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.
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$\begingroup$ $$\ce{KHCO3 -> KOH + CO2}$$ $\endgroup$– ZheJan 20, 2017 at 2:54
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$\begingroup$ 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 ChristmasJan 20, 2017 at 13:00
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$\begingroup$ Related: chemistry.stackexchange.com/questions/42130/… $\endgroup$– Nilay GhoshJan 20, 2017 at 14:48
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1 Answer
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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.
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$\begingroup$ Tried \DeltaH and it came out bad. What was wrong? $\endgroup$ May 21, 2017 at 9:58
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