# Question 542b5

Jan 21, 2016

Here's what I got.

#### Explanation:

The problem provides you with the number of moles of each gas and with the volume of the reaction vessel, so right from the start you can use this information to calculate the concentration of the three gases.

So, molarity is defined as moles of solute divided by liters of solution, so your $\text{1-L}$ vessel implies that you will have

["H"_2] = "1 mole"/"1 L" = "1 M"

["I"_2] = "2 moles"/"1 L" = "2 M"

["HI"] = "3 moles"/"1 L" = "3 M"

Next, use an ICE table to help you determine the equilibrium concentrations of the three species. Make sure that you write a balanced chemical equation for this equilibrium.

Before doing any calculations, take a look at the value of the equilibrium constant, ${K}_{c}$, which at that temperature is said to be equal to $49.9$.

Since ${K}_{c} > 1$, you know that the reaction will favor the product, i.e. the equilibrium will lie to the right.

This means that you can expect the equilibrium concentrations of the two reactants to decrease compared with their initial values.

So, an ICE table for this equilibrium would look like this

${\text{ ""H"_text(2(g]) " "+" " "I"_text(2(g]) " "rightleftharpoons" " color(red)(2)"HI}}_{\textrm{\left(g\right]}}$

color(purple)("I")" " " "1" " " " " " " " " "2" " " " " " " " "3
color(purple)("C")" "(-x)" " " " " "(-x)" " " " "(+color(red)(2)x)
color(purple)("E")" "1-x" " " " " "2-x" " " " " "3 + color(red)(2)x#

By definition, the equilibrium constant for this reaction will be

${K}_{c} = \left(\left[{\text{HI"]^color(red)(2))/(["H"_2] * ["I}}_{2}\right]\right)$

${K}_{c} = {\left(3 + \textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / \left(\left(1 - x\right) \left(2 - x\right)\right) = 49.9$

Rearrange this to get

$9 + 12 x + 4 {x}^{2} = 49.9 \cdot \left(2 - 3 x + {x}^{2}\right)$

$45.9 {x}^{2} - 161.7 x + 90.8 = 0$

This quadratic equation will produce two positive values for $x$

$\left\{\begin{matrix}\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}_{1} = 2.82}}} \\ {x}_{2} = 0.701 \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}\end{matrix}\right.$

Since equilibrium concentration (or any concentration, for that matter) cannot be negative, the first value of $x$ is eliminated.

This means that the equilibrium concentrations of the three species will be

$\left[\text{H"_2] = 1 - 0.701 = color(green)("0.3 M}\right)$

$\left[\text{I"_2] = 2 - 0.701 = color(green)("1.3 M}\right)$

$\left[\text{HI"] = 3 + 2 * 0.701 = color(green)("4.4 M}\right)$

Now, you should round these off to one sig fig, since that's how many sig figs you have for the initial number of moles of each species, but I'll leave them as-is.

Notice that the prediction is valid - the concentrations of the reactants decreased compared with their initial values.