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

only.by.me.these.are.my.personal.notes

3.6 Liter Container, 0.066mol (I2) at 981 Kelvin

Calculate the concentrations of I2 and I at equilibrium.

I used PolySolv for the Quadratic Equation at the bottom, and cannot come up with an appropriate value for X.

2 Answers
Truong-Son N.
Feb 25, 2018

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

"I"_2(g) rightleftharpoons 2"HI"(g)I2(g)⇌2HI(g)

K = (2x)^2/(0.0183 - x) = 2.04 xx 10^(-3)K=(2x)20.0183−x=2.04×10−3

  • Yes, the coefficient of 22 goes into the change in ["HI"][HI]. That also does go into the exponent.
  • The starting concentration of "I"_2(g)I2(g) is "0.0183 M"0.0183 M, that's right. "0.066 mols"/"3.6 L" = "0.0183 M"0.066 mols3.6 L=0.0183 M.

So it may just be a small algebraic misstep.

2.04 xx 10^(-3)(0.0183 - x) = 4x^22.04×10−3(0.0183−x)=4x2

3.73 xx 10^(-5) - 2.04 xx 10^(-3)x = 4x^23.73×10−5−2.04×10−3x=4x2

color(blue)(4x^2 + 2.04 xx 10^(-3)x - 3.73 xx 10^(-5) = 0)4x2+2.04×10−3x−3.73×10−5=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"x=0.00281 M, so see if you get that.

2.04 xx 10^(-3) stackrel(?" ")(=) (2cdot 0.00281)^2/(0.0183 - 0.00281)2.04×10−3? =(2⋅0.00281)20.0183−0.00281

= 2.039 xx 10^(-3) ~~ 2.04 xx 10^(-3) color(blue)(sqrt"")=2.039×10−3≈2.04×10−3√

anor277
Feb 25, 2018

We gots...I_2(g) rightleftharpoons2dotII2(g)⇌2.I

Explanation:

I_2(g) rightleftharpoons2dotI(g)I2(g)⇌2.I(g)..

K_"eq"=[dotI]^2/([I_2(g)])=2.04xx10^-3Keq=[.I]2[I2(g)]=2.04×10−3

Initially....[I_2]=(0.066*mol)/(3.6*L)=0.0183*mol*L^-1[I2]=0.066⋅mol3.6⋅L=0.0183⋅mol⋅L−1

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

K_"eq"=(4x^2)/(0.0183-x)=2.04xx10^-3Keq=4x20.0183−x=2.04×10−3...

x=sqrt(1/4{2.04xx10^-3xx(0.0183-x)})x=√14{2.04×10−3×(0.0183−x)}

Anf if xx IS SMALL compared to 0.01830.0183, an approx. we have to justify..then...

x~=sqrt(1/4{2.04xx10^-3xx(0.0183)})x≅√14{2.04×10−3×(0.0183)}

x_1=3.05xx10^-3*mol*L^-1x1=3.05×10−3⋅mol⋅L−1...and now that we got an approximation for xx...we can plug this back into the expression, and taken a second and third approximation....

x_2=2.79xx10^-3*mol*L^-1x2=2.79×10−3⋅mol⋅L−1...and keep plugging this in....the value of xx gets closer to the true value each time...

x_3=2.81xx10^-3*mol*L^-1x3=2.81×10−3⋅mol⋅L−1

x_4=2.81xx10^-3*mol*L^-1x4=2.81×10−3⋅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^-12.81×10−3⋅mol⋅L−1 diiodine have dissociated...

...at equilibrium

[I_2]={0.0183-2.81xx10^-3}*mol*L^-1=1.55xx10^-3*mol*^-1[I2]={0.0183−2.81×10−3}⋅mol⋅L−1=1.55×10−3⋅mol⋅−1.

[dotI]=5.62xx10^-3*mol*L^-1[.I]=5.62×10−3⋅mol⋅L−1...

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