Chemistry 3/4 2013 Thread

[clıppy]

Just another quick question that I can't wrap my head around.
Let's say I have a reaction with S and it releases 9kJ/g. If i wanted it in terms of kJ/mol I just have to multiply 9 * 32.1
I just can't seem to grasp what happened. Any help please?

[Conic]

It was multiplied by the molar mass of Sulfur. Multiplying by the molar mass cancels out the grams and makes it kJ/mol (ie,  ).

[clıppy]

It was multiplied by the molar mass of Sulfur. Multiplying by the molar mass cancels out the grams and makes it kJ/mol (ie,  ).

Ahhhh! Cheers Conic

[Edward21]

How is it that solid Li is produced from electrolysis on molten Li2Cl and KCl? At the cathode reduction occurs, Li+ and K+ can both be reduced, isn't the stronger oxidant the one that actually does get reduced?? ie. K+ over Li+ I'm really confused

[SocialRhubarb]

The electrochemical series is valid at a temperature of 25 degrees Celsius, a pressure of 1 atmosphere and with concentrations of 1 M. If you have molten LiCl and KCl, it's probably not at a temperature of 25 degrees Celsius or at a concentration of 1 M.

[Edward21]

The electrochemical series is valid at a temperature of 25 degrees Celsius, a pressure of 1 atmosphere and with concentrations of 1 M. If you have molten LiCl and KCl, it's probably not at a temperature of 25 degrees Celsius or at a concentration of 1 M.

Haha ok, that makes sense!! Molten at 25 degrees, whoops 

[clıppy]

Regarding efficiencies in calculating energy from reactions:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) ; Enthalpy = -890 kJ/mol
a) If a student used 2MJ (1MJ = 1 * 10^6 J) of heat energy to heat up a room with a natural gas heater, calculate the mass of natural gas (methane) used for this purpose if the conversion of chemical energy to heat energy was 85% efficient

2 Questions. Does the order (chemical energy to heat energy) matter?
How do I account for the 85%. I know I'm going to require more mass to get the same amount of heat, so how do I calculate it?

[Patches]

Does that seem right? Multiplying by 1/.85 should give you the amount used accounting for 85% efficiency.

[clıppy]

Does that seem right? Multiplying by 1/.85 should give you the amount used accounting for 85% efficiency.

The answer was 42.3g so yeah I think that's right.
I'm confused with bit, what did you do here?

[lzxnl]

The electrochemical series is valid at a temperature of 25 degrees Celsius, a pressure of 1 atmosphere and with concentrations of 1 M. If you have molten LiCl and KCl, it's probably not at a temperature of 25 degrees Celsius or at a concentration of 1 M.

Perhaps the electrode potentials of lithium and potassium ions NOT in aqueous solution are different than when they're molten. Remember, aqueous lithium and potassium ions are solvated by water; that may have an effect on the oxidising strengths of lithium and potassium

SocialRhubarb, using the Nernst equation, we'd see that if we had the same concentrations of lithium and potassium ions at the same temperature, as when oxidised they both only give up one electron, solutions containing lithium and potassium ions may have different electrode potentials, but it will not invert the order of the series.

I'm not sure how to define a reaction quotient for an ionic liquid...

[SocialRhubarb]

Are they necessarily at equal concentrations though?

This seems to me like a situation which would probably occur in the industrial production of lithium metal, where lithium metal is produced from lithium chloride, and the potassium chloride is added thereafter for practical purposes. I would guess that a small amount of potassium chloride is added to decrease the melting temperature of the compound, reducing the energy required for the process and hence reducing cost.

Obviously it wasn't specified in the question of the exact conditions, but it sounds to me like an industrial application rather than an experiment, say, in a lab.

[lzxnl]

They may not be, that is true. Of course, if you had that much more lithium ions, then the reduction of lithium ions would be favoured. I may need to see the question.
I'm hopeless with industrial applications of chemistry anyway.

[Edward21]

I'm hopeless with industrial applications of chemistry anyway.

I doubt that 

[clıppy]

In regards to equilibrium constants, how does temperature affect them?

For example in this question:

For a particular equilibrium system, it was found that as the temperature was increased the value of the equilibrium constant also increased. For the forward reaction of this system it would be true to say that
A - all the reactants are gases
B - it is an endothermic process
C - the number of mole of products must be less than the number of mole of reactants
D - it produces heat energy

The answer is B, but i'm not sure as to why because I can't recall how K values are affected by temperature. Any help?

[lzxnl]

In regards to equilibrium constants, how does temperature affect them?

For example in this question:
The answer is B, but i'm not sure as to why because I can't recall how K values are affected by temperature. Any help?

If we have a reaction that's at equilibrium, and we increase the temperature, the system will try to counter this with a reaction that reduces the temperature. Well, an endothermic reaction does that, so the endothermic reaction will be favoured by temperature rises.