VCE Chemistry Question Thread

[sweetcheeks]

You can use any set of values that are at equilibrium, as the temperature has not changed, so the equilibrium constant should be the same for whatever set you use (at equilibrium).

[larissaaa_]

If water is added to a reaction at equilibrium does it have an effect on the on the overall equilibrium? So like would it be a forward reaction or reverse?

[Swagadaktal]

Hey guys - is saying "favours the forward reaction" same as saying a "net forward reaction" ?

[jyce]

Hey guys - is saying "favours the forward reaction" same as saying a "net forward reaction" ?

Yep.

[Swagadaktal]

Yep.

jyce you're always there to make my day :')

[Sine]

CH3COOH + H2O <----> H3O+ + CH3COO-

if dilution occcure would this shift equilibrium to the right since reactants (1 aqeous species) and products ( 2 aqeous species) why can we / can't we include H2O (other than it's a liquid).

And more generally is it moreso only gaseous molecules and aqeous molecules when talking about changes of pressure and how equilibrium will change

[zsteve]

CH3COOH + H2O <----> H3O+ + CH3COO-

if dilution occcure would this shift equilibrium to the right since reactants (1 aqeous species) and products ( 2 aqeous species) why can we / can't we include H2O (other than it's a liquid).

And more generally is it moreso only gaseous molecules and aqeous molecules when talking about changes of pressure and how equilibrium will change

Dilution shifts the system to the right, your reasoning is correct. We don't include H2O because it's liquid (the reasons for this are rather complex).

Yes, only gaseous and aqueous species affect equilibrium position.

[Adequace]

http://imgur.com/5MssDNS, need some help with this q

For this question, I had to guess how to do but I got -0.8J/gC, when the answer is 0.8.

My working was find q(H2O) = m * L = 3kJ, then q = C * m * dT for the metal = 3000/25*-150 = -0.8.
I'm curious why the amount of energy from melting the 9g of H2O, is used for the amount of energy in the 2nd formula? I'm just entirely confused on the theory on why I can use the same q variable for both formulas.

Thanks

[Syndicate]

http://imgur.com/5MssDNS, need some help with this q

For this question, I had to guess how to do but I got -0.8J/gC, when the answer is 0.8.

My working was find q(H2O) = m * L = 3kJ, then q = C * m * dT for the metal = 3000/25*-150 = -0.8.
I'm curious why the amount of energy from melting the 9g of H2O, is used for the amount of energy in the 2nd formula? I'm just entirely confused on the theory on why I can use the same q variable for both formulas.

Thanks

Since the energy(lost) = energy(gain)

So the energy lost by the metal = the energy gained by the ice.

The first part of the question was done correctly (I haven't read the rest of it).

Q=mcT
3000 = 25 X C X 150 (change is temperature is positive as the system gains that heat)

Therefore c = 0.8

[Adequace]

Since the energy(lost) = energy(gain)

So the energy lost by the metal = the energy gained by the ice.

The first part of the question was done correctly (I haven't read the rest of it).

Q=mcT
3000 = 25 X C X 150 (change is temperature is positive as the system gains that heat)

Therefore c = 0.8

Okay thanks, but wouldn't the metal be losing energy, so -150C? Would I need to back the 3000, -3000 since the metal loses that energy?

[larissaaa_]

Can someone please explain the equilibrium constant and equilibrium position? Why does the equilibrium constant depend only on temperature but position of equilibrium can be changed by various factors like pressure and temp and dilution??

[lzxnl]

Can someone please explain the equilibrium constant and equilibrium position? Why does the equilibrium constant depend only on temperature but position of equilibrium can be changed by various factors like pressure and temp and dilution??

Equilibrium constant is a function of the equilibrium concentrations. For ANY system at equilibrium, the relevant function of equilibrium concentrations evaluates to the same thing. Let's suppose we have 2NO2(g) <-> N2O4(g). Then, the equilibrium constant, in terms of concentrations (makes more sense in terms of partial pressures but whatever): [N2O4]/[NO2]^2. This thing is always the same for any system at equilibrium, but the actual concentrations of N2O4 and NO2. The position of equilibrium is then the actual value of the concentrations. As you can see, if I quadruple the N2O4 concentration but only double the NO2 concentration, the system remains at equilibrium. However, you'd then have a larger proportion of N2O4 now. Equilibrium would be said to have shifted towards the right.

Now, this constant is only affected by temperature. Changing the pressure may change the equilibrium concentrations, but not the equilibrium constant. So, in the above case, you can change the volume, but at equilibrium [N2O4]/[NO2]^2 is the same as it was before, even though the individual concentrations may have changed.

[HopefulLawStudent]

The most appropriate analytical technique to detect the presence of particular proteins in cancerous body tissue are:

A. TLC
B. HPLC
C. GLC

Justify your answer.

I get why the answer is not A. But how would I determine whether it's B or C?

[sweetcheeks]

@HopefulLawStudent

Gas liquid chromatography involves heating a sample to turn it into a gas. What happens when you heat a protein?

HPLC involves separation of compounds, without heating them and is used as an alternative for analysis of things that cannot be analysed using GLC.

Could you explain why Thin Layer is not an appropriate answer?

[HopefulLawStudent]

@HopefulLawStudent

Gas liquid chromatography involves heating a sample to turn it into a gas. What happens when you heat a protein?

HPLC involves separation of compounds, without heating them and is used as an alternative for analysis of things that cannot be analysed using GLC.

Could you explain why Thin Layer is not an appropriate answer?

Ohhhhh. Denaturation! I'm an idiot.

Isn't TLC typically used for non-volatile compounds that are coloured or can be seen under UV light? I didn't think TLC falls under either category... right?