My book gives the following reaction: $$\ce{SiCl4 + H2O -> HCl + Si(OH)4}$$ Since equilibrium favours the formation of weaker acids. How is this reaction able to proceed if a strong acid like HCl is being formed? Am I missing some point here? Or is my interpretation or equilibrium in this case is faulty?
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3Rather than strong versus weak acid, it is affinity Si-Cl versus Si-O. – Poutnik Dec 01 '20 at 07:34
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1Related: here, here, here, here. – Nilay Ghosh Dec 01 '20 at 07:55
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"Since equilibrium favours the formation of weaker acids." Not a very fundamental rule. Also SiCl4 is a strong Lewis acid. – Karl Dec 01 '20 at 08:10
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1Thanks for the replies, so what i am to basically look for is the overall stability of the components and the favourability of one substance over another in a thermodynamical sense – Singularity Dec 01 '20 at 08:48
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1A good exercise would be to find out enthalpy/free energy of this reaction (using standard enthalpy of formation of these compounds) and you'll most likely notice that it's negative – Aniruddha Deb Dec 01 '20 at 09:47
1 Answers
As this video clearly confirms, the hydrolysis of SiCl4 is highly exothermic from breaking of chlorine bonds. The solution boils and solid insoluble Si(OH)4 is produced.
Also, per Wikipedia on the hydrolysis, to quote:
The reaction can be noticed on exposure of the liquid to air, the vapour produces fumes as it reacts with moisture to give a cloud-like aerosol of hydrochloric acid.[5]
As such, the balanced reaction is likely best expressed as:
$\ce{ SiCl4 (l) + 4 H2O (l) ⟶ 4 HCl (g) + Si(OH)4 (s)}$
where the gaseous and solid precipitates formed can push the reaction to the right.
There is an interesting comment in Atomistry, to quote:
Consider, for example, the two reactions:
$\ce{Ca(OH)2 + 2HCl → CaCl2 + 2H2O}$
$\ce{Si(OH)4 + 4HCl ← SiCl4 +4H2O}$
The difference between the metal and the non-metal is signalised by the direction of the reaction. This difference, however, is seldom absolute, and with elements of intermediate character it disappears. In the case of stannic tin, for example, the reaction is reversible:
$\ce{Sn(OH)4 + 4HCl ⇔ SnCl4 + 4H2O;}$
and stannic chloride can be prepared either by dissolving stannic hydroxide in hydrochloric acid, or by passing chlorine over heated tin or a mixture of stannic oxide and carbon.
So, part of the answer also relates to the non-metal characteristics of Silicon as well.
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