# Question d1102

Mar 30, 2017

${K}_{\textrm{c}}$ is LARGE.

#### Explanation:

The fact that ${\text{P}}_{4}$ spontaneously bursts into smoke (${\text{P"_4"O}}_{10}$) and releases a large amount of heat suggests three things:

1. ΔH is large and negative (highly exothermic).
2. ΔG is negative (a requirement for spontaneity).
3. ΔH must be so large that it overwhelms the unfavourable entropy change.

Let's assume some reasonable values for the formation of ${\text{P"_4"O}}_{10}$ and calculate an approximate value for ${K}_{\textrm{c}}$.

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#

${K}_{\textrm{c}} = {e}^{1200}$

That’s such a big number that my calculator gives an error message when I try to calculate it!