Chemistry 3/4 2013 Thread

[lolipopper]

question: Electrolysis of a molten metal chloride using a current of 6.50A for 1397 seconds deposits 1.41g of the metal at one of the electrodes. Identify the metal deposited.

 

[lzxnl]

Current * time = charge = 9080.5 C
One mole of electrons is 96485 C
So we have 9.4113 * 10^-2 mol electrons.

Let the charge on the metal cation be x. Then the mole ratio of electrons : metal cation = x : 1
We have 9.4113/x * 10^-2 mol metal
But this is 1.41 g, so the molar mass of the metal is 14.98x g/mol

No metal exists with molar mass of 14.98 g/mol in group one, so x is not 1. Similarly, x cannot be 2 as no metal has a molar mass of around 30 g/mol. However, x can be 3 as this leads to the metal scandium, the first of the transition metals which does have three valence electrons. This is your metal.

[lolipopper]

Current * time = charge = 9080.5 C
One mole of electrons is 96485 C
So we have 9.4113 * 10^-2 mol electrons.

Let the charge on the metal cation be x. Then the mole ratio of electrons : metal cation = x : 1
We have 9.4113/x * 10^-2 mol metal
But this is 1.41 g, so the molar mass of the metal is 14.98x g/mol

No metal exists with molar mass of 14.98 g/mol in group one, so x is not 1. Similarly, x cannot be 2 as no metal has a molar mass of around 30 g/mol. However, x can be 3 as this leads to the metal scandium, the first of the transition metals which does have three valence electrons. This is your metal.

so this had to be done by trial and error at the end? isn't there a straight method?

[lzxnl]

You don't know anything about the metal other than the fact that it is a metal and can form an ionic compound with chloride ions. That's all you know.

[lolipopper]

You don't know anything about the metal other than the fact that it is a metal and can form an ionic compound with chloride ions. That's all you know.

have you done the course??

[9_7]

Hello guys can someone please help me! I'm a bit confused with these questions

-When you are calculating the equilibrium constant, all the concentrations have to be at equilibrium right?
- when the it is an equilibrium reaction/equation or whatever, would the moles reactant 1 always equal to the moles of reactant 2 and vice versa?
- last of all, how do you improve the yield? Is it always a forward reaction? :s

I feel so stupid hahah, well if anyone can help I'd appreciate it!! Thanks

[lzxnl]

have you done the course??

I don't understand the purpose of your question. All I can say is that I've done a lot of the course.

Hello guys can someone please help me! I'm a bit confused with these questions

-When you are calculating the equilibrium constant, all the concentrations have to be at equilibrium right?
- when the it is an equilibrium reaction/equation or whatever, would the moles reactant 1 always equal to the moles of reactant 2 and vice versa?
- last of all, how do you improve the yield? Is it always a forward reaction? :s

I feel so stupid hahah, well if anyone can help I'd appreciate it!! Thanks

An equilibrium constant will only make sense at equilibrium as it only gives information about the concentrations at equilibrium. That's what it is defined for.

You'll have to be more specific for the second question.

As for the third, it's known as pushing the reaction more to the right. The main ways you could achieve this would be to either increase the volume (for gases if there are more moles of gas on the right hand side of the equation), decrease the volume (for gases if there are less moles of gas on the right hand side), dilute the solution with more solvent (if there are more dissolved particles on the right hand side), heat the system if the reaction is endothermic and cool it down if the reaction is exothermic.

[thushan]

- when the it is an equilibrium reaction/equation or whatever, would the moles reactant 1 always equal to the moles of reactant 2 and vice versa?

When the system is in equilibrium, the amount of the reactants don't need to be equal. They can be anything, so long as the concentration fraction (reaction quotient) is equal to K.

[Edward21]

When the system is in equilibrium, the amount of the reactants don't need to be equal. They can be anything, so long as the concentration fraction (reaction quotient) is equal to K.

So basically when all the concentrations come to a standstill, that's equilibrium yes, so once you calculate the reaction quotient and you get your value, that is equal to K which is your equilibrium constant at that specific temperature?

ALSO ok, I understand equlibrium with a reversible reaction is like a dynamic state in which the concentrations of each reactant or product don't change, that's cool, HOWEVER for the reaction of ethanoic acid in water, if only a little bit of the ethanoate ion is actually present, does that mean that the position of equilibrium for that reaction is to the left, where there's WAY more products than reactants formed??? I thought equilibrium was where the forward equals the back reaction, how the hell can this be when ethanoic acid only partially ionises to a tiny degree, that is totally not equal forward and backwards reactions!/? 

[9_7]

Oh okay!! thanks for clearing things up for me

[thushan]

ALSO ok, I understand equlibrium with a reversible reaction is like a dynamic state in which the concentrations of each reactant or product don't change, that's cool, HOWEVER for the reaction of ethanoic acid in water, if only a little bit of the ethanoate ion is actually present, does that mean that the position of equilibrium for that reaction is to the left, where there's WAY more products than reactants formed??? I thought equilibrium was where the forward equals the back reaction, how the hell can this be when ethanoic acid only partially ionises to a tiny degree, that is totally not equal forward and backwards reactions!/? 

It is actually. H+ and CH3COO- react with one another MUCH more readily than do H2O and CH3COOH. Remember that the reaction rate is dependent on more things than temperature and concentration - they are also dependent on things like the activation energy (which is specific to a specific reaction).

Hence, although H+ and CH3COO- are present in much lower concentrations, the rate of this reaction (the back reaction) at equilibrium is equal to the rate of ionisation of CH3COOH (although this, and water, are present in much higher concentrations).

[Edward21]

It is actually. H+ and CH3COO- react with one another MUCH more readily than do H2O and CH3COOH. Remember that the reaction rate is dependent on more things than temperature and concentration - they are also dependent on things like the activation energy (which is specific to a specific reaction).

Hence, although H+ and CH3COO- are present in much lower concentrations, the rate of this reaction (the back reaction) at equilibrium is equal to the rate of ionisation of CH3COOH (although this, and water, are present in much higher concentrations).

So the little amount of ionisation the CH3COO- does, it is the same rate as it is turned back into CH3COOH at equilibrium?

[Stick]

When we double/halve/do some other manipulation to the co-efficients of reactants and products in a chemical reaction, the value of the equilibrium constant changes accordingly. Does that mean its unit (powers of M) also changes?

[Alwin]

When we double/halve/do some other manipulation to the co-efficients of reactants and products in a chemical reaction, the value of the equilibrium constant changes accordingly. Does that mean its unit (powers of M) also changes?

No, unless you manipulate the equation by adding a new reactant or something. sorry, misread your q, I thought you said if we change the conditions like volume etc. Yes, it does change look at nliu's explanation for more detail.

Also, most exams qs ask for the value of the concentration fraction, so no units required

[lzxnl]

I am going to put it out here. VCAA has it wrong by having equilibrium constants with units. They DON'T have units. Otherwise the Gibbs free energy equation delta G = -RT ln K would make no sense. You can't take the natural log of something with units.
Strictly speaking, equilibrium constants relate chemical activities instead of concentrations, but we can usually make the approximation that the numerical value of the concentration in M or the partial pressure (yes, gas pressure not gas concentration) of the gas in bar (100 kPa) is equal to the chemical activity, which is dimensionless.
It's a pity VCE uses units -.-

In VCE language however, if you double the coefficients of reactants and products, the constant DOES change units. I'll give an example.
H+ + OH- <=> H2O
Let's assume that we have a neutral solution of [H+] = [OH-] = 10^-7 M
For this reaction (I don't need to state the states do I?), K = ([H+][OH-])^-1 = (10^-7 M * 10^-7 M)^-1 = 10^14 M^-2 at 298 K

Now look at 2H+ + 2OH- <=> 2H2O
Now K = ([H+]^2*[OH-]^2)^-1 = [10^-7 M]^-2 * [10^-7 M]*-2 = 10^28 M^4
And what do you know...the units have changed. It's because there's an exponential relationship between concentrations and reaction order, and if the sum of the indices of the top and bottom are different, then the units can change.