The enthalpy of formation of h2o(l) is -285kJ/mol and enthalpy of neutralisation is -56.7kJ/mol .What is the enthalpy of formation of OH ions? ∆Hf of H+(aq)= 0

To determine the enthalpy of formation of hydroxide ions (OH^-), we can use the given enthalpy values for the formation of water and the enthalpy of neutralization. Let's break this down step by step.

Understanding the Concepts

First, we need to clarify what enthalpy of formation and enthalpy of neutralization mean:

• Enthalpy of Formation (∆H_f): This is the heat change that results when one mole of a compound is formed from its elements in their standard states.
• Enthalpy of Neutralization: This is the heat change that occurs when an acid reacts with a base to form one mole of water.

Given Values

We have the following data:

• Enthalpy of formation of water (H₂O(l)): ∆H_f = -285 kJ/mol
• Enthalpy of neutralization: ∆H_neut = -56.7 kJ/mol
• Enthalpy of formation of H⁺(aq): ∆H_f = 0 kJ/mol

Setting Up the Reaction

The neutralization reaction can be represented as follows:

H⁺(aq) + OH^-(aq) → H₂O(l)

The enthalpy change for this reaction is the enthalpy of neutralization, which is -56.7 kJ/mol. This means that when one mole of water is formed from one mole of hydrogen ions and one mole of hydroxide ions, 56.7 kJ of energy is released.

Calculating the Enthalpy of Formation of OH^-

We can express the enthalpy change for the formation of water in terms of the enthalpies of formation of the reactants:

∆H_neut = ∆H_f(H₂O) - [∆H_f(H⁺) + ∆H_f(OH^-)]

Substituting the known values into this equation gives us:

-56.7 kJ/mol = -285 kJ/mol - [0 + ∆H_f(OH^-)]

Solving for ∆H_f(OH^-)

Now, we can rearrange the equation to solve for the enthalpy of formation of hydroxide ions:

∆H_f(OH^-) = -285 kJ/mol + 56.7 kJ/mol

Calculating this gives:

∆H_f(OH^-) = -228.3 kJ/mol

Final Result

Therefore, the enthalpy of formation of hydroxide ions (OH^-) is -228.3 kJ/mol. This value indicates that the formation of hydroxide ions from their elements is an exothermic process, releasing energy.

In summary, by using the enthalpy of formation of water and the enthalpy of neutralization, we were able to calculate the enthalpy of formation of hydroxide ions effectively. This approach highlights the interconnectedness of thermodynamic properties in chemical reactions.