# How can pressure build up in an equimolar gaseous reaction? [closed]

To make the question concrete, if you conduct a reaction like $\ce{H2 + Cl2 -> 2HCl}$ in a stoppered test tube, the stopper will aggressively pop off (or the glass will shatter) despite the fact that there shouldn't be any moles building up in the vessel because the reaction is equimolar. I'm sure there is an easy explanation for this but for whatever reason I can't think of it.

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• Ever heard about temperature? – Ivan Neretin Nov 13 '17 at 11:25
• @Ivan Almost every comment a bon mot. :D – Martin - マーチン Nov 13 '17 at 12:46

If we consider an idealised gas behaviour and the gas state equation $$pV = nRT$$

where

• $p$ is pressure

• $n$ is molar amount

• $R$ is universal gas constant

• $T$ is absolute temperature

• $V$ is gas volume.

..then it should be clear.

As the reaction is highly exothermic, temperature and therefore pressure significantly increase, as $$p= \left( \frac {nR}{V} \right) T$$

• en.wikipedia.org/wiki/Gas_laws#Gay-Lussac.27s_law – paracetamol Nov 13 '17 at 12:32
• Yes. But the ideal gas state equation covers all 3 basic gas laws ( for isothermic, isobaric and isochoric changes ). – Poutnik Nov 13 '17 at 12:34
• Yes, I just pointed out something more succinct ;) – paracetamol Nov 13 '17 at 12:43
• You are right. My approach elegantly avoids my issues with names. – Poutnik Nov 13 '17 at 12:55