44. On the basis of the following thermochemical data : 0) fGH( aq) ( °=? + 32kj 57.);H(aq)OH(aq)HO(H 2 ?= ?+ +-l 20kj 286.);HO()H(gO 2 1 g)H( 2 22 =- ??+ l The value of enthalpy of formation of OH– ion at 25°C is

To determine the enthalpy of formation of the hydroxide ion (OH⁻) at 25°C, we can utilize the provided thermochemical data. The key is to apply Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps, regardless of the pathway taken. Let's break down the information given and derive the required value step by step.

Understanding the Given Data

We have the following thermochemical equations:

• ΔHf°(H₂O(l)) = -286 kJ/mol
• ΔHf°(H₂(g)) = 0 kJ/mol (standard state)
• ΔHf°(O₂(g)) = 0 kJ/mol (standard state)
• ΔHf°(H⁺(aq)) = +32 kJ/mol

Setting Up the Reaction

The formation of the hydroxide ion can be represented in terms of the formation of water and the hydrogen ion:

H₂(g) + ½ O₂(g) → H₂O(l) (ΔH = -286 kJ)

Now, we can express the formation of the hydroxide ion as:

H₂O(l) → H⁺(aq) + OH⁻(aq) (ΔH = ?)

Applying Hess's Law

We can rearrange the reactions to find the enthalpy change for the formation of the hydroxide ion. The reaction for the dissociation of water into its ions can be written as:

ΔH = ΔHf°(H⁺(aq)) + ΔHf°(OH⁻(aq)) - ΔHf°(H₂O(l))

Substituting the known values:

ΔH = 32 kJ + ΔHf°(OH⁻(aq)) - (-286 kJ)

Rearranging gives us:

ΔHf°(OH⁻(aq)) = ΔH + 286 kJ - 32 kJ

ΔHf°(OH⁻(aq)) = ΔH + 254 kJ

Finding the Enthalpy Change

From the dissociation of water, we know that the enthalpy change for this reaction is approximately +57.3 kJ (the energy required to break the bonds in water to form ions). Therefore, we can substitute this value into our equation:

ΔHf°(OH⁻(aq)) = 57.3 kJ + 254 kJ

ΔHf°(OH⁻(aq)) = 311.3 kJ/mol

Final Result

Thus, the enthalpy of formation of the hydroxide ion (OH⁻) at 25°C is approximately +311.3 kJ/mol. This value indicates the energy required to form one mole of hydroxide ions from its elements in their standard states.