Question

Given the following data, how do I find the mols at equilibrium for this reaction?

`"BrCl"(g) rightleftharpoons 1/2"Br"_2(g) + 1/2"Cl"_2(g)`

`DeltaG_f^@ ("Br"_2(g)) = "3.11 kJ/mol"`
`DeltaG_f^@("BrCl"(g)) = -"0.98 kJ/mol"`

`A)` Find the mols of `"BrCl"(g)` at equilibrium if the volume of the container is fixed at `"1.0 L"`.
`B)` Find the mols of `"Br"_2(g)` at equilibrium.
`C)` Find the mols of `"Cl"_2(g)` at equilibrium.

Answer 1

`n_(BrCl(g),eq) = "0.81 mols"`
`n_(Br_2(g),eq) = "0.29 mols"`
`n_(Cl_2(g),eq) = "0.29 mols"`

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DISCLAIMER: LONG ANSWER!

Well, if we assume a rigid container, we can ignore the volume of the vessel and just use `"mols"`. The idea here is then:

1. Find `DeltaG_(rxn)^@`, i.e. at `25^@ "C"` and `"1 atm"` using standard Gibbs' free energies of formation.
2. Use the equilibrium condition to find `K`.
3. Use `K` to find `n_(Br_2(g),eq)` and `n_(Cl_2(g),eq)`.

`A)`

`DeltaG_(rxn)^@ = sum_P nu_P DeltaG_(f,P)^@ - sum_R nu_R DeltaG_(f,R)^@`,

where:

• `P` and `R` indicate products and reactants, respectively.
• `nu` is the stoichiometric coefficient.
• `DeltaG_f^@` is the standard change in Gibbs' free energy (of formation), due to forming each substance from its elements in their elemental state (that is, at `25^@ "C"` and `"1 atm"`).
• `DeltaG_f^@` for elements in their elemental state is thus `0`.

`=> ul(DeltaG_(rxn)^@) = overbrace((1/2 cdot "3.11 kJ/mol" + 1/2 cdot "0 kJ/mol"))^"Products" - overbrace((1 cdot -"0.98 kJ/mol"))^"Reactants"`

`=` `ul("2.535 kJ/mol")`

Normally, we could calculate `DeltaG_(rxn)` at nonequilibrium conditions with:

`DeltaG_(rxn) = DeltaG_(rxn)^@ + RTlnQ`,

where `RTlnQ` accounts for the shift away from standard conditions and `Q` would be the reaction quotient.

At chemical equilibrium, the reaction has no tendency to move either direction, so `DeltaG = 0` and `Q -= K`. Thus,

`DeltaG_(rxn)^@ = -RTlnK`,

and the equilibrium constant (by dividing by `-RT` and exponentiating both sides) is:

`ulK = "exp"(-DeltaG_(rxn)^@//RT)`

`= e^(-"2.535 kJ/mol"//("0.008314472 kJ/mol"cdot"K" cdot "298.15 K"))`

`= ul0.3597`

At this point, we can now determine the mols present of `"BrCl"(g)` at equilibrium. Construct an ICE table using `"mols"`:

`"BrCl"(g) rightleftharpoons 1/2"Br"_2(g) + 1/2"Cl"_2(g)`

`"I"" "1.40" "" "" "0" "" "" "" "0`
`"C"" "-x" "" "+x//2" "" "+x//2`
`"E"" "1.40 - x" "x//2" "" "" "x//2`

`K = ((x//2)^cancel(1//2))^cancel(2)/(1.40 - x) = (x//2)/(1.40 - x) = 0.3597`

This `K` is not small, but solving this is not that bad. Eventually we obtain the physical answer as:

`x = |Deltan_(BrCl(g))| = "0.5858 mols"`

`= 2n_(Br_2(g),eq) = 2n_(Cl_2(g),eq)`

And that means...

`color(blue)(n_(BrCl(g),eq)) = 1.40 - 0.5858 = ulcolor(blue)("0.81 mols")`

Everything follows from here. Now the rest is easy.

`B)`

Refer to the ICE table above to realize that:

`color(blue)(n_(Br_2(g),eq)) = x/2 ~~ ulcolor(blue)("0.29 mols")`

`C)`

Refer to the ICE table above to realize that:

`color(blue)(n_(Cl_2(g),eq)) = x/2 ~~ ulcolor(blue)("0.29 mols")`

And as a check, is `K` still correct?

`K = ((0.5858//2)^(1//2))^2/(1.40 - 0.5858) ~~ 0.3597` `color(blue)(sqrt"")`