[rcole23]

Hey All, this is the question and how far I have been able to take it.  I just cant see this as being realistic as nothing in the couse has even come close to expecting us to do this and yet again I have no lesson to assist me so I am learning what I can by looking online.  Does the below seem right to you so far?

Question 5) Consider the following reaction:

2H2O(l) <--> 2H2(g) + O2(g)    Kc = 7.3 X 10-18 at 1000 degrees Celsius


The initial concentration of water in a reaction vessel is 0.055 mol/L.  What is the equilibrium concentration of H2 at 1000 degrees Celsius? (6 marks)


Kc = 0.0000000000000000073


H2O[initial] = 0.055 mol/L


                                              2 H2O <>   2 H2 +    O2
Initial concentration (mol/L)                      0.055   0.00   0.00
Change in concentration (mol/L)           -2x               +2x   +x
Equilibrium concentration (mol/L)         0.055 – 2x   +2x   +x

Kc = products / reactants


Kc = [2H2] [O2] / [2H2O]

7.3 X 10-18 = [2x]² [X] / [0.055 – 2x]²

7.3 X 10-18 = 4X3 / [0.055 – 2x]²

4X3 = 7.3 X 10-18[0.055 – 2x] [0.055 – 2x]

4X3 = [4.015 X 10-19 – 1.46x X 10-17][0.055 – 2x]

FOIL

4X3 = 2.20825 X 10-20 – 2.92x X 10-17 – 8.0x X 10-19 + 2.92x2 X 10-17

0 = 2.20825 X 10-20 – 2.92x X 10-17 – 8.03x X 10-19 + 2.92x2 X 10-17 – 4X3

0 = -4X3 + 2.92x2 X 10-17 – 2.8937x X 10-17 + 2.20825 X 10 -20

[UG]

                                           2 H2O <>     2 H2 +    O2
Initial concentration (mol/L)             0.055       0.00      0.00
Change in concentration (mol/L)        -2x        +2x          +x
Equilibrium concentration (mol/L)     0.055 – 2x   +2x     +x

I think it would be 0.055-x rather than 2x and the concentration of O₂ at equilibrium would be 0.5x, and H₂ = x

[rcole23]

                                           2 H2O <>     2 H2 +    O2
Initial concentration (mol/L)             0.055       0.00      0.00
Change in concentration (mol/L)        -2x        +2x          +x
Equilibrium concentration (mol/L)     0.055 – 2x   +2x     +x

I think it would be 0.055-x rather than 2x and the concentration of O₂ at equilibrium would be 0.5x, and H₂ = x

Hey UG, thanks for taking a visit.  What makes you think that I should use different values?  Is it becasue of the line in the question What is the equilibrium concentration of H2 at 1000 degrees Celsius? .  It seems you have cut everything in half which is actually a good idea.  I have not run the numbers yet but I wanted to double check on what you are thinking. 

Let me know 

[rcole23]

well I completed the question and the self check works out so I figured I would post the answer to help anyone and everyone out there that may come across this question.

Question 5) Consider the following reaction:

2H2O(l)   2H2(g) + O2(g)    Kc = 7.3 X 10-18 at 1000 degrees Celsius


The initial concentration of water in a reaction vessel is 0.055 mol/L.  What is the equilibrium concentration of H2 at 1000 degrees Celsius? (6 marks)


Kc = 0.0000000000000000073


H2O[initial] = 0.055 mol/L


   H2O     H2 +    O2
Initial concentration (mol/L)   0.055   0.00   0.00
Change in concentration (mol/L)   -x   +x   +x
Equilibrium concentration (mol/L)   0.055 – x   x   (0.5)x

Kc = products / reactants


Kc = [H2] [1/2O2] / [H2O]

7.3 X 10-18 =

• / [0.055 – x]

7.3 X 10-18 = X2 / [0.055 – x]
X2 = 7.3 X 10-18 [0.055 – x]
X2 = 4.015 X 10-19 - 7.3 X 10-18 X
0 = -X2 - 7.3 X 10-18 X + 4.015 X 10-19
        A             B                      C

X = -b +- square root of [b2 – 4ac] / 2a
X = {-(- 7.3 X 10-18) +- SQRT [- 7.3 X 10-18 2 – 4(-1)( 4.015 X 10-19)} / 2(-1)
X = {7.3 X 10-18 +- SQRT [5.329 X 10-35 - -1.606 X 10-18]} / -2
X = 7.3 X 10-18 +- SQRT [1.606 X 10 18] / -2
X = 7.3 X 10-18 +- 1.26728 X 10-9 / -2

X = -6.3364 X 10-10 (impossible) or X = 6.3364 X 10-10 (correct)

H2O[equilibrium] = 0.055 – x = 0.055 - 6.3364 X 10-10 = 0.54999999

H2[equilibrium] = x = 6.3364 X 10-10

O2[equilibrium] = (0.5)x = 0.5 * = 6.3364 X 10-10 = 3.1682 X 10-10

Self Check

Kc= products / reactants
7.3 X 10-18 =

• / [0.055 – x]

7.3 X 10-18 = [6.3364 X 10-10] [6.3364 X 10-10] / [0.055 - 6.3364 X 10-10]
7.3 X 10-18 = 4.015 X 10-19 / 0.054999999
7.3 X 10-18 = 7.3 X 10-18 (answers are correct)