For :H2O(g) --> H(g) + OH(g); At 3000K and P(total)=1.00 atm, from your K value calculate system composition from 1.00 mol H2O(g) and no products initially.
The K I got (using Delta G superscript zero = [Delta H superscript zero]-[T* Delta S superscript zero)=]) was 9.5x10^-4 atm.
I know system composition is mole composition, which is proportional to mole fraction. I think Ptotal=P(H2O)+p(H)+P(OH) but where do I proceed from there?
If you are satisfied that dG is ok, then set up the ICE chart.
.......H2O ==> H^+ + OH^-
I......1.0.....0......0
C......-p......p......p
E......1.0-p...p......p
Then Kp = pH^+ * pOH^-/pH2O
and substitute the E line into Kp expression and solve for p. Then use p to calculate mole fraction or use PV = nRT to calculate mols and/or grams.
I looked in my text to find dGo for H^+ and OH^- but no luck. But if your text has those values it would be interesting to calculate dGo products - dGo reactants = dG rxn and plug into dG = -RTlnK and see how close the two K values are.
I agree that Ptotal initially is 1.0 atm from the problem. After the ICE chart it is what you have. It will be Ptotal = 1-p + p + p = 1+p so
XH2O = 1-p/1+p
XH^+ = p/1+p
XOH^- = p/1+p
To calculate the system composition from the given K value, you can use the equation
K = (P(H) * P(OH))/(P(H2O))
Where K is the equilibrium constant, P(H2O) is the partial pressure of water, P(H) is the partial pressure of hydrogen, and P(OH) is the partial pressure of hydroxide.
Given that the K value is 9.5x10^-4 atm and P(total) = 1.00 atm, we can assume that the partial pressures of hydrogen and hydroxide are very small compared to the partial pressure of water at equilibrium. Therefore, we can ignore the terms P(H) and P(OH) in the equation.
So, we have:
K = P(H) * P(OH) / P(H2O)
Since we initially have 1.00 mol of H2O(g) and no products, the initial partial pressure of water, P(H2O), is equal to 1.00 atm.
We can rearrange the equation to solve for P(H) and P(OH):
K * P(H2O) = P(H) * P(OH)
Substituting the given values:
(9.5x10^-4 atm) * (1.00 atm) = P(H) * P(OH)
Simplifying the expression:
9.5x10^-4 atm = P(H) * P(OH)
Since these are partial pressures and are proportional to mole fractions, you can also express this equation as:
K = [H] * [OH]
Where [H] represents the mole fraction of hydrogen and [OH] represents the mole fraction of hydroxide at equilibrium.
To find the system composition, you need to solve for the mole fraction of hydrogen ([H]) and the mole fraction of hydroxide ([OH]).
Since the mole fraction is proportional to partial pressure, you can write:
K = P(H) * P(OH) = [H] * [OH]
Given that the sum of the mole fractions is equal to 1 (since it represents the whole system), you can write:
[H] + [OH] = 1
Therefore, you have two equations and two unknowns:
K = [H] * [OH]
[H] + [OH] = 1
You can use algebraic methods to solve these equations simultaneously. One possible approach is to solve the second equation for one unknown and substitute it into the first equation.
For example, solving the second equation for [H]:
[H] = 1 - [OH]
Now substitute this into the first equation:
K = (1 - [OH]) * [OH]
You can solve this equation for [OH] using suitable numerical methods such as polynomial solvers or graphical methods. Once you find [OH], you can substitute it back into the equation [H] = 1 - [OH] to find [H]. These resulting mole fractions will represent the system composition at equilibrium.