ThermochemistryHard

Question

Study the following thermodynamic data given by E. H. P. Cordfunke, A. S. Booji and M. Y. Furkalionk.

(i) DyCl3(s) → DyCl3 (aq., in 4.0 M-HCl): ΔH0 = −180.06 kJmol−1

(ii) Dy(s) + 3HCl(aq, 4.0 M) → DyCl3 (aq, in 4.0 M-HCl) + 3/2 H2(g): ΔH° = –699.43 kJ mol–1

(iii) ½ H2(g) + ½ Cl2(g) → HCl (aq, 4.0 M): ΔH° = –158.31 kJ mol–1

What is ΔHf0 of DyCl3(s) from these data?

Options

A.−248.58 kJ mol–1
B.−994.30 kJ mol–1
C.−3977.2 kJ mol–1
D.−1469.2 kJ mol–1

Solution

Required thermochemical equation is

$Dy(s) + \frac{3}{2}Cl_{2}(g) \rightarrow DyCl_{3}(s) $$${\text{From }(ii) + 3 \times (iii) - (i),\text{ we get:} }{\Delta H = ( - 699.43) + 3( - 158.31) - ( - 180.06) = - 994.3\text{ kJ/mol}}$$

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