Equilibrium questions (2 Viewers)

[adomad]

hey world, got this question...

swimming pools can be sterlised by adding calcium hypochlorite Ca(OCl)_2 or sodium hypochlorite NaOCl

the equilibrium involved is
OCl^- (aq) + H_2 O(l) <-> HOCl(aq) + OH^-(aq)

the specices that is best at destroying bacteria and at resisting decomposistion by sun light is HOCl

Identify the reaction conditions that will favour the formation of HOCl
(a) adding NaOH
(b) adding HCl
(C) adding water
(d) both B and C

answers say B but i say D. i know that is it B is correct because we increased the conc. of cl^- ions, but if we increase the water, wouldn't the equilibrium also shift to the right? making D the better option?

 

[gurmies]

I think increasing concentration of species shifts to the other side if in gaseous state

 

[study-freak]

hey world, got this question...

swimming pools can be sterlised by adding calcium hypochlorite Ca(OCl)_2 or sodium hypochlorite NaOCl

the equilibrium involved is
OCl^- (aq) + H_2 O(l) <-> HOCl(aq) + OH^-(aq)

the specices that is best at destroying bacteria and at resisting decomposistion by sun light is HOCl

Identify the reaction conditions that will favour the formation of HOCl
(a) adding NaOH
(b) adding HCl
(C) adding water
(d) both B and C

answers say B but i say D. i know that is it B is correct because we increased the conc. of cl^- ions, but if we increase the water, wouldn't the equilibrium also shift to the right? making D the better option?

If you add water, the concentration of water remains the same (although its amount increases) and that of all other species drops in equal ratio.

Hence adding water does not favour the formation of HOCl.

 

[adomad]

why doesn't the conc. of water increase? we are adding more...

 

[study-freak]

why doesn't the conc. of water increase? we are adding more...

Well, you cannot concentrate water, can you?
Its density won't change even if you add more.

Edit: take it this way.
When you add more water, you are not sqeezing water into a small volume but added water occupies larger volume.
So concentration won't increase.

 

[adomad]

ohh ok makes sense. ty

 

[omgmynameislong]

What. So adding water has no effect or...???

 

[adomad]

it just makes the system wetter lol

 

[study-freak]

What. So adding water has no effect or...???

I'd guess it doesn't affect in this case.

 

[gurmies]

What if water were in gaseous form? I think in this case it would favour products...

 

[adomad]

doesn't pressure affect gaseous substances.

 

[shaon0]

What if water were in gaseous form? I think in this case it would favour products...

Yeah but says liquid state in equ.

 

[gurmies]

Yeah but says liquid state in equ.

I am aware, but what if it were in gaseous state?

 

[shaon0]

I am aware, but what if it were in gaseous state?

I think you'd be right if water were in gaseous form.

 

[annabackwards]

Aha i remember gawking at this question just before my trials - it's just as study-freak said

If you added more water in the gaesous form, it'd have to specify that it was the concentration of steam increasing not just steam adedd because steam would just escape....

 

[mitochondria]

I think the answer from the book is actually wrong - the correct answer should be (d), that is, both adding water and HCl should favour the formation of hypochloric acid.

I can't actually remember if the following is part of the HSC. I have a feeling that it's not - but the idea is simple and is similar to the idea of pK_a for acids.

For a system in equilibrium, an equilibrium constant can be expressed as the quotient of the concentrations of the products by those of the reactants:

K^o = [Products] / [Reactants]


And in our case:

K^o = [HOCl][OH^-] / [OCl^-][H₂O]


Clearly, increasing the concentration of water would favour the formation HOCl. This is more obvious for a moderately concentrated solution.

Now, there is the argument that the change in water concentration is negligible at a relatively dilute solution, such as in the case of pool chlorine in a swimming pool! In fact, for this reactions, equilibrium constants are often written with the concentration of water being absorbed into the equilibrium constant term:

K' = K^o = [HOCl][OH^-] / [OCl^-]


So what happens if you add more water to this system? Well, according to the equation the equilibrium constant stays the same but the concentrations of everything has actually been reduced. If you substitute = n_s/V into all of them, you end up with an expression in terms of moles and an 1/V term on the R.H.S. and if you do the simple algebra the concentration of the products have actually increased relative to that of the starting material.

I think it's a case where the question writer and/or the answer just stuffed up.

 

[adomad]

i didn't understand a thing u just said

 

[mitochondria]

Umm... I'll try to edit it when I have time.

 

[Pwnage101]

I think the answer from the book is actually wrong - the correct answer should be (d), that is, both adding water and HCl should favour the formation of hypochloric acid.

I can't actually remember if the following is part of the HSC. I have a feeling that it's not - but the idea is simple and is similar to the idea of pK_a for acids.

For a system in equilibrium, an equilibrium constant can be expressed as the quotient of the concentrations of the products by those of the reactants:

K^o = [Products] / [Reactants]


And in our case:

K^o = [HOCl][OH^-] / [OCl^-][H₂O]


Clearly, increasing the concentration of water would favour the formation HOCl. This is more obvious for a moderately concentrated solution.

Now, there is the argument that the change in water concentration is negligible at a relatively dilute solution, such as in the case of pool chlorine in a swimming pool! In fact, for this reactions, equilibrium constants are often written with the concentration of water being absorbed into the equilibrium constant term:

K' = K^o = [HOCl][OH^-] / [OCl^-]


So what happens if you add more water to this system? Well, according to the equation the equilibrium constant stays the same but the concentrations of everything has actually been reduced. If you substitute = n_s/V into all of them, you end up with an expression in terms of moles and an 1/V term on the R.H.S. and if you do the simple algebra the concentration of the products have actually increased relative to that of the starting material.

I think it's a case where the question writer and/or the answer just stuffed up.

Equilibrium Constant is only covered in The Option Module Industrial Chemistry.

 

[Miss Sunshine]

Equilibrium Constant is only covered in The Option Module Industrial Chemistry.

What is the difference between equilibrium constant and position?