# 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)"#

(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#

Reaction:

(I'm trying to upload my work, but the camera doesn't want to cooperate. I got as far as the

Not really sure how to work out the

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

From trial and error, I assume the range would be:

##### 1 Answer

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

The partial pressures you were given are initial, so I would first see what

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

Since

#"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

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

#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

#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")#