Question #d1102

1 Answer
Mar 30, 2017

Answer:

#K_text(c)# is LARGE.

Explanation:

The fact that #"P"_4# spontaneously bursts into smoke (#"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 #"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!