Chemistry - Thermochemistry Question with Solution | TestHub
Calculate the enthalpy of formation (in kcal/mol) of HI(g) from the following data:
H2(g) + Cl2(g) → 2HCl(g); ΔH = -44.20 kcal
HCl(g) + aq → HCl(aq); ΔH = -17.31 kcal
HI(g) + aq → HI(aq); ΔH = -19.21 kcal
KOH(aq) + HCl(aq) → KCl(aq); ΔH = -13.74 kcal
KOH(aq) + HI(aq) → KI(aq); ΔH = -13.67 kcal
Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s); ΔH = -52.42 kcal
Options:
Answer:
Solution:
The target reaction is:
H2(g) + I2(s) tarrow 2HI(g)
To determine the enthalpy change for this reaction, we can sum the enthalpy changes of a series of known reactions:
1. H2(g) + Cl2(g) tarrow 2HCl(g) (Δ H = -44.20 kcal)
2. 2HCl(g) tarrow 2HCl(aq) (Δ H = 2 × -17.31 = -34.62 kcal)
3. 2HCl(aq) + 2KOH(aq) tarrow 2KCl(aq) + 2H2O(l) (Δ H = 2 × -13.74 = -27.48 kcal)
4. 2KCl(aq) + I2(s) tarrow Cl2(g) + 2KI(aq) (Δ H = +52.42 kcal)
5. 2KI(aq) + 2H2O(l) tarrow 2KOH(aq) + 2HI(aq) (Δ H = +27.34 kcal)
6. 2HI(aq) tarrow 2HI(g) (Δ H = +38.42 kcal)
Summing these enthalpy changes gives the total enthalpy change for the formation of 2 moles of HI:
Δ Htotal = -44.20 - 34.62 - 27.48 + 52.42 + 27.34 + 38.42 = +11.88 kcal
This value is for 2 moles of HI. To find the enthalpy change per mole of HI, we divide by 2:
Δ Hper mole = (+11.88 kcal)/2 mol = +5.94 kcal/mol
This corresponds to option (c).
