In a dry powder fire extinguisher, there is a chemical known as $\ce{KHCO3}$ and I was wondering how $\ce{KHCO3}$ works in a chemical equation to extinguish the fire.

  • $\begingroup$ $$\ce{KHCO3 -> KOH + CO2}$$ $\endgroup$ – Zhe Jan 20 '17 at 2:54
  • $\begingroup$ Please also see this answer which contains a subsection and a reference for how $\ce{NaHCO3}$ works. It is more that just the decomposition yielding $\ce{CO2}$; surpressing chain reactions tying up radicals is also key here (and heat of vaporisation and heat capacity). $\endgroup$ – Linear Christmas Jan 20 '17 at 13:00
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/42130/… $\endgroup$ – Nilay Ghosh Jan 20 '17 at 14:48

According to this Wikipedia entry:

Decomposition of the bicarbonate occurs between 100 and 120 °C (212 and 248 °F):

$$\ce{2 KHCO3 → K2CO3 + CO2 + H2O}, \Delta H>0$$

So, you produce a nonflammable powder and two nonflammable gases, and you absorb heat from the fire in an endothermic reaction.

The picture here is more complex than just a chemical equation however. The $\ce{KHCO3}$ also acts as an energy-absorbing material and provides a solid surface on which free radicals can be destroyed.

| improve this answer | |
  • $\begingroup$ Tried \DeltaH and it came out bad. What was wrong? $\endgroup$ – Oscar Lanzi May 21 '17 at 9:58
  • 1
    $\begingroup$ Thx @OscarLanzi . $\endgroup$ – airhuff May 21 '17 at 18:59
  • 1
    $\begingroup$ Thx @NilayGhosh. $\endgroup$ – airhuff May 21 '17 at 19:00

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