The fact that "P"_4 spontaneously bursts into smoke ("P"_4"O"_10) and releases a large amount of heat suggests three things:
- ΔH is large and negative (highly exothermic).
- ΔG is negative (a requirement for spontaneity).
- ΔH must be so large that it overwhelms the unfavourable entropy change.
Let's assume some reasonable values for the formation of "P"_4"O"_10 and calculate an approximate value for K_text(c).
Assume that ΔH^° = "-3000 kJ·mol"^"-1" and ΔS^° = "+200 J·K"^"-1""mol"^"-1"
Then
ΔG^° = ΔH^° - TΔS^° = "-3000 kJ·mol"^"-1" - 298 color(red)(cancel(color(black)("K"))) × "0.200 kJ"·color(red)(cancel(color(black)("K"^"-1")))"mol"^"-1" = "-3000 kJ·mol"^"-1" - "60 kJ·mol"^"-1" ≈ "-3000 kJ·mol"^"-1"
ΔG^° = -RTlnK_text(c)
lnK_text(c) = -(ΔG^°)/(RT) = -("-3000" color(red)(cancel(color(black)("kJ·mol"^"-1"))))/(8.314 × 10^"-3" color(red)(cancel(color(black)("kJ·K"^"-1""mol"^"-1"))) × 298 color(red)(cancel(color(black)("K")))) ≈ 1200
K_text(c) = e^1200
That’s such a big number that my calculator gives an error message when I try to calculate it!
