# Question ee888

Jun 25, 2016

Rather than reveal it immediately, I will have the student read through the explanation.

#### Explanation:

To begin with, the problem itself is not accurately stated. ${\text{K}}_{c}$ should be a combination of concentrations, not total moles, so a volume should have been given for the system. Fortunately, since this reaction involves only gases that we may take to be ideal, the equilibrium composition does not depend on volume and we can choose a convenient volume to calculate with. Let us go with $1 {\text{ m}}^{3}$.

We then have:

"K"_c={["HI"]^2}/{["H"_2]["I"_2]}

"K"_c={(0.26" mol/m"^3)^2}/{("X mol/m"^3)("Y mol/m"^3)}

$= \frac{0.26}{\text{(X)(Y)}}$

$\text{X}$ is the number of moles of hydrogen left at equilibrium. This is less than the $0.20$ moles you put in because some of the hydrogen had to react to make the hydrogen iodide. Similarly $\text{Y}$ is the number of moles of iodine remaining which is less than the $0.15$ moles you statted with. Note that all the units cancelled out. You get that with a gas phase reaction where moles are constant, that's why the result is independent of volume. Lucky us.

So if $\text{X}$ is less than $0.20$, what is it? We formed $0.26 \text{ mol"" HI}$ via the reaction whose balanced equation is given as

$\text{H"_2+"I"_2=2" HI}$

Clearly one mole of hydrogen is consumed for every two moles of hydrogen iodide formed. So the remaining amount of hydrogen, in moles, is

$\text{X} = 0.20 - \frac{0.26}{2} = 0.07$

And for the iodine we have:

$\text{Y} = 0.15 - \frac{0.26}{2} = 0.02$

And then

"K"_c={0.26}/{"(X)(Y)"}#
$= \frac{0.26}{\left(0.07\right) \left(0.02\right)}$

Compare that with the listed answers and you find it matches answer C.