Question

Find ∆H◦ for the reaction P4(s) + 2O2(g) + 6Cl2(g) 4POCl3(l). ∆H◦ = −2168.8kJ for the reaction: P4(s) + 2O2(g) + 6Cl2(g) 4POCl3(g) The vaporization of POCl3, POCl3(l) POCl3(g), ∆H◦vap = 38.6kJmol−1. Note that POCl3 is a liquid in the first reaction and a gas in the second.

Answer 1

Solution:

P₄(s) + 2O₂(g) + 6Cl2(g)-------> 4POCl₃(l); ∆H^◦ =?

Given: P4(s) + 2O₂(g) + 6Cl2(g) --------> 4POCl₃(g); ∆H₁^◦ = −2168.8 kJ

POCl₃(l)------> POCl₃(g), ∆H^◦vap = 38.6 kJmol⁻¹

Therefore molar enthalpy of condensation will be as follows:

POCl₃(g) ------> POCl₃(l), ∆H₂^◦ = - 38.6 kJmol⁻¹

4POCl₃(g) ------> 4POCl₃(l), ∆H₂^◦ = - 4x38.6 kJ = - 154.4 kJ

Now the overall reaction can be rewritten as:

P4(s) + 2O₂(g) + 6Cl2(g) --------> 4POCl₃(g) --------> 4POCl₃(l); ∆H^◦=∆H₁^◦ + ∆H₂^◦ = −2168.8-154.4 kJ

; ∆H^◦ = - 2323.2 kJ