# Question 38276

Jan 3, 2018

Here's what I got.

#### Explanation:

Right from the start, the fact that equilibrium constant is so small compared with $1$ tells you that, at this temperature, most of the hydrogen bromide present in the reaction vessel will not dissociate to produce hydrogen gas and bromine gas.

${K}_{c} \text{ << " 1 " "=>" " "the equilibrium lies far to the left}$

In other words, at this temperature, the equilibrium concentration of hydrogen bromide will be significantly higher than the equilibrium concentrations of the two products.

The balanced chemical equation that describes this equilibrium looks like this

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

By definition, the equilibrium constant is equal to

${K}_{c} = \left({\left[\text{H"_2] * ["Br"_2])/(["HBr}\right]}^{2}\right)$

Now, you know that you're starting with

["HBr"]_ 0 = 1.2 * 10^(-3)color(white)(.)"M"

As you can see, in order for the reaction to produce $1$ mole of hydrogen gas and $1$ mole of bromine gas, it must consume $2$ moles of hydrogen bromide.

This means that if you take $x$ $\text{M}$ to be the equilibrium concentration of hydrogen gas, you can say that

["Br"_2] = ["H"_2] = xcolor(white)(.)"M"

and

${\left[\text{HBr"] = ["HBr}\right]}_{0} - 2 x$

This means that the equilibrium constant can be written as

${K}_{c} = \frac{x \cdot x}{1.2 \cdot {10}^{- 3} - 2 x}$

$7.7 \cdot {10}^{- 11} = {x}^{2} / \left(1.2 \cdot {10}^{- 3} - 2 x\right)$

Now, because the equilibrium constant is so small compared to the initial concentration of hydrogen bromide, you can use the approximation

$1.2 \cdot {10}^{- 3} - 2 x \approx 1.2 \cdot {10}^{- 3}$

This means that you have

7.7 * 10^(-11) = x^2/(1.2 * 10^(-3)

which gets you

$x = \sqrt{7.7 \cdot {10}^{- 11} \cdot 1.2 \cdot {10}^{- 3}} = 3.0 \cdot {10}^{- 7}$

You can thus say that the equilibrium concentrations of the three species will be

["H"_2] = 3.0 * 10^(-7)color(white)(.)"M"

["Br"_2] = 3.0 * 10^(-7)color(white)(.)"M"

["HBr"] = 1.2 * 10^(-3)color(white)(.)"M" - 2 * 3.0 * 10^(-7)color(white)(.)"M"

color(white)(["HBr"])= 1.1994 * 10^(-3)color(white)(.)"M" ~~ 1.2 * 10^(-3)color(white)(.)"M"

The equilibrium concentrations of hydrogen gas and hydrogen gas are rounded to two sig figs and the equilibrium concentration of hydrogen bromide is rounded to one decimal place.

To find the percent decomposition of hydrogen chloride, simply compare the initial concentration of the compound and the concentration that dissociated, i.e. $2 \cdot 3.0 \cdot {10}^{- 3}$ $\text{M}$.

"% decomposition" = (2 * 3.0 * 10^(-7)color(red)(cancel(color(black)("M"))))/(1.2 * 10^(-3)color(red)(cancel(color(black)("M")))) xx 100%

"% decomposition" = 0.050%#

The answer is rounded to two sig figs.

This tells you that for every $\text{10,000}$ molecules of hydrogen bromide present in the reaction vessel, only $5$ molecules will dissociate to produce hydrogen gas and bromine gas.