VCE Chemistry Question Thread

[ArtyDreams]

Hey guys,

I received this question from my teacher and gave this answer: https://snipboard.io/CU8IiZ.jpg
The correct answer was this: https://snipboard.io/FroaA7.jpg
My intuition told me that this would be the case, but I couldn't explain to myself why.

Do all lone atoms of an element group together in twos? Why is this pattern important and what makes it occur?

Many thanks,
Corey

There are 7 elements on the periodic table that occur naturally as diatomic elements. These are: Hydrogen, Nitrogen, Fluorine, Oxygen, Iodine, Chlorine and Bromine. (You can remember these by the acronym Have No Fear of Ice Cold Beer).

There, Iodine only exists as an diatomic molecule.

The reason that diatomic elements occur is beyond the scope of VCE Chemistry, and there are lots of articles on google that go into a lot of detail on it. But basically, it is due to the fact that for these diatomic elements, they are much more electronically stable when they exist in pairs. Binding to another molecule helps them exist in a state of lower energy - this is the aim for a lot of atoms

[Chocolatepistachio]

if someone could help with this question

[Snow Leopard]

if someone could help with this question

You would do an ICE (one row for the initial concentrations, one row for change in concentrations and a final row in equilibrium concentrations) table for that question and then sub in your values into the Kc equation = [products]/[reactants]

[Chocolatepistachio]

How would I work out the equilibrium concentration in the question it only gives initial

[wingdings2791]

Hi Chocolatepistachio,
I suggest using a variable to represent the unknown concentrations. As the experiment is investigating the decomposition of NOCl_(g) and the initial amount of NOCl_(g) is given, the initial:



As it is unknown how much NOCl_(g) is reacted to form NO_(g) and Cl_2(g), we can use a variable to represent the change in mole. This change will be proportional to the NOCl:NO:Cl₂ ratio of 2:2:1.



meaning that since the volume of the flask is 2.00L, equilibrium moles will be



Now the equilibrium constant. Following

and the given equilibrium constant (at 35^oC, presumably the temperature of this experimental setup)
we can construct the equation
from which we can get

Applying polynomials, you would eventually come to
which can be substituted back into the original equilibrium mole values to get our final answers of:



at equilibrium.

Hope this helps, please let me know if there are any errors

[Chocolatepistachio]

Isn’t the equilibrium concentration for NOCl 0.500-2x , And NO 2x

[fun_jirachi]

Isn’t the equilibrium concentration for NOCl 0.500-2x , And NO 2x

No, since the entire numerator has to be divided by the denominator. \(\frac{1-2x}{2} = 0.5-x\), not \(0.5-2x\). This also applies for the concentration of NO.

[Chocolatepistachio]

but itsn't the equilibrium concentration just the initial + change like this

[Corey King]

Hey guys,

https://snipboard.io/26fX9O.jpg

Appreciate your help

[wingdings2791]

Hey guys,

https://snipboard.io/26fX9O.jpg

Appreciate your help

Hi Corey King,
Electrodes can be positive or negative depending on each electrode's proportion of electrons. It's the same case with electrolytic cells; whilst galvanic cells have a positive cathode (electrodes consumed in reduction, therefore net decrease in free electrons) and negative anode (electrodes produced in oxidation, therefore net increase in free electrons), electrolytic cells have a negative cathode (electrons pushed towards the cathode by power supply) and a positive anode (electrons produced by oxidation are transferred to the cathode by the power supply, leaving the positive ions from oxidation at the anode).

As you've described, the oxidation of Cu_(s) will release Cu²⁺_(aq) and electrons into the solution at the anode, which on its own does not have an effect on overall charge. The continuous transport of electrons from the anode to the cathode by the power supply maintains the positive charge of the anode. If you like, you can think about it as the electrons produced by oxidation being transferred to the cathode so they can be consumed in reduction.

[Corey King]

So why is the charge positive from the anode if it's producing e-? Which are negative

[ArtyDreams]

So why is the charge positive from the anode if it's producing e-? Which are negative

If you take a standard electric circuit, the current moves out of the positive side into the negative side. This is the same case here, where the electrons need to flow out of the positive side into the negative anode. This is why the anode is positive in an electrolytic cell, as the current always flows from anode to cathode.

[wingdings2791]

but itsn't the equilibrium concentration just the initial + change like this

What you've labeled as equilibrium concentration is actually equilibrium mole (\(n\) of each reactant/product), not concentration. You need to divide by the volume (which is given as 2.00L in the question) to obtain equilibrium concentrations.

[Corey King]

If you take a standard electric circuit, the current moves out of the positive side into the negative side. This is the same case here, where the electrons need to flow out of the positive side into the negative anode. This is why the anode is positive in an electrolytic cell, as the current always flows from anode to cathode.

Ohh I see, so positive refers to Direction of electron flow, not the Charge of the current that is flowing?

The the anode has negative charge (is this polarity) and positive charge in its electrode for a primary cell,

But in a secondary cell the anode is switched to the opposite electrode, and the electrode it now exists with is still is still providing electron Current so the current is positive?

[Chocolatepistachio]

For this question at point A would the volume have decreased (increasing pressure) and reactions shifts to the left

At point B would the temperature have decreased, and the reaction shifts to the left

At point c would some C2H6 have been added and the reactions shifts to the left