Chemistry Equilibrium ICE tables

[Coco12]

Homework Statement

consider the following reaction: 2H2O(i) <-> 2H2(g) + O2(g)

kc=7.3*10^-18 at 1000oC

The initial concentration of water in a reaction vessel is 0.055 mol/L. What is the equilibrium concentration of H2(g) at 1000oC?

Homework Equations

h20 h2 02
I 0.055 0 0
C -2x +2x +x
E0.055-2x 2x x

The Attempt at a Solution

The equation I got is:

-4x^3+2.92*(10^-17) x^2 -1.606*(10^-18) x +2.20825*10^-20 =0

How do I solve it if it is a cubic function?

 

[SteamKing]

1. You can guess a solution.
2. You can plot the equation and use that to estimate a solution, which can be refined with iteration.
3. You can use the cubic formula, which is much more involved than 1. or 2.

Given the relative magnitudes of the coefficients, roundoff will be a problem in evaluating possible solutions.

 

[Saitama]

Homework Statement

consider the following reaction: 2H2O(i) <-> 2H2(g) + O2(g)

kc=7.3*10^-18 at 1000oC

The initial concentration of water in a reaction vessel is 0.055 mol/L. What is the equilibrium concentration of H2(g) at 1000oC?

Homework Equations

h20 h2 02
I 0.055 0 0
C -2x +2x +x
E0.055-2x 2x x

The Attempt at a Solution

The equation I got is:

-4x^3+2.92*(10^-17) x^2 -1.606*(10^-18) x +2.20825*10^-20 =0

How do I solve it if it is a cubic function?

Assume that x amount of H₂O gets used up. x amount of H₂ and x/2 amount of O₂ is formed.

You can assume that x is far smaller than 0.055. So you get the following equilibrium expression:

$$K_{eq}=\frac{(x)^2(x/2)}{(0.055)^2}$$

The assumption is based on the fact that the equilibrium constant is in powers of 10^(-13).

 

[Coco12]

In my ice tables I used 2x but u used 1/2x.. Did I do it right?

 

[Borek]

You have chosen different substance for x, so your have different equations, They should be equivalent though.

 

[Coco12]

Ok thanks