What is the Partial Pressure of H2S at Equilibrium?

[Sonny101]

Homework Statement

H2(g) + S(s) = H2S(g) Kc= 6.8x10^-2 If 0.2 moles of H2 and 1.0 mole of S are heated in a 1L vessel upto 90C, what will be the partial pressure of H2S at equilibrium? Can someone help me with this step by step?

Homework Equations

The Attempt at a Solution

Kc = 6.8X10^-2 = x / (0.2-x)
x = 0.013 mol/L

PV = n RT
P (1L) = 0.013 mol (0.0821 Latm/molK) (273+90)
P = 0.38 atm

ALthought the answer is 0.44 :/

Homework Statement

Lead iodide, PbI2, is a sparingly soluble solute in water. At 25C it is found that at equilibrium,
PbI2(s) = Pb2+(aq) + 2 I-(aq)
the solubility of PbI2 in water is 1.2×10-3 M.
One liter of solution containing 12.00g NaI with solid PbI2 present at the bottom of the
solution is prepared at 25C. The NaI completely dissolves and fully dissociates,
according to the chemical equation,
NaI(s) = Na+(aq) + I-(aq)

Calculate the concentration of Pb2+
(aq) in this solution containing NaI

Homework Equations

The Attempt at a Solution

I have no idea where to start with this one :/

 

[Borek]

P = 0.38 atm

Could be I am missing something - but it looks OK to me.

I have no idea where to start with this one :/

Do you know what ICE table is?

 

[Sonny101]

Here is the original question. As for the ICE table, I am not too good with it, can only perform the basic generic operations. Would be helpful if you could show me and I will learn from it.

 

[Borek]

Here is the original question.

I still get 0.38.

As for the ICE table, I am not too good with it, can only perform the basic generic operations. Would be helpful if you could show me and I will learn from it.

That's not how the forum works. Try, show what you've got.

 

[Sonny101]

Hi I have attached the method I tried to solve these questions, along with additional question. I am not sure what I did wrong, but the quadratic equations i get give out two completely possible values of the variable. The question and the my tries are labelled respectively. Thanks

 

[Borek]

the solubility of PbI2 in water is 1.2×10-3 M.

Your ICE table is correct, but solubility is not Ksp.

 

[Sonny101]

so what do you suppose i do? 2x + 0.08 = 1.2x10^-3 would yield a negative value? And what about the other two?

 

[Borek]

You are supposed to calculate Ksp from the solubility.

 

[Sonny101]

So how do I make the equation after the table?

 

[Borek]

You did the ICE table correctly, your mistake was to plug solubility into the equation, instead of a correct Ksp value.

 

[Sonny101]

Oh right right you mean

Ksp= (1.2x10^-3)(2.4x10^-3)^2

and then equate this to the final solubilities. Thanks.

And what is my mistake with questions 4 and 5?

 

[Borek]

4th problem: why do you think both answers are correct?

 

[Sonny101]

i dont, i just followed the standard procedure and got 2 positive values (ususally in this type of cases i get 1 -ve and +ve so i can discard the -ve value), which can't be right, but cannot figure out my mistake :/

 

[Borek]

What is x? Hint: it is not the final concentration.

 

[Sonny101]

yeah i know, it would be (2 + 2x), but like I said which value of x should I use, seeing that both are possible. Anyways I don't get the right answer by using either value :/

 

[Borek]

which value of x should I use, seeing that both are possible

Calculate final concentrations of ALL species involved.

 

[Sonny101]

Doesn't the question only asks for the final concentration of HF only? Plus that's what I tried doing, but got confused by the value of 'x'.

PS: This is not my homework, but I am preparing for the finals, so this is the past paper with only the answers, so showing me the correct way would only help me.

 

[Borek]

Doesn't the question only asks for the final concentration of HF only?

Yes, that's what the question asks, but let's ignore that for a moment. Please try to calculate concentrations of all species involved (and I am not going to answer your further posts if you ignore the request).

In ICE table last line contains equilibrium concentrations, all you have to do is to plug calculated x into formulas you derived by yourself.

 

[Sonny101]

OK, so using the value of x to be 1.53

H2 = F2= 0.47
2HF= 5.06

if X= 2.69

H2=F2= -0.69
2HF = 7.38

Oh so THATS what you meant. Sorry my bad :P yeah one of the values get -ve so that will be discarded, and the other will be the correct value. Thanks.

And what do you think about Q5?

 

[Borek]

And what do you think about Q5?

Nice question.

Increase in pressure means some of the I₂ dissociated, try to find the dependence between pressure and dissociation fraction.