Study the following thermodynamic data for yttrium chloride. (i) YCl 3 (s) → YCl 3 (aq., in 4.0 M-HCl); ΔH 0 = -170.20 kJ/mol (ii) Y(s) + 3HCl(aq, 4.0 M) → YCl 3 (aq, in 4.0 M-HCl) + 3/2 H 2 (g); ΔH° = -680.10 kJ/mol (iii) 1/2 H 2 (g) + 1/2 Cl 2 (g) → HCl(aq, 4.0 M); ΔH° = -158.31 kJ/mol What is ΔH f 0 of YCl 3 (s) from these data?

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Chemistry - Thermochemistry Question with Solution | TestHub

ChemistryThermochemistryEnthalpy of solution/dilution/hydrationHard2 minQB

ChemistryHardsingle choice

Study the following thermodynamic data for yttrium chloride.

(i) YCl₃(s) → YCl₃(aq., in 4.0 M-HCl); ΔH⁰ = -170.20 kJ/mol

(ii) Y(s) + 3HCl(aq, 4.0 M) → YCl₃(aq, in 4.0 M-HCl) + 3/2 H₂(g); ΔH° = -680.10 kJ/mol

(iii) 1/2 H₂(g) + 1/2 Cl₂(g) → HCl(aq, 4.0 M); ΔH° = -158.31 kJ/mol

What is ΔH_f⁰ of YCl₃(s) from these data?

Options:

Answer:

Solution:

Target: Y(s) + 3/2 Cl₂(g) → YCl₃(s) = (ii) + 3 × (iii) - (i). Δ_f H = -680.10 + 3(-158.31) - (-170.20) = -680.10 - 474.93 + 170.20 = -984.83 kJ/mol. This is option (d).

Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of solution/dilution/hydration

⏱ 2mℹ️ Source: QB

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