VCE Chemistry Question Thread

[jyce]

Thanks Jyce 

Just wondering whats wrong with using "forward reaction" ?

Because chemical equilibrium is a dynamic state, which in this context means that the forward and backward reactions are both still occurring, it's just that their rates are equal to one another and so in this way the concentrations of the species involved don't change. What's significant is when the rate of either the forward or backward reaction becomes greater than the other - this is how an equilibrium system will attempt to (partially) oppose a change in conditions. Simply stating a system has responded to a change with a "forward reaction" isn't really correct - there was always a forward reaction, but now the rate of the forward reaction has become greater than that of the backward reaction, meaning that there is a NET forward reaction.

Does this make sense?

[knightrider]

Because chemical equilibrium is a dynamic state, which in this context means that the forward and backward reactions are both still occurring, it's just that their rates are equal to one another and so in this way the concentrations of the species involved don't change. What's significant is when the rate of either the forward or backward reaction becomes greater than the other - this is how an equilibrium system will attempt to (partially) oppose a change in conditions. Simply stating a system has responded to a change with a "forward reaction" isn't really correct - there was always a forward reaction, but now the rate of the forward reaction has become greater than that of the backward reaction, meaning that there is a NET forward reaction.

Does this make sense?

yep this makes perfect sense, Thanks jyce 

defs gonna try get back my mark tomorrow haha i am pretty sure i used net forward reaction 

[Adequace]

Just a question, do we need to memorise the solubility table for chem 3/4?

[zsteve]

Just a question, do we need to memorise the solubility table for chem 3/4?

Not per se, but you'll need to have some general knowledge about what's soluble and what's not, e.g. nitrates (NO3) always soluble, group I metals e.g. LiOH, LiCl, etc.
Also, hydroxides like Ba(OH)2, Mg(OH)2, Fe(OH)3, Zn(OH)2 and the like aren't soluble.

This gets picked up more when you pay attention throughout the course than through rote-learning tables.

[Adequace]

Not per se, but you'll need to have some general knowledge about what's soluble and what's not, e.g. nitrates (NO3) always soluble, group I metals e.g. LiOH, LiCl, etc.
Also, hydroxides like Ba(OH)2, Mg(OH)2, Fe(OH)3, Zn(OH)2 and the like aren't soluble.

This gets picked up more when you pay attention throughout the course than through rote-learning tables.

Ah okay, thanks zsteve.

One more thing, would we need to memorise the charges of cations and anions?

[zsteve]

Ah okay, thanks zsteve.

One more thing, would we need to memorise the charges of cations and anions?

Yes, of most of the common polyatomic anions/cations, e.g. NO3-, SO4(2-), PO4(3-), and so on

[blacksanta62]

Hey guys, prepping for a SAC and I have some questions which I was given today:
1) In this reaction:
Fe³⁺(aq) + SCN^-(aq) <----> Fe(SCN)²⁺(aq) (0.50L)
We add 0.060 mol of Fe³⁺(aq) and 0.13 mol of SCN^-(aq). Once equilibrium is reached, 0.030 mol of Fe(SCN)²⁺(aq) has been formed
Find the equilibrium constant for the reaction
Is my reasoning correct (hence the answer)?
Because 0.030 mol of product has been formed and there is an increase in the amount of mol present, a net forward reaction has occured. Therefore, we have lost 0.030 mol from each of the reactants.
Fe³⁺(aq) + SCN^-(aq) <----> Fe(SCN)²⁺(aq)
0.060                         0.13                                  0 mol
-0.030 mol                 -0.030 mol                         0.030 mol

c = 0.030/0.50           c = 0.1/0.50                       c = 0.030/0.50
c = 0.06M                  c = 0.2M                            c = 0.06

K = 0.06/0.2 * 0.06 ===> K = 5M^-1
2) With the equilibrium reaction: N₂O₄(g) <------> 2NO₂(g)
We have to explain what happens when the volume of the sealed vessel is increased
Is this correct:
Initially we increase the volume of the sealed vessel, decreasing the pressure and decreasing the concentration of particles (will this decrease the concentration of both gases or just one?). This will result in a lower concentration fraction/reaction quotient
Then, the system will partially oppose this it will try to increase the pressure ===> net forward reaction will be favoured and the concentration of NO₂ increasing and the concentration of N₂O₄ decreasing. This will increase the reaction quotient/concentration fraction until it is equal to the equilibrium constant.

I appreciate any help

[sweetcheeks]

What are the appropriate units for ka? Is it mole/litre?

[blacksanta62]

Yep, you can work it out using index laws: Ka = M^2/M
Ka = M

All of the questions I've done sugget that it'll always be M. Hopefully the above example shows you why
Hope I helped

[sweetiepi]

What are the appropriate units for ka? Is it mole/litre?

Yep, you can work it out using index laws: Ka = M^2/M
Ka = M

All of the questions I've done sugget that it'll always be M. Hopefully the above example shows you why
Hope I helped

It's best if you work the units out separately, I've seen/done examples with M^-1, although BlackSanta is right about most of the time the units M.

[blacksanta62]

Hey guys, prepping for a SAC and I have some questions which I was given today:
1) In this reaction:
Fe³⁺(aq) + SCN^-(aq) <----> Fe(SCN)²⁺(aq) (0.50L)
We add 0.060 mol of Fe³⁺(aq) and 0.13 mol of SCN^-(aq). Once equilibrium is reached, 0.030 mol of Fe(SCN)²⁺(aq) has been formed
Find the equilibrium constant for the reaction
Is my reasoning correct (hence the answer)?
Because 0.030 mol of product has been formed and there is an increase in the amount of mol present, a net forward reaction has occured. Therefore, we have lost 0.030 mol from each of the reactants.
Fe³⁺(aq) + SCN^-(aq) <----> Fe(SCN)²⁺(aq)
0.060                         0.13                                  0 mol
-0.030 mol                 -0.030 mol                         0.030 mol

c = 0.030/0.50           c = 0.1/0.50                       c = 0.030/0.50
c = 0.06M                  c = 0.2M                            c = 0.06

K = 0.06/0.2 * 0.06 ===> K = 5M^-1
2) With the equilibrium reaction: N₂O₄(g) <------> 2NO₂(g)
We have to explain what happens when the volume of the sealed vessel is increased
Is this correct:
Initially we increase the volume of the sealed vessel, decreasing the pressure and decreasing the concentration of particles (will this decrease the concentration of both gases or just one?). This will result in a lower concentration fraction/reaction quotient
Then, the system will partially oppose this it will try to increase the pressure ===> net forward reaction will be favoured and the concentration of NO₂ increasing and the concentration of N₂O₄ decreasing. This will increase the reaction quotient/concentration fraction until it is equal to the equilibrium constant.

I appreciate any help

Bump

[zsteve]

Hey guys, prepping for a SAC and I have some questions which I was given today:
1) In this reaction:
Fe³⁺(aq) + SCN^-(aq) <----> Fe(SCN)²⁺(aq) (0.50L)
We add 0.060 mol of Fe³⁺(aq) and 0.13 mol of SCN^-(aq). Once equilibrium is reached, 0.030 mol of Fe(SCN)²⁺(aq) has been formed
Find the equilibrium constant for the reaction
Is my reasoning correct (hence the answer)?
Because 0.030 mol of product has been formed and there is an increase in the amount of mol present, a net forward reaction has occured. Therefore, we have lost 0.030 mol from each of the reactants.
Fe³⁺(aq) + SCN^-(aq) <----> Fe(SCN)²⁺(aq)
0.060                         0.13                                  0 mol
-0.030 mol                 -0.030 mol                         0.030 mol

c = 0.030/0.50           c = 0.1/0.50                       c = 0.030/0.50
c = 0.06M                  c = 0.2M                            c = 0.06

K = 0.06/0.2 * 0.06 ===> K = 5M^-1

Haven't checked your calculations, but can verify that your method is clearly correct.

2) With the equilibrium reaction: N₂O₄(g) <------> 2NO₂(g)
We have to explain what happens when the volume of the sealed vessel is increased
Is this correct:
Initially we increase the volume of the sealed vessel, decreasing the pressure and decreasing the concentration of particles (will this decrease the concentration of both gases or just one?) - This will decrease concentration of BOTH gases. This will result in a lower concentration fraction/reaction quotient use 'reaction quotient'
Then, the system will partially oppose this it will try to increase the pressure Need to invoke Le Chatelier's Principle ===> net forward reaction will be favoured and the concentration of NO₂ increasing and the concentration of N₂O₄ decreasing. This will increase the reaction quotient/concentration fraction until it is equal to the equilibrium constant.

I appreciate any help

Overall, looking good. For (2), you want to specifically refer to Le Chatelier's Principle as the driving mechanism responsible for pushing the system back to equilibrium.

For instance,
As the volume is increased, the concentration of the chemical species N2O4 and NO2 drop and so does the pressure. By Le Chatelier's principle, the system will shift in the direction that opposes this change. In this case, the system shifts to the right, as more gaseous particles are produced (2NO2) than consumed (1xN2O4). This maximises the number of particles at equilibrium, (hence maximising the equilibrium pressure), thus partially opposing the initial pressure decrease.

Take the above with a grain of salt, not sure exactly how much detail is required for your SAC response

[blacksanta62]

Overall, looking good. For (2), you want to specifically refer to Le Chatelier's Principle as the driving mechanism responsible for pushing the system back to equilibrium.

For instance,
As the volume is increased, the concentration of the chemical species N2O4 and NO2 drop and so does the pressure. By Le Chatelier's principle, the system will shift in the direction that opposes this change. In this case, the system shifts to the right, as more gaseous particles are produced (2NO2) than consumed (1xN2O4). This maximises the number of particles at equilibrium, (hence maximising the equilibrium pressure), thus partially opposing the initial pressure decrease.

Take the above with a grain of salt, not sure exactly how much detail is required for your SAC response

Thank you so much zsteve, legend!!

[mary1911997]

need help please

5. A galvanic cell consists of one half cell that is made up of a copper electrode in a solution containing 1 M Cu2+ (aq) at 25 C.
Which one of the following could be used as the second half cell so that the polarity of the electrode in this second half cell is negative
   • A lead electrode in a solution of 1.0M Pb2+ (aq)
   • A silver electrode in a solution of 1.0M Ag+ (aq)
   • An inert electrode in a solution of 1.0M Fe2+ (aq) and 1M Fe 3+ (aq)
   • An inert graphite electrode in a solution of 1.0M Sn4+ (aq)


6. The rechargeable lithium iron phosphate cell is used to power small appliances such as portable computers. When the cell is being charged, the electrode reactions are
   LiFePO4 ----->FePO4   +    Li+    +   e-
   Li+   +   e-   +   6C  ----->  LiC6
   Which one of the following occur
   I. LiC6 is produced at the positive electrode
   II. The concentration of Li+ in the electrolyte increases
   III. The direction of electron flow in the external circuit is from the anode to the cathode

[zsteve]

need help please

5. A galvanic cell consists of one half cell that is made up of a copper electrode in a solution containing 1 M Cu2+ (aq) at 25 C.
Which one of the following could be used as the second half cell so that the polarity of the electrode in this second half cell is negative
   • A lead electrode in a solution of 1.0M Pb2+ (aq)
   • A silver electrode in a solution of 1.0M Ag+ (aq)
   • An inert electrode in a solution of 1.0M Fe2+ (aq) and 1M Fe 3+ (aq)
   • An inert graphite electrode in a solution of 1.0M Sn4+ (aq)


6. The rechargeable lithium iron phosphate cell is used to power small appliances such as portable computers. When the cell is being charged, the electrode reactions are
   LiFePO4 ----->FePO4   +    Li+    +   e-
   Li+   +   e-   +   6C  ----->  LiC6
   Which one of the following occur
   I. LiC6 is produced at the positive electrode
   II. The concentration of Li+ in the electrolyte increases
   III. The direction of electron flow in the external circuit is from the anode to the cathode

Hi there! Happy to help, but please specify what exactly you're not getting with the above questions. Once you've described that, happy to help out indeed