How can you calculate a new Keq when a reaction is reversed? [closed]








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> 
> Consider the following equilibrium:
>  \begin{align}
>  \ce{H2 (g) + I2 (g) &<--> 2HI (g)}& K\_\mathrm{eq} &= 50.0
> \end{align}
>  What is the value $K\_\mathrm{eq}$ for the reaction rewritten as:
>  \begin{align}
>  \ce{2HI (g) &<--> H2 (g) + I2 (g)}& K\_\mathrm{eq} &= \mathbf{?}
> \end{align}
>  Choices
>  \begin{align}
>  \mathrm{A}&:-50.0 &
>  \mathrm{B}&: 0.0200 &
>  \mathrm{C}&: 25.0 &
>  \mathrm{D}&: 50.0
> \end{align}
> 
> 
> 


The answer is $0.0200$, but I'm confused as to why? How can you figure it out?




First let's use a conceptual approach, as this multiple choice problem is answerable in that way.


$K\_\mathrm{eq}$ cannot be negative. Therefore $\mathrm{A}$ is incorrect.


$K\_{\mathrm{eq}}$ is greater than one when the $\ce{left -> right}$ reaction goes farther; it is less than one when the reverse reaction goes farther (by going farther I mean end in greater concentration at equilibrium).  

The 
$$\ce{H2 + I2 <=> 2HI}$$
reaction has $K\_{\mathrm{eq}}$ over $1$. Therefore the 
$$\ce{H2 + I2 -> 2HI}$$ 
happens faster than the 
\begin{align}
&&\ce{H2 + I2 &<- 2HI}\\
\text{or}&&
\ce{2HI &-> H2 + I2}.
\end{align}


So in 
$$\ce{2HI <=> H2 + I2}$$
or the reverse, the $\ce{left->right}$ reaction occurs slower; the $K\_{\mathrm{eq}}$ must be less than $1$. $\mathrm{B}$ is the only answer.


Another way to approach the problem is using algebraic reasoning.


Initially, 
$$K\_{eq} = \frac{\ce{[HI]^2}}{\ce{[H2][I2]}} = 50.$$
Then, when you reverse the reaction, 
$$K\_{eq} = \frac{\ce{[H2][I2]}}{\ce{[HI]^2}} 
\frac{1}{\left(\frac{\ce{[HI]^2}}{\ce{[H2][I2]}}\right)}= \frac{1}{50.0} = 0.0200.$$


In general when you reverse the equation
$$K\_{\text{eq (new)}} = \frac{\text{products (new)}}{\text{reactants (new)}} = \frac{\text{reactants (old)}}{\text{products (old)}} = \frac{1}{K\_{\text{eq (old)}}}.$$


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