# Question 072cf

Apr 13, 2016

Here's what I got.

#### Explanation:

All you have to do here is make sure that you write the correct expression for the equilibrium constant, ${K}_{c}$, associated with this equilibrium reaction.

The first thing to do here is write the balanced chemical equation

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

Notice that one mole of hydrogen gas, ${\text{H}}_{2}$, reacts with one mole of iodine gas, ${\text{I}}_{2}$, to form $\textcolor{red}{2}$ moles of hydrogen iodide, $\text{HI}$.

By definition, the equilibrium constant for this reaction, which uses equilibrium concentrations raised to the power of their associated stoichiometric coefficients, looks like this

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

The problem tells you that at ${430}^{\circ} \text{C}$, the equilibrium constant is equal to $54.3$. Since you have

${K}_{c} > 1$

you can say that the equilibrium will lie mostly to the right, i.e. the forward reaction will be favored.

You can thus expect the equilibrium concentration of hydrogen iodide to be significantly higher than the equilibrium concentrations of the two reactants.

Now, the problem also provides initial concentrations for the three chemical species. In order to determine in which direction will the equilibrium shift once the reaction starts, you need to calculate the reaction quotient, ${Q}_{c}$.

The reaction quotient takes the same form as the equilibrium constant ${K}_{c}$

${Q}_{c} = \left({\left[{\text{HI"]_0^color(red)(2))/(["H"_2]_0 * ["I}}_{2}\right]}_{0}\right)$

with the important difference being that ${Q}_{c}$ uses the concentrations of the three chemical species at any given moment in reaction.

In your case, this moment will be right before the reaction starts and ${Q}_{c}$ will use initial concentrations. By calculating ${Q}_{c}$ and comparing its value to that of ${K}_{c}$, you can determine if the equilibrium will shift to the left or to the right.

So, plug in the values given to you for the initial concentrations of the three chemical species to get

${Q}_{c} = \left({0.0424}^{\textcolor{red}{2}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"^color(red)(2)))))/(0.00623 color(red)(cancel(color(black)("M"))) * 0.00414color(red)(cancel(color(black)("M}}}}\right) = 69.7$

Since you have ${Q}_{c} > {K}_{c}$, you know that system has gone past its equilibrium point. As a result, a shift to the left will occur, i.e. the reverse reaction will be favored.

Use an ICE table to find the equilibrium concentrations - remember, the equilibrium shifts to the left, which implies that hydrogen iodide will be consumed and hydrogen gas and iodine gas will be produced

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

color(purple)("I")color(white)(aaacolor(black)(0.00623)aaaaacolor(black)(0.00414)aaaaaacolor(black)(0.0424)
color(purple)("C")color(white)(aaacolor(black)((+x))aaaaaacolor(black)((+x))aaaaaacolor(black)((-color(red)(2)x))
color(purple)("E")color(white)(acolor(black)(0.00623+x)aacolor(black)(0.00414+x)aacolor(black)(0.0424-color(red)(2)x)

The equilibrium constant will be equal to

${K}_{c} = \frac{{\left(0.0424 - \textcolor{red}{2} x\right)}^{\textcolor{red}{2}}}{\left(0.00623 + x\right) \cdot \left(0.00414 + x\right)}$

SIDE NOTE This is equivalent to saying that

$2 {\text{HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I}}_{2 \left(g\right)}$

with the equilibrium constant being

${K}_{\text{c reverse}} = \frac{1}{K} _ c$

To keep things simple, I used the forward reaction as the base for the ICE table.

END OF SIDE NOTE

This will be equivalent to

54.3 = (4x^2 - 0.0848x + 0.001798)/((x^2 + 0.01037x + 0.0000258)

Rearrange to get

$50.3 {x}^{2} + 0.6479 x - 0.000397 = 0$

This quadratic equation will produce two values for $x$, but only one will be positive. Since the concentration of hydrogen gas and iodine gas must increase and that of hydrogen iodide must decrease, pick

$x = 0.000586$

The equilibrium concentrations for the three chemical species will be

["HI"] = 0.0424 - color(red)(2) * 0.00568 = color(green)(|bar(ul(color(white)(a/a)"0.0412 M"color(white)(a/a)|)))

["H"_2] = 0.00623 + 0.000586 = color(green)(|bar(ul(color(white)(a/a)"0.00682 M"color(white)(a/a)|)))

["I"_2] = 0.00414 + 0.000586 = color(green)(|bar(ul(color(white)(a/a)"0.00473 M"color(white)(a/a)|)))#

I'll leave the answers rounded to three sig figs.

Notice that despite the fact that the concentration of hydrogen iodide decreased and the concentrations of hydrogen gas and iodine gas increased, the value of ${K}_{c}$ still determines the difference in magnitude between the equilibrium concentrations of the three chemical species.