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$\ce{AgNO3 + NaCl}$ gives $\ce{NaNO3 + AgCl}$ but why it's vice versa is not possible. My question is how to predict if a double displacement reaction will occur naturally or not.

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    $\begingroup$ Strictly speaking, vice versa is possible, it's just way far from completion. In general, one remembers which compounds posses good solubility and which don't. If in doubt, have a look a their solubility products. Note, however, that solubility (and reactivity) may change drastically by introducing third-party component such as complexing agent or another solvent. $\endgroup$ – andselisk Jun 24 at 17:30
  • $\begingroup$ @andselisk couldn't you possibly write an answer based on this? $\endgroup$ – William R. Ebenezer Jun 24 at 17:51
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On a microscopic level, you still have «a forward» reaction described by

$\ce{AgNO3 + NaCl -> NaNO3 + AgCl}$

and a «backward reaction», which may be described by

$\ce{AgCl + NaNO3 -> AgNO3 + NaCl}$.

However, reaching the thermodynamic equilibrium of these two reactions is heavily influenced by the low solubility product of $\ce{AgCl}$ at about $\sqrt{1.1 \times 10^{-7}}\,\mathrm{mol}$ (or $1.9\,\mathrm{mg}/\mathrm{L}$) in pure water at room temperature (reference).

Consequently, you may either replace the «balanced» double arrow about chemical equilibria

$\ce{AgNO3 + NaCl <=> NaNO3 + AgCl}$

by one tilted in favour to one side

$\ce{AgNO3 + NaCl <=>> NaNO3 + AgCl v}$

where $\ce{v}$ is used to highlight the preciptation of solid $\ce{AgCl}$ shifting the equilibrium.

Or write the reaction equation as if there were no microscopic reversibility at all:

$\ce{AgNO3 + NaCl -> NaNO3 + AgCl v}$


Depending on school and what you want to put emphasis on, the microscopic reversibility often is «omitted» (i.e. the double arrow is dropped in favour of the single arrow) for reactions with a $K > 10^4$ or $K < 10^{-4}$. On the other hand, you see in many textbooks

$\ce{2 H2O <=> H3O^+ + OH^-}$

despite a $K_w \approx 10^{-14}$, too.

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