Help! With my solution to this Equilibrium problem. Please help me solve the quadratic for this problem and Are my written steps correct?

The constant term in the quadratic is supposed to be negative.

#"I"_2(g) rightleftharpoons 2"HI"(g)#

#K = (2x)^2/(0.0183 - x) = 2.04 xx 10^(-3)#

• Yes, the coefficient of #2# goes into the change in #["HI"]#. That also does go into the exponent.
• The starting concentration of #"I"_2(g)# is #"0.0183 M"#, that's right. #"0.066 mols"/"3.6 L" = "0.0183 M"#.

So it may just be a small algebraic misstep.

#2.04 xx 10^(-3)(0.0183 - x) = 4x^2#

#3.73 xx 10^(-5) - 2.04 xx 10^(-3)x = 4x^2#

#color(blue)(4x^2 + 2.04 xx 10^(-3)x - 3.73 xx 10^(-5) = 0)#

You almost got to this point, but you have a sign error. The last term is supposed to be negative.

The solution is supposed to be #x = "0.00281 M"#, so see if you get that.

#2.04 xx 10^(-3) stackrel(?" ")(=) (2cdot 0.00281)^2/(0.0183 - 0.00281)#

#= 2.039 xx 10^(-3) ~~ 2.04 xx 10^(-3) color(blue)(sqrt"")#

#I_2(g) rightleftharpoons2dotI(g)#..

#K_"eq"=[dotI]^2/([I_2(g)])=2.04xx10^-3#

Initially....#[I_2]=(0.066*mol)/(3.6*L)=0.0183*mol*L^-1#

And so if #x*mol*L^-1# diiodine dissociate....we gots...

#K_"eq"=(4x^2)/(0.0183-x)=2.04xx10^-3#...

#x=sqrt(1/4{2.04xx10^-3xx(0.0183-x)})#

Anf if #x# IS SMALL compared to #0.0183#, an approx. we have to justify..then...

#x~=sqrt(1/4{2.04xx10^-3xx(0.0183)})#

#x_1=3.05xx10^-3*mol*L^-1#...and now that we got an approximation for #x#...we can plug this back into the expression, and taken a second and third approximation....

#x_2=2.79xx10^-3*mol*L^-1#...and keep plugging this in....the value of #x# gets closer to the true value each time...

#x_3=2.81xx10^-3*mol*L^-1#

#x_4=2.81xx10^-3*mol*L^-1#

And since the approximations have converged, I am prepared to accept this as the true equilibrium value. I cannot see what you got on your paper....so please be discriminating with my answer...

And so if #2.81xx10^-3*mol*L^-1# diiodine have dissociated...

...at equilibrium

#[I_2]={0.0183-2.81xx10^-3}*mol*L^-1=1.55xx10^-3*mol*^-1#.

#[dotI]=5.62xx10^-3*mol*L^-1#...