An equilibrium mixture of "28 g" "H"_2 and "140 g" "Br"_2 is heated in a "4.01-L" vessel at "1400 K". At equilibrium, the vessel is found to contain "1.1 g" of "HBr". Calculate K_c ?

$\text{H"_2(g) + "Br"_2(g) rightleftharpoons 2"HBr} \left(g\right)$

Feb 26, 2018

Answer:

${K}_{c} = 7.7 \cdot {10}^{- 6}$

Explanation:

The first thing that you need to do here is to calculate the initial concentrations of the two reactants and the equilibrium concentration of the product.

To do that, convert the masses to moles by using the respective molar masses of the three chemical species.

28 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "13.89 moles H"_2

140 color(red)(cancel(color(black)("g"))) * "1 mole Br"_2/(79.904color(red)(cancel(color(black)("g")))) = "1.752 moles Br"_2

1.1 color(red)(cancel(color(black)("g"))) * "1 mole Br"/(80.912color(red)(cancel(color(black)("g")))) = "0.01360 moles HBr"

Next, use the volume of the reaction vessel to determine the concentrations.

["H"_ 2]_ 0 = "13.89 moles"/"4.01 L" = "3.464 M"

["Br"_ 2]_ 0 = "1.752 moles"/"4.01 L" = "0.4369 M"

["HBr"] = "0.01360 moles"/"4.01 L" = "0.003392 M"

Now, you know that the balanced chemical equation that describes this equilibrium reaction looks like this

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

Notice that for every $1$ mole of hydrogen gas and $1$ mole of bromine gas that react, the reaction produces $2$ moles of hydrogen bromide.

In your case, you know that, at equilibrium, the reaction vessel contains $\text{0.003392 M}$ of hydrogen bromide. This implies that the equilibrium concentration of hydrogen gas will be

["H"_ 2] = ["H"_ 2]_ 0 - 1/2 * "0.003392 M"

["H"_ 2] = "3.464 M" - "0.001696 M"

["H"_ 2] = "3.4623 M"

This is the case because in order for the reaction to produce $\text{0.003392 M}$ of hydrogen bromide, the initial concentration of hydrogen gas must decrease by half this amount.

Similarly, the equilibrium concentration of bromine gas will be

["Br"_ 2] = ["Br"_ 2]_ 0 - 1/2 * "0.003392 M"

["Br"_ 2] = "0.4369 M" - "0.001696 M"

["Br"_ 2] = "0.4352 M"

This is the case because in order for the reaction to produce $\text{0.003392 M}$ of hydrogen bromide, the initial concentration of bromine gas must decrease by half this amount.

By definition, the equilibrium constant will take the form

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

Plug in your values to get--I'll leave the expression for the equilibrium constant without added units!

${K}_{c} = {\left(0.003392\right)}^{2} / \left(3.4323 \cdot 0.4352\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{7.7 \cdot {10}^{- 6}}}}$

The answer is rounded to two sig figs.