Chemistry - Thermochemistry Question with Solution | TestHub
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) + 1/2 O2(g) → H2O(l); ΔH = -285.0 kJ
(viii) 1/2H2(g) + 1/2 Cl2(g) → HCl(g); ΔH = -92.0 kJ
(ix) HCl(g) + aq. → HCl(aq); ΔH = -75.0 kJ
Options:
Answer:
Solution:
The target reaction is:
I2(s) + 5/2O2(g) tarrow I2O5(s)
This equation is exactly reproduced by the combination 2(ii) - (i) - 6(iii) - 6(iv) + 6(v) - (vi) + 5(vii) - 10(viii) - 10(ix).
To verify, let's check the coefficients for I2O5: -c1 = 1. All other species, such as HIO3, KI, KCl, HClO, NaOCl, NaCl, NaOH, HCl, Cl2, H2, and H2O, cancel out.
The enthalpy change (Δ H) for this reaction is calculated as follows:
Δ H = 2(-322) - (4) - 6(-44) - 6(-57) + 6(-100) - (-224) + 5(-285) - 10(-92) - 10(-75)
Δ H = -644 - 4 + 264 + 342 - 600 + 224 - 1425 + 920 + 750
Δ H = -173 kJ
This corresponds to option (a).
