CHEM1901/3
2003-J-2
June 2003
 Use the thermochemical data provided to calculate the heat of reaction of the
following reaction:
Marks
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PCl3(l) + Cl2(g)  PCl5(s)
Data:
P4(s) + 6Cl2(g)  4PCl3(l)
H = –1280 kJ mol–1
P4(s) + 10Cl2(g)  4PCl5(s)
H = –1774 kJ mol–1
The reactions given in the data correspond to the enthalpy of formation of PCl 3(l)
and PCl5(s). The reactions as written correspond to the formation of 4 moles of
PCl3(l) and of 4 moles of PCl5(s).
The fHo values for these two substances are therefore:
fHo (PCl3(l)) = (-1280 × ¼) kJ mol-1
fHo (PCl5(l)) = (-1774 × ¼) kJ mol-1
As ΔrH° = Σ mΔfH°(products) - Σ mΔfH°(reactants) and fHo (Cl2(g)) = 0,
ΔrH° = ((-1774 × ¼) – ((-1280 × ¼)) kJ mol-1 = -124 kJ mol-1
ANSWER: 124 kJ mol1
 Identify one property of a molecule necessary for it to be considered a “greenhouse
gas”.
To be a greenhouse gas, a molecule must be able to absorb infrared light: it
must undergo a vibration which causes a change in the dipole moment.
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