Question
Calculate the enthalpy of formation of I2O5(s) from the following data.
(i) I2O5(s) + H2O(l) → 2HIO3(aq): ΔH = +4.0 kJ
(ii) KI(aq) + 3HClO(aq) → HIO3(aq) + 2HCl(aq) + KCl(aq): ΔH = −322.0 kJ
(iii) NaOH(aq) + HClO(aq) → NaOCl(aq) + H2O(l): ΔH = −44.0 kJ
(iv) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l): ΔH = −57.0 kJ
(v) 2NaOH(aq) + Cl2(g) → NaOCl(aq) + NaCl(aq) + H2O(l): ΔH = −100.0 kJ
(vi) 2KI(aq) + Cl2(g) → 2KCl(aq) + I2(s): ΔH = −224.0 kJ
(vii) H2(g) + ½ O2(g) → H2O(l): ΔH = −285.0 kJ
(viii) ½ H2(g) + ½ Cl2(g) → HCl(g): ΔH = −92.0 kJ
(ix) HCl(g) + aq → HCl(aq): ΔH = −75.0 kJ
Options
Solution
The required thermochemical equation is
$I_{2}(s) + \frac{5}{2}O_{2}(g) \rightarrow I_{2}O_{5}(s)$
From 2 × (ii) + 6 × (v) + 5 × (vii) – (i) – 6 × (iii) – 6 × (iv) – (vi) – 10 × (viii) – 10 × (ix), we get,
$\Delta H_{\text{Required}} = 2( - 322) + 6( - 100) + 5( - 255) - (4.0) - 6( - 44) - 6( - 57) - ( - 224) - 10( - 92) - 10( - 75) $$$ = - 169\text{ kJ}$$
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