# Question #072cf

##### 1 Answer

#### Answer:

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**,

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

#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons color(red)(2)"HI"_((g))#

Notice that **one mole** of hydrogen gas, **one mole** of iodine gas, **moles** of hydrogen iodide,

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 = (["HI"]^color(red)(2))/(["H"_2] * ["I"_2])#

The problem tells you that at

#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**,

The reaction quotient takes the same form as the equilibrium constant

#Q_c = (["HI"]_0^color(red)(2))/(["H"_2]_0 * ["I"_2]_0)#

with the **important difference** being that **at any given moment** in reaction.

In your case, this moment will be right before the reaction starts and **initial concentrations**. By calculating *comparing* its value to that of **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 = (0.0424^color(red)(2)color(red)(cancel(color(black)("M"^color(red)(2)))))/(0.00623 color(red)(cancel(color(black)("M"))) * 0.00414color(red)(cancel(color(black)("M")))) = 69.7#

Since you have *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**

#" ""H"_ (2(g)) " "+" " "I"_ (2(g)) " "rightleftharpoons" " color(red)(2)"HI"_((g))#

The equilibrium constant will be equal to

#K_c = ((0.0424 - color(red)(2)x)^color(red)(2))/(( 0.00623+x) * (0.00414 + x))#

**SIDE NOTE** *This is equivalent to saying that*

#2"HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I"_(2(g))#

*with the equilibrium constant being*

#K_"c reverse" = 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

#50.3x^2 + 0.6479x - 0.000397 = 0#

This quadratic equation will produce two values for *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 *magnitude* between the **equilibrium concentrations** of the three chemical species.