Question

For the reaction below, `K_p=81.9`. A reaction mixture contains `P_(I_2)=0.114` atm, `P_(Cl_2)=0.102` atm, and `P_(ICl)=0.355` atm. Find the equilibrium values for `P_(I_2)`, `P_(Cl_2)`, and `P_(ICl)`?

Reaction: `"I"_2"(g)"+"Cl"_2"(g)"\rightleftharpoons2"ICl (g)"`

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(I'm trying to upload my work, but the camera doesn't want to cooperate. I got as far as the `K_p` result shown below:)
`K_p=81.9=\frac{[ICl]^2}{[I_2][Cl_2]}=\frac{(0.355+2x)^2}{(0.114-x)(0.102-x)`

Not really sure how to work out the `x`.
I wanted to do trial-and-error (I'm told in our class that it's okay as long as the result is within 10% of the actual answer, and we only have to try 4 values of `x`.)

From trial and error, I assume the range would be:
`0.102\ltx\lt0.178`

Answer 1

Well, I would never recommend this "trial-and-error" method... I don't see why you should ever "have" to do it... just use the quadratic formula on the resulting quadratic equation.

I get:

`P_(ICl) ~~ 0.355 + 2(0.0575) = "0.470 atm"`

`P_(Cl_2) ~~ 0.102 - 0.0575 = "0.045 atm"`

`P_(I_2) ~~ 0.114 - 0.0575 = "0.057 atm"`

To check:

`(0.470)^2/((0.045)(0.057)) stackrel(?" ")(=) 81.9`

`= 86.1` `color(blue)(sqrt"")`

Close enough. Within `5.15%` error even with the approximations we made.

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The partial pressures you were given are initial, so I would first see what `Q_p` is:

`Q_p = (P_[ICl]^2)/(P_[I_2]P_[Cl_2]) = (0.355)^2/((0.114)(0.102)) = 10.84`

Since `Q_p < K_p`, the reaction should proceed forward to increase `P_(ICl)` and thus increase `Q_p` so that `Q_p = K_p` at equilibrium.

`"I"_2"(g)" " "+" " "Cl"_2"(g)" " "rightleftharpoons" " 2"ICl (g)"`

`"I"" "0.114" "" "" "0.102" "" "" "" "0.355`
`"C"" "-x" "" "" "-x" "" "" "" "" "+2x`
`"E"" "0.114-x" "0.102-x" "" "0.355+2x`

(If the reaction actually was supposed to go backwards, then the `x` you solve for will simply be negative instead of positive.)

Set it up, making sure that you include all the coefficients and exponents:

`81.9 = (0.355+color(red)(2)x)^color(red)(2)/((0.114-x)(0.102-x))`

The exact solution is around `"0.0560 atm"`%5E2%2F((0.114-x)(0.102-x)), but this is readily solvable by hand if we make some approximations with zero trial and error.

We approximate by using average partial pressures among the reactants so that

`81.9 ~~ (0.355 + 2x)^2/(0.108-x)^2`

The exact solution in this case is `"0.0563 atm"`, which is again still close to the original. The square root of `81` is `9`, so `sqrt(81.9)` is, say, `9.05` or so.

`9.05 ~~ (0.355 + 2x)/(0.108 - x)`

Now we can solve by hand (yes, use your calculator).

`9.05(0.108 - x) ~~ 0.355 + 2x`

`0.890 - 9.05x ~~ 0.355 + 2x`

`11.05x ~~ 0.635`

`x ~~ 0.635/11.05 = "0.0575 atm"`

And this is quite close, only off by `2.62%`. As a result (remember to put the initial pressures back to how they were... they're not the same):

`color(blue)(P_(ICl)) ~~ 0.355 + 2(0.0575) = color(blue)("0.470 atm")`

`color(blue)(P_(Cl_2)) ~~ 0.102 - 0.0575 = color(blue)("0.045 atm")`

`color(blue)(P_(I_2)) ~~ 0.114 - 0.0575 = color(blue)("0.057 atm")`