# 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?

## 1 Answer

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: $K_2=\frac{1}{K}$

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.

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

• Thank you. In regards to your bonus question, is my following logic correct: Gibbs free energy is extensive. so one of the factors on the RH side must relate to the amount of substance. R is a constant, while T depends on the average kinetic energy of particles. Therefore K must change depending on amount of substance, hence verifying your rule? – George Tian Jan 4 '18 at 8:16
• @GeorgeTian Yes, if you multiply $n$ on both sides, you'll get $n\Delta G^\circ=-nRT\ln(K)$ which is equivalent to $\Delta G'^\circ=-RT\ln(K^n)$ by properties of logarithm. Thus, for the new reaction, we have $\Delta G'^\circ=-RT\ln(K')$ with $K'=K^n$. Verified. – Gaurang Tandon Jan 4 '18 at 8:34