Chemistry - Thermochemistry Question with Solution | TestHub
Calculate the enthalpy of formation (in kcal/mol) of gaseous HCl using following data:
Substance | NH3(g) | HCl(g) | NH4Cl(s)
Heat of formation | -11 | X | -75
Heat of solution | -8.5 | -17.5 | +3.9 kcal
and, NH3(aq) + HCl(aq) → NH4Cl(aq); ΔH = -12 kcal
Options:
Answer:
Solution:
To determine the enthalpy change for the reaction NH3(g) + HCl(g) tarrow NH4Cl(s), we can consider a two-step process:
1. **Solution Phase:**
* Dissolving NH3(g): Δ H = -8.5 kcal
* Dissolving HCl(g): Δ H = -17.5 kcal
* Reaction in solution: Δ H = -12 kcal
* Crystallization of NH4Cl(aq) to NH4Cl(s): Δ H = -3.9 kcal (This is the reverse of its heat of solution).
The sum of these enthalpy changes is:
Δ Htotal = -8.5 - 17.5 - 12 - 3.9 = -41.9 kcal
2. **Direct Calculation using Heats of Formation:**
The enthalpy change for the reaction can also be calculated using the standard heats of formation (Δ Hf) of the reactants and products:
Δ H = Δ Hf(NH4Cl(s)) - Δ Hf(NH3(g)) - Δ Hf(HCl(g))
Given Δ Hf(NH4Cl(s)) = -75 kcal/mol and Δ Hf(NH3(g)) = -11 kcal/mol, let Δ Hf(HCl(g)) = X.
So, the equation becomes:
Δ H = -75 - (-11) - X = -75 + 11 - X = -64 - X
Equating the two calculated values for Δ H:
-64 - X = -41.9
X = -64 + 41.9
X = -22.1 kcal/mol
Therefore, the heat of formation of HCl(g) is -22.1 kcal/mol, which corresponds to option (b).
