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 $P_(ICl)$?

Md Ariful Islam

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Md Ariful Islam

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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.

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:

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

, 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):

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

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

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

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