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Topic: Equilibrium constant problem  (Read 1151 times)

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Offline Fish200398

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Equilibrium constant problem
« on: January 24, 2020, 03:25:32 AM »
reaction CO + H2O <=> CO2 + H2
Volume = 1 L

amount at equilibrium : 0.1 mole CO, H2O and H2, 0.2 mole CO2.

added 0.1 mole H2 and what is the amount of H2O reach new equilibrium
i am confused at, how to count the new equilibrium?
is it
CO      +               H2O <=>     CO2               +      H2
(0.1 + x) mole  (0.1 + x) mol    (0.2- x) mole        (0.2 - x)mole

new Eq = (0.2 - x) M . (0.2 - x) M / [(0.1 + x) M . (0.1 + x)M]

while new Eq = old Eq = [0.2 . 0.1]/[0.1 . 0.1] ?

Offline AWK

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Re: Equilibrium constant problem
« Reply #1 on: January 24, 2020, 03:57:07 AM »
Set ICE table correctly.
AWK

Offline Fish200398

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Re: Equilibrium constant problem
« Reply #2 on: January 24, 2020, 04:06:24 AM »
yes. i think

CO      +               H2O <=>     CO2               +      H2
(0.1 + x) mole  (0.1 + x) mol    (0.2- x) mole        (0.2 - x)mole

the reaction goes left because added H2 right?

Offline AWK

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Re: Equilibrium constant problem
« Reply #3 on: January 24, 2020, 04:19:48 AM »
Values at equilibrium give you K
Just solve quadratic equation.
AWK

Offline Fish200398

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Re: Equilibrium constant problem
« Reply #4 on: January 24, 2020, 04:38:46 AM »
is it
0.02 = (0.2 - x) M . (0.2 - x) M / [(0.1 + x) M . (0.1 + x)M]
-> (0.1 + x)^2 . 0.02 = (0.2 - x)^2 ???

Offline AWK

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Re: Equilibrium constant problem
« Reply #5 on: January 24, 2020, 05:02:32 AM »
check K
AWK

Offline Fish200398

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Re: Equilibrium constant problem
« Reply #6 on: January 24, 2020, 05:09:31 AM »
oh then is it

2 = (0.2 - x) M . (0.2 - x) M / [(0.1 + x) M . (0.1 + x)M]
-> (0.1 + x)^2 . 2 = (0.2 - x)^2 ??

Offline AWK

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Re: Equilibrium constant problem
« Reply #7 on: January 24, 2020, 05:33:48 AM »
solve this equation
AWK

Offline Fish200398

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Re: Equilibrium constant problem
« Reply #8 on: January 24, 2020, 05:39:14 AM »
then H2O = 0.125 mole

Offline AWK

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Re: Equilibrium constant problem
« Reply #9 on: January 24, 2020, 05:51:13 AM »
Insert the value of x into the equation and check if obtained value of K is close to 2. If not - solve the equation again
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Offline Fish200398

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Re: Equilibrium constant problem
« Reply #10 on: January 24, 2020, 07:42:22 AM »
i think theres a problem..
1. i take the sqrt of bothside
(0.1 + x) \sqrt 2 = (0.2 - x)
x = 0.06/2.4 = 1/40 = 0.025 mole
because K = [0.2 -x]^2/[0.1 + x]^2 = [0.175]^2/[0.125]^2 = sqrt 1.4

2. i didnt take the sqrt
(0.01 + 0.2x + x^2) . 2 = 0.04 - 0.4x + x^2
-0.02 + 0.8x + x^2 = 0
yields x = -0.4 + \sqrt{0.18} = -0.4 + ( 0.3 \sqrt 2) = - 0.4 + 0.42 = 0.02

but still x = 0.02 doesnt give K = 2 because K = [0.2 -x]^2/[0.1 + x]^2 = 0.18^2 / 0.12^2 = sqrt 3/2

if the ICE table is wrong, i dont know where is wrong

Offline AWK

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Re: Equilibrium constant problem
« Reply #11 on: January 24, 2020, 08:34:07 AM »
I didn't say it was the wrong answer.
If we already have some solution to the computational problem, we can check for ourselves whether this is a good result. In this case (0.175) ^ 2 / (0.125) ^ 2 = 1.96. If you can make a 2% error this is a good solution.
 The problem with this task is different and absurd. Your data (0.1 and 0.2) has one significant digit. If the final result had one significant digit, it would be absurd (0.1).
Therefore, the data for the task should be given as 0.100 and 0.200. And then your result is perfect because after rounding your result to three 3 significant that's what it should be.
AWK

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