The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H ions;that is, delta H(f) [H+(aq)]=0

A.for this reaction: calculate delta H(f) for the Cl- ions.
HCl(g) ==H2O==>H+(aq)+Cl-(aq)
delta H= -74.9 kJ/mol
(answer has to be in kJ/mol)

B.given that delta H(f) for OH- ions is -229.6 kJ/mol, calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25^C.
(answer has to be in kJ/mol)

A. To calculate the standard enthalpy of formation (ΔHf) for Cl- ions, we need to use the enthalpy change for the reaction and the known values for the enthalpy of formation of other species.

Given:
ΔH = -74.9 kJ/mol

We can break down the reaction into its constituent steps and use the known ΔHf values for those species:

1. HCl(g) → H+(aq) + Cl-(g) ΔH1 = ΔHf(H+(aq)) + ΔHf(Cl-(g))
2. HCl(g) + H2O(l) → H3O+(aq) + Cl-(g) ΔH2 = ΔHf(H3O+(aq)) + ΔHf(Cl-(g))

Since ΔH1 + ΔH2 = ΔH, we can rearrange the equation to solve for ΔHf(Cl-(g)):

ΔHf(Cl-(g)) = ΔH - ΔHf(H+(aq)) - ΔHf(H3O+(aq))

Given that ΔHf(H+(aq)) = 0 (from the given statement) and ΔHf(H3O+(aq)) ≈ 0 (since H3O+ is essentially the same as H+), we can simplify the equation:

ΔHf(Cl-(g)) = ΔH - 0 - 0
ΔHf(Cl-(g)) = ΔH = -74.9 kJ/mol

Therefore, the standard enthalpy of formation of Cl- ions is -74.9 kJ/mol.

B. The enthalpy of neutralization (ΔHneutralization) is defined as the enthalpy change when 1 mole of a strong monoprotic acid (like HCl) reacts with 1 mole of a strong base (like KOH) to form 1 mole of water at 25°C.

Given:
ΔHf(OH-(aq)) = -229.6 kJ/mol

The neutralization reaction can be represented as follows:
HCl(aq) + KOH(aq) → H2O(l) + KCl(aq)

Since the enthalpy of formation of ions in aqueous solutions is defined relative to the enthalpy of formation of H+(aq) (which is assigned a value of zero), the ΔHf values for H+, Cl-, K+, and OH- ions will be zero. Therefore, their enthalpies of formation do not contribute to the enthalpy of neutralization.

The overall ΔHneutralization will be equal to the enthalpy change observed during the reaction, which is given as -74.9 kJ/mol. Hence:

ΔHneutralization = -74.9 kJ/mol

So, the enthalpy of neutralization when 1 mole of HCl is titrated with 1 mole of KOH at 25°C is -74.9 kJ/mol.

A. To calculate the enthalpy of formation (ΔHf) for Cl- ions, we can use the enthalpy change (ΔH) of the given reaction and the enthalpies of formation for the other species involved in the reaction.

The enthalpy change of the provided reaction is ΔH = -74.9 kJ/mol, which represents the overall enthalpy change when 1 mole of HCl(g) reacts with H2O to form 1 mole of H+(aq) and 1 mole of Cl-(aq).

In this reaction, H2O is not a product or a reactant, so we don't need its enthalpy of formation. We are given that ΔHf[H+(aq)] = 0, as H ions were arbitrarily assigned a value of zero. Therefore, the enthalpy change (ΔH) only includes the ΔHf for Cl- ions.

The enthalpy change (ΔH) can be written as ΔH = ΣΔHf(products) - ΣΔHf(reactants), where the Σ symbol indicates the sum over all species.

Since H+ has a ΔHf of zero, the ΔHf(Cl-) = ΔH can be calculated as follows:
ΔHf(Cl-) = ΔH - ΔHf(H+) = -74.9 kJ/mol - 0 kJ/mol = -74.9 kJ/mol

Therefore, the enthalpy of formation for Cl- ions (ΔHf[Cl-]) is -74.9 kJ/mol.

B. The enthalpy of neutralization when 1 mole of a strong monoprotic acid (HCl) is titrated by 1 mole of a strong base (KOH) can be calculated using the enthalpies of formation for the ions involved.

In the neutralization reaction, H+(aq) reacts with OH-(aq) to form H2O(l). Since H2O is a product, we don't need its enthalpy of formation. We are given that ΔHf[OH-(aq)] = -229.6 kJ/mol.

The overall enthalpy change (ΔH) for the neutralization reaction can be written as:
ΔH = ΣΔHf(products) - ΣΔHf(reactants)

Here, the products are H2O(l) and the reactants are H+(aq) and OH-(aq).

Since 1 mole of HCl is reacting with 1 mole of KOH, which produces 1 mole of H2O, the enthalpy change (ΔH) can be calculated as follows:
ΔH = Σ(ΔHf[H2O]) - Σ(ΔHf[H+] + ΔHf[OH-])
ΔH = 0 kJ/mol - (0 kJ/mol + 0 kJ/mol) (ΣΔHf[H2O] is zero since it has a ΔHf of zero)

Therefore, the enthalpy of neutralization is:
ΔH = 0 kJ/mol - (0 kJ/mol + 0 kJ/mol) = 0 kJ/mol

Hence, the enthalpy of neutralization when 1 mole of a strong monoprotic acid is titrated by 1 mole of a strong base at 25°C is 0 kJ/mol.

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