Chemistry - Thermochemistry Question with Solution | TestHub
Carefully observe the given diagram which indicates standard enthalpy of formation of different states of one mole of Mg and two moles of Cl atoms and match the entries in Column I and II provided.
Column I | Column II |
|---|---|
P) Δf H [Mg2+(aq)] | i) -1900 kJ/mol |
Q) Δf H [Cl-(aq)] | ii) +460 kJ/mol |
R) ΔHHydration [Mg2+(g)] | iii) -625 kJ/mol |
S) Lattice enthalpy of MgCl2(s) | iv) +2510 kJ/mol |
(Column I = Quantity; Column II = Value)
Options:
Answer:
Solution:
To determine the enthalpy changes, read the level differences from the provided diagram.
For S: The lattice enthalpy is calculated as the energy difference between gaseous ions and the solid compound:
Lattice enthalpy = [Mg2+(g) + 2Cl-(g)] - [MgCl2(s)]
= 1870 - (-640) = +2510 kJ/mol
This corresponds to S-iv.
For Q: The enthalpy change from [Mg2+(g) + Cl2(g)] to [Mg2+(g) + 2Cl-(aq)] is:
Δ H = 1110 - 2360 = -1250 kJ
This value is for 2 moles of Cl-(aq). Therefore, the standard enthalpy of formation for Cl-(aq) is:
Δf H[Cl-(aq)] = (-1250)/2 = -625 kJ/mol
This corresponds to Q-iii.
For R: The hydration enthalpy of Mg2+ is calculated as:
Hydration of Mg2+ = [Mg2+(aq) + 2Cl-(aq)] - [Mg2+(g) + 2Cl-(aq)]
= -790 - 1110 = -1900 kJ/mol
This corresponds to R-i.
For P: The standard enthalpy of formation for Mg2+(aq) is:
Δf H[Mg2+(aq)] = -790 - 2(-625) = +460 kJ/mol
This corresponds to P-ii.
The final answer is: P-ii; Q-iii; R-i; S-iv.
