# After the second equilibrium was reached a student added an unknown amount of HBr gas. He found that the concentration of H2 gas at equilibrium increased by 0.050 M. How much HBr gas did the student add to the system?

## This question is related to [Calculate the concentration of each species at equilibrium? ](https://socratic.org/questions/calculate-the-concentration-of-each-species-at-equilibrium) The 2nd equilibrium is reached after 0.150 moles of Br is added. However, for this question, I got the answer of 0.202M. Can someone help verify this?

##### 1 Answer
Sep 4, 2016

I got Delta["HBr"]_(i,3) = "0.266 M".

I think I'm interpreting you correctly:

1. The linked question has the first established equilibrium, for which Stefan found the equilibrium concentrations to be ${\text{0.389 M H}}_{2}$, ${\text{0.293 M Br}}_{2}$, and $\text{1.02 M HBr}$.
2. A second equilibrium is established by adding $\text{0.150 mols}$ ${\text{Br}}_{2} \left(g\right)$, and the equilibrium concentrations in step 1 are the initial concentrations in step 2. The equilibrium concentrations due to step 2 are not known yet, though we are probably assuming the volume remained constant.
3. Then, an unknown amount of $\text{HBr}$ is added again, and the equilibrium shifts to the left to produce $\text{0.050 M}$ for the equilibrium concentration of ${\text{H}}_{2}$ on this third established equilibrium.

The goal is to find the $\Delta {\left[\text{HBr}\right]}_{i , 3}$ that caused Delta["H"_2]_(i,3) = "0.050 M" to be produced. That's what I'm getting from this question/discussion.

STEP 2

So the first thing I would do is draw out the ICE table to see what we know or can figure out. We added ${\text{Br}}_{2} \left(g\right)$, so more of it should react to form more $\text{HBr}$, i.e. the reaction is pushed forward.

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

$\text{I"" "0.389" "" "" "0.293+0.150" } 1.02$
$\text{C"" "-x" "" "" "-x" "" "" "" "" } + 2 x$
$\text{E"" "0.389-x" "0.443 - x" "" } 1.02 + 2 x$

Then we use ${K}_{e q}$ to determine the extent of equilibration, as Stefan had done earlier.

${K}_{e q} = 11.2 = \left({\left[{\text{HBr"]_(eq,2)^2)/(["H"_2]_(eq,2)["Br}}_{2}\right]}_{e q , 2}\right)$

$= {\left(1.02 + 2 x\right)}^{2} / \left(\left(0.389 - x\right) \left(0.443 - x\right)\right)$

$= \frac{{1.02}^{2} + 4.08 x + 4 {x}^{2}}{0.172327 - 0.832 x + {x}^{2}}$

$\implies 1.93006 - 9.3184 x + 11.2 {x}^{2} = 1.0404 + 4.08 x + 4 {x}^{2}$

$\implies 0.88966 - 13.3984 x + 7.2 {x}^{2} = 0$

$x = \text{0.0689556 M", "1.79193 M}$

The only one that makes sense is the first $x$, so that you don't get negative concentrations.

So the equilibrium concentrations due to step 2 should be:

$\left[\text{H"_2]_(eq,2) = 0.389 - 0.0689556 = color(green)("0.320 M}\right)$
$\left[\text{Br"_2]_(eq,2) = 0.443 - 0.0689556 = color(green)("0.374 M}\right)$
$\left[\text{HBr"]_(eq,2) = 1.02 + 2*0.0689556 = color(green)("1.16 M}\right)$

STEP 3

Now, we add an unknown quantity of $\text{HBr}$ so that the ${\left[{\text{H}}_{2}\right]}_{i , 3}$ increased by $\text{0.050 M}$ to some final, ${\left[{\text{H}}_{2}\right]}_{e q , 3}$. Since $\text{HBr}$ is a product, the equilibrium shifts left and ${\text{H}}_{2}$ and ${\text{Br}}_{2}$ get produced.

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

"I"" "0.320" "" "" "0.374" "" "" "" "1.16 + Delta["HBr"]_(i,3)
$\text{C"" "+0.050" "" "+0.050" "" } - 2 \left(0.050\right)$
"E"" "0.370" "" "" "0.424" "" "" "1.06 + Delta["HBr"]_(i,3)

Note that since we knew that $\text{0.050 M}$ was the increase in ${\text{H}}_{2}$'s initial concentration during the third to-be-established equilibrium, the mole ratio of ${\text{H}}_{2}$ with $\text{HBr}$ tells us that ${\left[\text{HBr}\right]}_{i , 3}$ decreased by twice that much to reach ${\left[\text{HBr}\right]}_{e q , 3}$ when it was consumed to shift the equilibrium leftwards.

Now, this unknown quantity is $\Delta {\left[\text{HBr}\right]}_{i , 3}$, since it adds to ${\left[\text{HBr}\right]}_{i , 3}$ and causes the equilibrium to shift left.

${K}_{e q} = 11.2 = \left({\left[{\text{HBr"]_(eq,3)^2)/(["H"_2]_(eq,3)["Br}}_{2}\right]}_{e q , 3}\right)$

= ("1.06 M" + Delta["HBr"]_(i,3))^2/(("0.370 M")("0.424 M"))

$\sqrt{1.757056}$ "M" = "1.06 M" + Delta["HBr"]_(i,3)

color(blue)(Delta["HBr"]_(i,3)) = 0.26554 => color(blue)("0.266 M")

Now let's check that the equilibrium constant remained constant.

(1.06 + "0.266 M")^2/(("0.370 M")("0.424 M")) = 11.2 color(blue)(sqrt"")