# Chemistry Equilibrium ICE tables

1. Nov 1, 2013

### Coco12

1. The problem statement, all variables and given/known data

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?

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

3. 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?

2. Nov 1, 2013

### SteamKing

Staff Emeritus
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.

3. Nov 1, 2013

### Pranav-Arora

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).

Last edited: Nov 2, 2013
4. Nov 3, 2013

### Coco12

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

5. Nov 3, 2013

### Staff: Mentor

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

6. Nov 3, 2013

Ok thanks