# 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?

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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?

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

I got

I think I'm interpreting you correctly:

- The linked question has the
**first**established equilibrium, for which Stefan found the equilibrium concentrations to be#"0.389 M H"_2# ,#"0.293 M Br"_2# , and#"1.02 M HBr"# . - A
**second**equilibrium is established by adding#"0.150 mols"# #"Br"_2(g)# , and theconcentrations in step*equilibrium**1*are theconcentrations in step*initial**2*. The equilibrium concentrations due to step 2 are not known yet, though we are probably assuming the volume remained constant. - Then, an unknown amount of
#"HBr"# is added again, and the equilibrium shifts to the left to produce#"0.050 M"# for the equilibrium concentration of#"H"_2# on this**third**established equilibrium.

The goal is to find the

**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

#"H"_2(g)" " + " ""Br"_2(g)" " rightleftharpoons" " 2"HBr"(g)#

#"I"" "0.389" "" "" "0.293+0.150" "1.02#

#"C"" "-x" "" "" "-x" "" "" "" "" "+2x#

#"E"" "0.389-x" "0.443 - x" "" "1.02 + 2x#

Then we use *extent of equilibration*, as Stefan had done earlier.

#K_(eq) = 11.2 = (["HBr"]_(eq,2)^2)/(["H"_2]_(eq,2)["Br"_2]_(eq,2))#

#= (1.02 + 2x)^2/((0.389 - x)(0.443 - x))#

#= (1.02^2 + 4.08x + 4x^2)/(0.172327 - 0.832x + x^2)#

#=> 1.93006 - 9.3184x + 11.2x^2 = 1.0404 + 4.08x + 4x^2#

#=> 0.88966 - 13.3984x + 7.2x^2 = 0#

#x = "0.0689556 M", "1.79193 M"#

The only one that makes sense is the first

So the **equilibrium concentrations due to step 2** should be:

#["H"_2]_(eq,2) = 0.389 - 0.0689556 = color(green)("0.320 M")#

#["Br"_2]_(eq,2) = 0.443 - 0.0689556 = color(green)("0.374 M")#

#["HBr"]_(eq,2) = 1.02 + 2*0.0689556 = color(green)("1.16 M")#

**STEP 3**

Now, we add an unknown quantity of **left** and

#"H"_2(g)" " + " ""Br"_2(g)" " rightleftharpoons" " 2"HBr"(g)#

#"I"" "0.320" "" "" "0.374" "" "" "" "1.16 + Delta["HBr"]_(i,3)#

#"C"" "+0.050" "" "+0.050" "" "-2(0.050)#

#"E"" "0.370" "" "" "0.424" "" "" "1.06 + Delta["HBr"]_(i,3)#

Note that since we knew that

Now, this unknown quantity is

#K_(eq) = 11.2 = (["HBr"]_(eq,3)^2)/(["H"_2]_(eq,3)["Br"_2]_(eq,3))#

#= ("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"")#