# Question 557b5

Mar 22, 2015

The equilibrium concentrations of all the species involved in your reaction will be

$\left[{H}_{2}\right] = \text{0.00424 M}$
$\left[{I}_{2}\right] = \text{0.00215 M}$
$\left[H I\right] = \text{0.0264 M}$

First, you need to determine the direction in which the equilibrium will proceed once all the species are placed in the flask. To do this, you must calculate the value of the reaction quotient, ${Q}_{c}$, and compare it to the value of the equilibrium constant, ${K}_{c}$.

${Q}_{c} = \frac{{\left[H I\right]}_{0}^{2}}{{\left[{H}_{2}\right]}_{0} \cdot {\left[{I}_{2}\right]}_{0}} = \text{0.0224"^(2)/("0.00623" * "0.00414") = "19.4}$

Because ${Q}_{c} < {K}_{c}$, the equilibrium will proceed to the right, i.e. more product will be formed. This means that the concentrations of the reactans must decrease, and the concentration of the product must increase.

To determine the equilibrium concentrations you must use the ICE chart method (more here: http://en.wikipedia.org/wiki/RICE_chart).

....${H}_{2 \left(g\right)}$ $\text{ "+" }$ ${I}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s \textcolor{red}{2} H {I}_{\left(g\right)}$
I..0.00623..........0.00414.......0.0224
C.....(-x)...................(-x)..............$\textcolor{red}{+ 2 x}$
E..0.00623-x.....0.00414-x.....0.0224 + 2x

Use the definition of the equilibrium constant to solve for $x$.

${K}_{c} = {\left(0.0244 + 2 x\right)}^{\textcolor{red}{2}} / \left(\left(0.00623 - x\right) \cdot \left(0.00414 - x\right)\right)$

SIDE NOTE I won't show you all the steps in solving this equation because it would make for a very long answer.

You'll eventually get to something like this

$50.3 {x}^{2} - 0.6527 x + 0.001098 = 0$

This will produce two positive values for $x$, but only one of them is acceptable - the one smaller than the concentrations of the reactants, since you can't have a negative equilibrium concentration for the reactants.

Thus, $x = 0.001986$.

The equilibrium concentrations will be

[H_2] = "0.00623 M" - "0.001986 M" = color(red)("0.00424 M")
$\left[{I}_{2}\right] = \text{0.00414 M" - "0.001986 M" = color(red)("0.00215 M}$
[HI] = "0.0224 M" - "0.001986 M" = color(red)("0.0264 M")#