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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: RMGC on March 05, 2007, 03:52:24 PM

Title: Equilibrium Calculations from Kc with a cubic equation
Post by: RMGC on March 05, 2007, 03:52:24 PM

Im really lost on this one:  1 mol of CO and 3mol of H₂ are placed in a 10L vessel at 1200K. 
The reaction is CO(g) + 3H₂(g) <-----> CH₄(g) + H₂O(g).   The Kc is 3.92.

Using Stochiochemistry and the formula for Kc I determined 3.92=(x²)/(.1-x)(.3-3x)³.  Im not the greatest at math and was having trouble solving for x.  Please help

Title: Re: Equilibrium Calculations from Kc with a cubic equation
Post by: xiankai on March 06, 2007, 07:27:16 AM

it is easiest to take volume out of the equation first, so that you can deal with moles only.

since c = n/v,

K_c = [CH₄][H₂O]/[H₂]³[CO]
= {CH₄}{H₂O}/{H₂}³{CO} * (1/V)^-2

where let us take {A} to be the number of moles of A. and V to be the volume of the system.

so we have 3.92 = x²/(1-x)(3-3x)³.

im not sure if your method works fine, though since i never particularly paid much attention to my physical chemistry lectures and i prefer my method any day, its your choice.  :P

now for the mathematical part... (im using my formula, you could change it to yours without much difficulty i should think)

3.92 = x²/(1-x)(3-3x)³
3.92 = x²/(1-x)(1-x)³(3)³
105.84 = x²/(1-x)⁴
10.29 = x/(1-x)²
10.29 = x/(1-2x+x²)
10.29 -20.58x + 10.29x² = x
10.29x² -21.58x +10.29 = 0

you can solve the quadratic equation henceforth...