HSC Chemistry Question Thread

[wesadora]

Got a really short question: (yet surprisingly complicated, I find  )

"Using the tabulated reduction potentials, calculate a cell emf for the disproportionation of H₂O₂ and into O₂ and H₂O.

thanks!
-Wes

[RuiAce]

Hi humans 
i did an experiment in class the other day- and i have a validation on it in the next couple of days...i was wandering if someone could please help me out? please
The experiment was on equilibrium and Le CHatelier's principle
first we got K2CrO7 and added HCl solution to it...then to the same solution we added the base NaOH. originally the K2CrO7 was yellow, with added HCl it went to a darker colour (around orange), then when we added NaOH to the solution it went back to yellow..
i did the first part of the experiment ok...but i need help with this bit
the question is: By referring to the collision theory, account for the observed colour change that occurred when NaOH solution and the HCl solution were added to the K2CrO7 solution
I'd appreciate any help   

Before I proceed with your question, do you mean K₂CrO₄ or K2Cr₂O₇? Because CrO7 does not exist whereas CrO₄^2- and Cr₂O₇^2- are the chromate and dichromate ions respectively.

Either way, the only disadvantage here is that I cannot formulate an equation. I will just call it potassium compound for now.

It would appear, as though the HCl reacted with the potassium compound in an equilibrium reaction. The addition of HCl shifts this equilibrium to the right according to LCP, favouring the forward reaction which produces this new orange colour. But when we introduced NaOH, because NaOH will react with the HCl in a non-equilibrium reaction (we assert that the reaction goes to completion: NaOH_(aq) + HCl_(aq) -> NaCl_(aq) + H2O_(l)), we remove the amount of HCl present in the mixture with the potassium compound. As the HCl got reduced, the equilibrium shifted to the left according to LCP, favouring the reverse reaction and reintroducing new potassium compound (yellow colour remade).

[RuiAce]

Got a really short question: (yet surprisingly complicated, I find  )

"Using the tabulated reduction potentials, calculate a cell emf for the disproportionation of H₂O₂ and into O₂ and H₂O.

thanks!
-Wes

Your question seems rather peculiarly worded and I can't make sense out of it. So these are the only tips I can provide at this point:
a) Hydrogen peroxide is a weak acid
b) This reaction in the standard reduction potential might be useful:
1/2 O_2(g) + 2 H⁺ + 2 e^- ⇌ H2O_(l)
c) I found this equation on Wikipedia just now:
2 H₂O_2(l) -> 2 H2O_(l) + 2 H_2(g)
Note that H+ is an acid

[Maz]

hey
im a little confused on how to do concentration vs time graphs...could you please help me out?
if a reaction is 2NO₂ N₂O₄ (g) + 57kJ
how do you sketch the concentration vs time graphs for the effect of changing volume on the system and
the effect of changing temperature

how would these be different for rate vs time graphs for the same things
thankyou so much 

[RuiAce]

hey
im a little confused on how to do concentration vs time graphs...could you please help me out?
if a reaction is 2NO₂ N₂O₄ (g) + 57kJ
how do you sketch the concentration vs time graphs for the effect of changing volume on the system and
the effect of changing temperature

how would these be different for rate vs time graphs for the same things
thankyou so much 

So at this point, because you don't have an exact amount of info on the concentrations of NO₂ OR N₂O₄, we are unable to draw very precise concentration-time graphs. We can only make our graphs relative.

We will assume that the mixture is already at equilibrium. Consider the reaction: N₂O_{4 (g)} ⇌ 2 NO_{2 (g)}. Note that technically I have flipped your equation around.

According to secondary sources, the equilibrium naturally lies WELL to the right at lower temperatures. This means, initially the concentration of NO₂ is far superior to that of N₂O₄.
Because heat is released when dinitrogen tetroxide is produced, we realise that this reaction above is endothermic. Higher temperatures favour the production of NO₂, and lower temperatures favour the production of N₂O₂. Diagram 1 will therefore show what happens when heat is added to the system.

According to Le Chatelier's Principle, the equilibrium will shift to the right to minimise the changes, thereby increasing the concentration of NO₂.
Firstly, all of the reactants given are gaseous. Hence, changes in pressure (and by consequence, volume) will affect ALL the substances in the equation. Therefore, we immediately proceed to the equation.

Note that on the left of the reaction, we only have 1 mol of gas. Whereas, on the right, we have 2. This means that if we increase the pressure, the system wants to eliminate the amount of gas present and thus the equilibrium will shift to the left, producing more N₂O₄. On the contrary, if we decrease the pressure, there is more room for gas to exist so the equilibrium will shift to the right and produce more NO₂.

By convention, when we talk about changing the pressure we mean adding more NO₂ AND N₂O₄ in, or taking some of BOTH out. But another way to change the pressure is to change the volume of the vessel. Note that if we DECREASE the volume of the vessel, we give LESS room for the gas to occupy. Hence, DECREASING the volume of the vessel is essentially the same as INCREASING the pressure. On the contrary, this means when we INCREASE the volume of the vessel, we DECREASE the pressure.

Thus, by decreasing the volume of the vessel we promote yield of N₂O₄. And vice versa for increase.

Now, let's consider the formula C=n/V. If you look at this formula, you will realise that if we decrease the volume of the vessel, we effectively INCREASE the concentration of EVERYTHING momentarily! This happens because obviously the amount mof moles present already can't be changed.

Analogy: Say at equilibrium we had 1 mol of NO₂ in a 1L container. The concentration of NO₂ is 1mol L^-1.
But then we decrease the volume of the vessel to 500mL (1/2 of a litre). The concentration, at the time we decreased it, effectively becomes 2mol L^-1 now, because we still have 1 mol of nitrogen dioxide present!

So if we decrease the volume of the vessel, momentarily we will have a SPIKE in the concentration graph. THEN, the system will try to adjust its equilibrium according to LCP.

This can be seen in diagram 2.
Now, the rate of reaction is dependent on two factors. We will now assume that a system has NOT YET achieved equilibrium, and is trying to.

1. Add a catalyst.
As you know, the catalyst speeds up the rate of reaction by providing an alternate pathway requiring a lower activation energy. Hence, if we add a catalyst, we are going to force the equilibrium to be achieved more quickly. Note that both the forward and reverse reaction use the same catalyst.

2. Add an inert impurity.
Say we pumped some argon gas into the reaction chamber. You could say that this is increasing the pressure as well. BUT, there is a difference.
Argon will not react, so it will not cause havoc (oxygen being introduced can cause explosions in all sorts of reactions e.g. Haber process for production of ammonia). But argon has a counter effect. When we increase the pressure by pumping in actual equilibrium mixture (NO₂ and N₂O₄) we are promoting reactions on one side. But pumping in argon favours NEITHER side. Hence, if a system is not at equilibrium, the presence of argon is only going to SLOW DOWN the rate of reaction.

HOWEVER... In both situations, where the equilibrium is actually AT is unaffected. This only changes the rate, not the final result.
See diagram 3.

- Diagrams are coming. I don't have pen and paper with me beside me but what I do have is an ortho appointment so I'm running short of time right now

[RuiAce]

[Maz]

(Image removed from quote.)

So at this point, because you don't have an exact amount of info on the concentrations of NO₂ OR N₂O₄, we are unable to draw very precise concentration-time graphs. We can only make our graphs relative.

We will assume that the mixture is already at equilibrium. Consider the reaction: N₂O_{4 (g)} ⇌ 2 NO_{2 (g)}. Note that technically I have flipped your equation around.

According to secondary sources, the equilibrium naturally lies WELL to the right at lower temperatures. This means, initially the concentration of NO₂ is far superior to that of N₂O₄.
Because heat is released when dinitrogen tetroxide is produced, we realise that this reaction above is endothermic. Higher temperatures favour the production of NO₂, and lower temperatures favour the production of N₂O₂. Diagram 1 will therefore show what happens when heat is added to the system.

According to Le Chatelier's Principle, the equilibrium will shift to the right to minimise the changes, thereby increasing the concentration of NO₂.
Firstly, all of the reactants given are gaseous. Hence, changes in pressure (and by consequence, volume) will affect ALL the substances in the equation. Therefore, we immediately proceed to the equation.

Note that on the left of the reaction, we only have 1 mol of gas. Whereas, on the right, we have 2. This means that if we increase the pressure, the system wants to eliminate the amount of gas present and thus the equilibrium will shift to the left, producing more N₂O₄. On the contrary, if we decrease the pressure, there is more room for gas to exist so the equilibrium will shift to the right and produce more NO₂.

By convention, when we talk about changing the pressure we mean adding more NO₂ AND N₂O₄ in, or taking some of BOTH out. But another way to change the pressure is to change the volume of the vessel. Note that if we DECREASE the volume of the vessel, we give LESS room for the gas to occupy. Hence, DECREASING the volume of the vessel is essentially the same as INCREASING the pressure. On the contrary, this means when we INCREASE the volume of the vessel, we DECREASE the pressure.

Thus, by decreasing the volume of the vessel we promote yield of N₂O₄. And vice versa for increase.

Now, let's consider the formula C=n/V. If you look at this formula, you will realise that if we decrease the volume of the vessel, we effectively INCREASE the concentration of EVERYTHING momentarily! This happens because obviously the amount mof moles present already can't be changed.

Analogy: Say at equilibrium we had 1 mol of NO₂ in a 1L container. The concentration of NO₂ is 1mol L^-1.
But then we decrease the volume of the vessel to 500mL (1/2 of a litre). The concentration, at the time we decreased it, effectively becomes 2mol L^-1 now, because we still have 1 mol of nitrogen dioxide present!

So if we decrease the volume of the vessel, momentarily we will have a SPIKE in the concentration graph. THEN, the system will try to adjust its equilibrium according to LCP.

This can be seen in diagram 2.
Now, the rate of reaction is dependent on two factors. We will now assume that a system has NOT YET achieved equilibrium, and is trying to.

1. Add a catalyst.
As you know, the catalyst speeds up the rate of reaction by providing an alternate pathway requiring a lower activation energy. Hence, if we add a catalyst, we are going to force the equilibrium to be achieved more quickly. Note that both the forward and reverse reaction use the same catalyst.

2. Add an inert impurity.
Say we pumped some argon gas into the reaction chamber. You could say that this is increasing the pressure as well. BUT, there is a difference.
Argon will not react, so it will not cause havoc (oxygen being introduced can cause explosions in all sorts of reactions e.g. Haber process for production of ammonia). But argon has a counter effect. When we increase the pressure by pumping in actual equilibrium mixture (NO₂ and N₂O₄) we are promoting reactions on one side. But pumping in argon favours NEITHER side. Hence, if a system is not at equilibrium, the presence of argon is only going to SLOW DOWN the rate of reaction.

HOWEVER... In both situations, where the equilibrium is actually AT is unaffected. This only changes the rate, not the final result.
See diagram 3.

- Diagrams are coming. I don't have pen and paper with me beside me but what I do have is an ortho appointment so I'm running short of time right now

Thankyou so so so much 

[katherine123]

why does acetic acid produce the least concentration of H+ ? doesnt it produce the same amount as HCl since the formula is 
CH3COOH --> CH3COO- + H+
HCl --> H+ + Cl-

[RuiAce]

why does acetic acid produce the least concentration of H+ ? doesnt it produce the same amount as HCl since the formula is 
CH3COOH --> CH3COO- + H+
HCl --> H+ + Cl-

No. Because acetic acid is weak and hydrochloric acid is strong.
HCl → H⁺ + Cl^-
CH3COOH ⇌ H⁺ + CH3COO^-

Hydrochloric acid fully ionises in solution. (The reality is that it's about 90-99%, but we assume it goes to 100%).
However, with acetic acid, only about 1.3% or so ionises.

[jakesilove]

why does acetic acid produce the least concentration of H+ ? doesnt it produce the same amount as HCl since the formula is 
CH3COOH --> CH3COO- + H+
HCl --> H+ + Cl-

Just to build on RuiAce's answer, basically the way you assess the answer to this multiple choice question is with an understanding of "Weak" and "Strong" Acids.

Strong acids ionise completely in solution. Like RuiAce says, that means that approximately 90-99% of the HCl converts into the products (ie. Hydrogen ions and Chloride ions).

Weak acids do not ionise completely in solution. Like RuiAce says, that means that approximately 1.3% of the Acetic acid converts into the products.

In the HSC, you sort of just need to know which acids are weak, and which are strong. From memory, the only strong acids you work with are HCl and Sulfuric acid. The rest, you can assume, are weak! Therefore, as HCl is the only Strong acid in the list, it will produce the highest concentration of Hydrogen ions, and therefore have the lowest pH.

Jake

[RuiAce]

Just to build on RuiAce's answer, basically the way you assess the answer to this multiple choice question is with an understanding of "Weak" and "Strong" Acids.

Strong acids ionise completely in solution. Like RuiAce says, that means that approximately 90-99% of the HCl converts into the products (ie. Hydrogen ions and Chloride ions).

Weak acids do not ionise completely in solution. Like RuiAce says, that means that approximately 1.3% of the Acetic acid converts into the products.

In the HSC, you sort of just need to know which acids are weak, and which are strong. From memory, the only strong acids you work with are HCl and Sulfuric acid. The rest, you can assume, are weak! Therefore, as HCl is the only Strong acid in the list, it will produce the highest concentration of Hydrogen ions, and therefore have the lowest pH.

Jake

A tiny bit to add on again: Nitric acid is strong.

But yep, the list does not go beyond HCl, H₂SO₄ and HNO₃ with regards to the HSC.

(However, you can assume that HBr and HI are also strong. Beause Cl, Br and I are all halogens (group A VII elements) There is a trap though - HF is weak!)
To fully answer the question though: To justify why citric acid is stronger than acetic obviously depends on the fact that citric acid (2-hydroxypropane-1,2,3-tricarboxylic acid) is triprotic and has three protons to donate, whereas acetic acid (ethanoic acid) only has one. But I think we all already knew that as the answer was disregarded immediately.

[amandali]

do you round of answer according to the given data with lowest sigfig
eg.Student mixed 20ml of 0.02mol/L of H2SO4 with 500ml of 0.001mol/L of Ca(OH)2 find the ph
so you will round the answer to 1 sig fig because of "0.02ml/0.001ml"
Will i lose marks if i don't round them enough  eg. 4.1231x10^-2 when its meant to be 4.12x10^-2

[RuiAce]

do you round of answer according to the given data with lowest sigfig
eg.Student mixed 20ml of 0.02mol/L of H2SO4 with 500ml of 0.001mol/L of Ca(OH)2 find the ph
so you will round the answer to 1 sig fig because of "0.02ml/0.001ml"
Will i lose marks if i don't round them enough  eg. 4.1231x10^-2 when its meant to be 4.12x10^-2

Correct. In the question you provided, because 1 sig. fig. is the lowest amount of figures given, you must round it to that.

In terms of losing marks? Some examiners care, and others don't. Hence, not rounding correctly is a risk that you can't take.

[katherine123]

How do you solve this? thanks ^^
The solubility of calcium hydroxide is 0.12 g per 100ml of water at 25 degrees. Calculate maximum pH of a solution of Calcium hydroxide assuming that the addition of the solid changes the volume only negligibly.   ans: 11.5

[jakesilove]

How do you solve this? thanks ^^
The solubility of calcium hydroxide is 0.12 g per 100ml of water at 25 degrees. Calculate maximum pH of a solution of Calcium hydroxide assuming that the addition of the solid changes the volume only negligibly.   ans: 11.5

Hey Katherine!

I actually get a different answer (12.5) and have looked through my working a few times and am not sure where I've gone wrong. Perhaps your answers are wrong? Otherwise, hopefully someone in the community can help me out!



Jake