Chemistry - Thermochemistry Question with Solution | TestHub
Calculate proton affinity of NH3(g) from the following data:
ΔHdissociation H2 = 218 kJ mole-1
ΔHdissociation Cl2 = 124 kJ mole-1
ΔHf0 of NH3(g) = -46 kJ mole-1
ΔHf0 of NH4Cl(s) = -314 kJ mole-1
Ionization energy of H = 1310 kJ mole-1, E.A. of Cl(g) = -348 kJ mole-1
Lattice energy of NH4Cl(s) = -683 kJ mole-1
Options:
Answer:
Solution:
The proton affinity (PA) is defined as the enthalpy change (Δ H) for the reaction:
NH3(g) + H+(g) tarrow NH4+(g)
To determine the PA, we can use a thermochemical cycle involving the formation of ammonium chloride (NH4Cl(s)). The overall reaction for the formation of NH4Cl(s) from NH3(g), H2(g), and Cl2(g) is:
NH3(g) + 1/2H2(g) + 1/2Cl2(g) tarrow NH4Cl(s)
The enthalpy change for this reaction is Δ H = -314 - (-46) = -268 kJ.
Now, let's break down the cycle into individual steps and their corresponding enthalpy changes:
1. Atomization of 1/2H2: 218/2 = 109 kJ
2. Atomization of 1/2Cl2: 124/2 = 62 kJ
3. Ionization of H (formation of H+): +1310 kJ
4. Electron affinity of Cl (formation of Cl-): -348 kJ
5. Proton affinity (PA) of NH3: PA
6. Lattice energy for NH4Cl(s): -683 kJ
According to Hess's Law, the total enthalpy change for the overall reaction is the sum of the enthalpy changes for these individual steps:
-268 = 109 + 62 + 1310 - 348 + PA - 683
-268 = 450 + PA
Solving for PA:
PA = -268 - 450 = -718 kJ mol-1
This corresponds to option (b).
