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morpeh [17]
3 years ago
5

The standard enthalpy of formation of Ca2+ (g) (∆H = 1934 kJ/mol)

Chemistry
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

see explanation

Explanation:

The process of ionization to produce cations is endothermic. For formation of Ca⁺² two ionization steps need be illustrated as follows...

1st ionization step: Ca° + 590Kj => Ca⁺ + e⁻

2nd ionization step: Ca⁺ + 1151Kj => Ca⁺² + e⁻

__________________________________-

Net Ionization Rxn: Ca° + 1741Kj => Ca⁺² + 2e⁻

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Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2
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Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

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Answer:

Explanation:

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