Finding equilibrium constant from solubility product constant


> 
> Given
>  $$\ce{CuI(s) <=> Cu+(aq) + I-(aq)}\, ;\qquad K\_\mathrm{sp} = 1.2 \cdot 10^{-12}$$
>  write a reasonable chemical reaction that describes the decomposition of $\ce{CuI2}$ in aqueous solution, and show that this is a spontaneous reaction under standard conditions.
> 
> 
> 


The answer found the Gibbs free energy change of the precipitation reaction (backward reaction of the one shown above) using the formula $\Delta G = -RT\ln{K\_\mathrm{eq}}$ and substituted the value $1/K\_\mathrm{sp}$ for $K\_\mathrm{eq}$.


I do not understand why the inverse of the solubility constant can be used as the equilibrium constant of the backwards reaction.


Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?





> 
> Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?
> 
> 
> 


Yes. The general rule actually is: For a given chemical reaction with equilibrium constant $K$:


$$\ce{aA\+bB->cC\+dD}$$


If we multiply the chemical reaction by $n$, we get $K\_1$ for the following equilibrium:


$$\ce{\mathit{na} A\+\mathit{nb} B->\mathit{nc} C\+\mathit{nd} D}$$


as $K\_1=K^n$.


Reversing the reaction is equivalent to taking $n=-1$, in which case we get: $K\_2=\frac{1}{K}$




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You must be wondering that:



> 
> Why does the new equilibrium constant $K'=K^n$?
> 
> 
> 


The answer lies in a simple observation on the Law of Mass Action. Recall that for a reaction: $\ce{A +B —> C + D}$, $K\_\mathrm c = \frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}$


Now, notice that $n\ce{A}$ can be written as $\ce{A}+\ce{A}+...+\ce{A}$ ($n$ times) Same applies on $n\ce{B}, n\ce{C}$ and $n\ce{D}$.


So, if we multiply our reaction with $n$, it actually can be rewritten as:


$$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—>}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$$
and, thus, its new $K\_\mathrm{c}'$ will be
$$K\_\mathrm c' = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^n.[\ce{D}]^n}{[\ce{A}]^n.[\ce{B}]^n}=\left(\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}\right)^n=K\_\mathrm{c}^n$$


Hence, proved.




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**PS:** The previous answer proved the same result, but incorrectly. That proof has now been corrected thanks to the comment by @DavePhD below.


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