# 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: $\text{I"_2"(g)"+"Cl"_2"(g)"\rightleftharpoons2"ICl (g)}$ (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: $0.102 \setminus < x \setminus < 0.178$

Jul 18, 2018

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:

${P}_{I C l} \approx 0.355 + 2 \left(0.0575\right) = \text{0.470 atm}$

${P}_{C {l}_{2}} \approx 0.102 - 0.0575 = \text{0.045 atm}$

${P}_{{I}_{2}} \approx 0.114 - 0.0575 = \text{0.057 atm}$

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:

${Q}_{p} = \frac{{P}_{I C l}^{2}}{{P}_{{I}_{2}} {P}_{C {l}_{2}}} = {\left(0.355\right)}^{2} / \left(\left(0.114\right) \left(0.102\right)\right) = 10.84$

Since ${Q}_{p} < {K}_{p}$, the reaction should proceed forward to increase ${P}_{I C l}$ and thus increase ${Q}_{p}$ so that ${Q}_{p} = {K}_{p}$ at equilibrium.

$\text{I"_2"(g)" " "+" " "Cl"_2"(g)" " "rightleftharpoons" " 2"ICl (g)}$

$\text{I"" "0.114" "" "" "0.102" "" "" "" } 0.355$
$\text{C"" "-x" "" "" "-x" "" "" "" "" } + 2 x$
$\text{E"" "0.114-x" "0.102-x" "" } 0.355 + 2 x$

(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:

$81.9 = {\left(0.355 + \textcolor{red}{2} x\right)}^{\textcolor{red}{2}} / \left(\left(0.114 - x\right) \left(0.102 - x\right)\right)$

The exact solution is around $\text{0.0560 atm}$, 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

$81.9 \approx {\left(0.355 + 2 x\right)}^{2} / {\left(0.108 - x\right)}^{2}$

The exact solution in this case is $\text{0.0563 atm}$, 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.

$9.05 \approx \frac{0.355 + 2 x}{0.108 - x}$

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

$9.05 \left(0.108 - x\right) \approx 0.355 + 2 x$

$0.890 - 9.05 x \approx 0.355 + 2 x$

$11.05 x \approx 0.635$

$x \approx \frac{0.635}{11.05} = \text{0.0575 atm}$

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):

$\textcolor{b l u e}{{P}_{I C l}} \approx 0.355 + 2 \left(0.0575\right) = \textcolor{b l u e}{\text{0.470 atm}}$

$\textcolor{b l u e}{{P}_{C {l}_{2}}} \approx 0.102 - 0.0575 = \textcolor{b l u e}{\text{0.045 atm}}$

$\textcolor{b l u e}{{P}_{{I}_{2}}} \approx 0.114 - 0.0575 = \textcolor{b l u e}{\text{0.057 atm}}$