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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?

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  • $\begingroup$ Do you know how to write the equation for the equilibrium constant in terms of the concentrations of products and reactants? Doing that might shed some light on what's going here. Hint: notice that 0.02 = 1/50 $\endgroup$ – airhuff Mar 1 '17 at 2:17
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    $\begingroup$ The first Keq equation would be [HI]^2/[H2][I2]= 50. The reaction written backwards is the reciprocal, so would Keq just equal the reciprocal of 50? $\endgroup$ – Kara Mar 1 '17 at 2:43
  • $\begingroup$ Yep, you got it ;) $\endgroup$ – airhuff Mar 1 '17 at 3:11
  • $\begingroup$ For homework like questions, please use the tag homework-and-exercises $\endgroup$ – Aaron John Sabu Mar 1 '17 at 4:05
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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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  • $\begingroup$ It would be easier, more conceptual, I feel to do in terms of gibbs free energy. The log term is exactly negative of initial, indication new K is inverse of original. $\endgroup$ – Black Jack 21 Mar 1 '17 at 4:17

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