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If in the following reaction

$$\ce{H2 + I2 <=> 2HI}$$

the initial concentrations of $\ce{H2}$ and $\ce{I2}$ are $\pu{1 M}$ and $\pu{5 M}$, respectively, and at equilibrium $x$ moles of each have combined. Then what will be the $K_\mathrm{c}$ for the reaction?

What I think is that the $K_\mathrm{c}$ will not change because it does not depend on the initial concentrations of reactants and depends only on the stoichiometric equation, but the one thing I can't reason out is that those extra $4$ moles of $\ce{I2}$ should also have an affect on its equilibrium concentration (since those $4$ moles are still present in the vessel) and thus affect $K_\mathrm{c}$.

Please help me find the correct explanation for the aforesaid question.

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  • $\begingroup$ can you just clear up what you mean by "x moles of each have combined" do you mean the total moles at equilibrium is x or at equilibrium there is x moles of each species. $\endgroup$ – H.Linkhorn Dec 17 '18 at 10:54
  • $\begingroup$ Lookup the definition of equilibrium constant and fill in the equilibrium concentrations. Then you'll get the value of the equilibrium constant. $\endgroup$ – aventurin Dec 17 '18 at 19:31
  • $\begingroup$ Just how the very word "change" came about? To speak of a change, you must have some "before" and some "after". Here you don't have any "before". $\endgroup$ – Ivan Neretin Dec 17 '18 at 20:22
  • $\begingroup$ If $x$ moles have combined, then there must be $2x$ moles of $\ce{HI}$, $1-x$ moles of $\ce{H2}$ and $5-x$ moles of $\ce{I2}$, so therefore $K_c$ is ... $\endgroup$ – Curt F. Dec 18 '18 at 1:25
  • $\begingroup$ @IvanNeretin By "change" I mean the deviation from the ideal condition where moles of $\ce{H2}$ and $\ce{I2}$ are given according to the balanced equation ( 1 each in this case). $\endgroup$ – Shashwat Saxena Dec 18 '18 at 5:54

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