How do I calculate the number of molecules in an equilibrium depending on a change in the equilibrium constant? [closed]








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> 
> Consider 1000 molecules of $\ce{A}$ and 1000 molecules of $\ce{B}$ in a eucaryotic cell. The concentration of both will be about $\pu{10^-9 M}$.  
> 
>  If the equilibrium constant $K$ for $\ce{A + B <--> AB}$ is $\pu{10^10}$, then at equilibrium there will be
>  \begin{array}{ccc}
>  \bbox[1ex,lightgreen]{270} 
>  & \bbox[1ex,lightgreen]{270} 
>  & \bbox[1ex,yellow]{730} \\
>  \ce{A} & \ce{B} & \ce{AB}\\
>  \text{molecules}&\text{molecules}&\text{molecules}\\
> \end{array}
>  If the equilibrium constant is a little weaker at $\pu{10^8}$, which represents a loss of $\pu{2.8 kcal/mol}$ of binding energy from the example above, or 2-3 fewer hydrogen bonds, then there will be
>  \begin{array}{ccc}
>  \bbox[1ex,lightgreen]{915} 
>  & \bbox[1ex,lightgreen]{915} 
>  & \bbox[1ex,yellow]{\phantom{0}85} \\
>  \ce{A} & \ce{B} & \ce{AB}\\
>  \text{molecules}&\text{molecules}&\text{molecules}\\
> \end{array}
> 
> 
> 
> 
> ---
> 
> 
> **[Figure 3-44 Molecular Biology of the cell 5/e (© Garland Science 2008)](https://i.stack.imgur.com/rCiF0.jpg)**
> 
> 
> 


I don't understand how in the above figure, the number of molecules is calculated. 




Setting out the problem like this using an ICE table: 
\begin{array}{cccc}
 \text{species} & \ce{A} & \ce{B} & \ce{AB} \\\hline
 \text{I} & 10^{-9} & 10^{-9} & 0 \\
 \text{C} & -x & -x & +x \\
 \text{E} & 10^{-9}-x & 10^{-9}-x & x \\
\end{array}


Where $\text{I}$ is the initial concentration, $\text{E}$ is the final concentration (concentration at equilibrium) and $\text{C}$ is the change in concentration.


This ICE table (Initial, Change, Equilibrium), is a standard method for solving these types of problems. 


The equilibrium constant is defined as 
$$K = \frac{[\ce{AB}]}{[\ce{A}][\ce{B}]},$$ 
where these are the concentrations at equilibrium.


We have our equilibrium concentrations as set out by the problem. Both $\ce{A}$ and $\ce{B}$ start out at a concentration of $10^{-9}M$ with their final concentrations being $\pu{10^-9 M} - x$ where $x$ is some change in the concentration.


Note, it mentions that 1000 molecules corresponds to a concentration of $\pu{10^-9 M}$. Therefore one can see that this corresponds to $10^{12}~\frac{\text{molecules}}{\pu{M}}$.  

($\pu{1000 molecules} = \beta \cdot \pu{10^-9 M}$ where one can see $\beta$ has units $\frac{\text{molecules}}{\pu{mol}}$)


We can substitute into the expression for $K$ and noting that $[\ce{A}]=[\ce{B}]$


$$10^{10} = \frac{x}{(10^{-9}-x)^2}$$


I could not be bothered to work this out by hand, so Mathematica gives two values for $x$ as this is a [quadratic](https://i.stack.imgur.com/qDvZem.png):
\begin{align}
 x\_1 &= \pu{7.3e-10 M}\\
 x\_2 &= \pu{1.4e-9 M}
\end{align}


One can immediately see that $x\_2$ is not a valid solution as it corresponds to a concentration of $\ce{AB}$ higher than 100% conversion. 


Taking $x\_1$ as the solution: $$\pu{7.3e-10 M} \times 10^{12}~\frac{\text{molecules}}{\pu{M}} = \pu{730 molecules}\text{ of }\ce{AB}$$


The same can be repeated for the second case where the equilibrium constant is given as $K=10^8$.


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