G^{\circ} \):\[ \Delta G^{\circ} = -(8.314)(298.15)(-32.236)\]Calculate the result:\[ \Delta G^{\circ} = (8.314)(298.15)(32.236) \]\[ \Delta G^{\circ} = 1.84 \times 10^5 \, \text{J/mol} \, (or \, 184 \, \text{kJ/mol})\]
06

Interpret the Result

The calculated \(\Delta G^{\circ}\) value is positive, indicating that the autoionization of water is non-spontaneous under standard conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, often represented by the letter \( K \), is central to understanding reactions that have reached a state of balance. It quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction. In the case of water autoionization, the equilibrium constant \( K_w \) is \( 1.0 \times 10^{-14} \) at \( 25^{\circ}C \).

This small value indicates that the concentration of ions formed from water is very low, hence the reaction strongly favors the reactants in this condition.

When calculating Gibbs free energy change, the equilibrium constant is crucial because it relates to how far the reaction lies towards the products or reactants, influencing the spontaneity of the process.
Gibbs Free Energy Change
Gibbs free energy change, represented as \( \Delta G^{\circ} \), tells us whether a reaction is spontaneous under standard conditions. This energy change can be calculated using the formula:
\[ \Delta G^{\circ} = -RT \ln K \]where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin. If \( \Delta G^{\circ} \) is negative, the reaction is spontaneous. If positive, like in water's autoionization, the reaction is non-spontaneous.

The magnitude of \( \Delta G^{\circ} \) indicates the amount of energy required to drive the reaction. In our exercise, a calculated \( \Delta G^{\circ} \) of 184 kJ/mol reflects significant non-spontaneity under standard conditions.
Kelvin Temperature Conversion
Temperature plays a key role in calculations involving thermodynamics. Often, temperatures are provided in Celsius but need to be converted to Kelvin for accurate thermodynamic calculations, since Kelvin is the SI unit for temperature.

Conversion is simple: add 273.15 to the Celsius temperature to get Kelvin.
For example, \( 25^{\circ}C = 25 + 273.15 = 298.15 \, K \).

This unit is essential as it ensures a uniform standard that is widely accepted, leading to accurate and reliable thermodynamic results.
Universal Gas Constant
The universal gas constant, symbolized as \( R \), is a key fundamental constant in chemistry and appears in many equations, including those involving energy calculations like Gibbs free energy.

It relates energy scale in chemistry to temperature and amounts of substance.

The value used in thermodynamics is \( 8.314 \, J/mol\cdot K \).

When calculating \( \Delta G^{\circ} \), this constant acts as a bridge between the temperature in Kelvin, the equilibrium constant, and Gibbs free energy, thereby integrating different aspects of the reaction dynamics into one coherent calculation.

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Most popular questions from this chapter

Consider the following facts: Water freezes spontaneously at \(-5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and ice has a lower entropy than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

Many hydrocarbons exist as structural isomers, which are compounds that have the same molecular formula but different structures. For example, both butane and isobutane have the same molecular formula of \(\mathrm{C}_{4} \mathrm{H}_{10}\) Calculate the mole percent of these molecules in an equilibrium mixture at \(25^{\circ} \mathrm{C}\), given that the standard free energy of formation of butane is \(-15.9 \mathrm{~kJ} / \mathrm{mol}\) and that of isobutane is \(-18.0 \mathrm{~kJ} / \mathrm{mol}\). Does your result support the notion that straight-chain hydrocarbons (i.e. hydrocarbons in which the \(\mathrm{C}\) atoms are joined along a line) are less stable than branch-chain hydrocarbons?

The reaction \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) proceeds spontaneously at \(25^{\circ} \mathrm{C}\) even though there is a decrease in entropy in the system (gases are converted to a solid). Explain.

Hydrogenation reactions (e.g., the process of converting \(\mathrm{C}=\mathrm{C}\) bonds to \(\mathrm{C}-\mathrm{C}\) bonds in the food industry) are facilitated by the use of a transition metal catalyst, such as \(\mathrm{Ni}\) or \(\mathrm{Pt}\). The initial step is the adsorption, or binding, of hydrogen gas onto the metal surface. Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) when hydrogen gas is adsorbed onto the surface of Ni metal.

When a native protein in solution is heated to a high enough temperature, its polypeptide chain will unfold to become the denatured protein. The temperature at which a large portion of the protein unfolds is called the melting temperature. The melting temperature of a certain protein is found to be \(46^{\circ} \mathrm{C},\) and the enthalpy of denaturation is \(382 \mathrm{~kJ} / \mathrm{mol}\). Estimate the entropy of denaturation, assuming that the denaturation is a twostate process; that is, native protein \(\longrightarrow\) denatured protein. The single polypeptide protein chain has 122 amino acids. Calculate the entropy of denaturation per amino acid. Comment on your result.

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