# Question #41629

Dec 4, 2017

The scope of this problem is beyond molarity calculations, I've included them in the ICE table below. Although to effectively solve these problems you must be meticulous with them: if you make a trivial mistake here you get the wrong answer to a complex problem.

${H}_{2} \left(g\right) + {I}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 H I \left(g\right)$

$2 x = 0.188 M \implies x \approx 0.0942 M$

Now,

${K}_{c} = \frac{{\left[H I\right]}^{2}}{\left[{H}_{2}\right] \left[{I}_{2}\right]} = \frac{{0.188}^{2}}{{0.0078}^{2}} \approx 581$

is the equilibrium constant under these conditions.