Stankovic123's chem q's

[lzxnl]

Slight rephrasing; there was an error in my previous argument.

The nickel forms cations. This repels the hydrogen cations, so the hydrogen cations go over to the neutral copper where there is less repulsion. The hydrogen ions, being positive charged, attract electron density from both the copper and nickel, but the nickel gives up electrons more readily, so the nickel pushes electron density through the copper, which then pushes electron density to the hydrogen ions.



As for the energy efficiency, I'll provide an example. Suppose you had fuel A with a molar mass of 100 g/mol and it produced 100 kJ of energy per mol that reacted. Suppose fuel B had a molar mass of 1000 g/mol and it produced 200 kJ of energy per mol that reacted. Now, the only reason fuel B has a higher enthalpy of combustion is because its molecules are larger than fuel A's molecules, so there are more bonds reforming. However, B is ten times as massive as A, yet it only produces twice as much energy as A. If A were of the same amount as B, i.e. we had 1000g of A and 1000g of B, you can see that A will produce more energy than B. Although we have different molar quantities of A and B, the masses of the atoms that make up these amounts of A and B are the same. So enthalpy per gram gives a better measurement of how much energy is released per atom and thus how efficient these fuels are.

[zvezda]

Slight rephrasing; there was an error in my previous argument.

The nickel forms cations. This repels the hydrogen cations, so the hydrogen cations go over to the neutral copper where there is less repulsion. The hydrogen ions, being positive charged, attract electron density from both the copper and nickel, but the nickel gives up electrons more readily, so the nickel pushes electron density through the copper, which then pushes electron density to the hydrogen ions.



As for the energy efficiency, I'll provide an example. Suppose you had fuel A with a molar mass of 100 g/mol and it produced 100 kJ of energy per mol that reacted. Suppose fuel B had a molar mass of 1000 g/mol and it produced 200 kJ of energy per mol that reacted. Now, the only reason fuel B has a higher enthalpy of combustion is because its molecules are larger than fuel A's molecules, so there are more bonds reforming. However, B is ten times as massive as A, yet it only produces twice as much energy as A. If A were of the same amount as B, i.e. we had 1000g of A and 1000g of B, you can see that A will produce more energy than B. Although we have different molar quantities of A and B, the masses of the atoms that make up these amounts of A and B are the same. So enthalpy per gram gives a better measurement of how much energy is released per atom and thus how efficient these fuels are.

Ahhhhhh. I see now.

However, with the oxidation of nickel by itself in the acid solution without the copper, wouldnt the nickel cations just as easily repel the hydrogen cations?

[lzxnl]

Yes, without the copper, the nickel cations would repel the hydrogen ions. The reaction rate would decrease without the copper.

[zvezda]

Yes, without the copper, the nickel cations would repel the hydrogen ions. The reaction rate would decrease without the copper.

Oh yeah. You were saying though how the hydrogen ions attract electron density from BOTH copper and nickel, but why not just nickel??

Also, when we talk about equilibrium having been reached in a primary cell, what is necessarily meant by this? If products remain in contact with the electrodes (as heinemann describes equilibrium as the build up of products on the electrodes), why cant primary cells be recharged?

[lzxnl]

Oh yeah. You were saying though how the hydrogen ions attract electron density from BOTH copper and nickel, but why not just nickel??

Also, when we talk about equilibrium having been reached in a primary cell, what is necessarily meant by this? If products remain in contact with the electrodes (as heinemann describes equilibrium as the build up of products on the electrodes), why cant primary cells be recharged?

Well...electrons are negative, and hydrogen ions are positive...logically H+ attracts electron density from both metals; nickel is just better at giving them up.

Equilibrium refers to when the voltage of the cell is zero. Primary cells generally DON'T remain in contact with the electrodes, as the products generally involve at least one aqueous product. Then, an equilibrium expression can be written in terms of its concentration.

[zvezda]

Hey,
There's this one question ive come across:
If the heat of combustion if alkanes is greater than their corresponding alcohols, what does this suggest about the bonding between alkanes and their corresponding alcohols?
Cheers in advance

[jgoudie]

Interesting question: This is what i would say.

The products of both combustions are Carbon Dioxide and Water.  this allows you to compare the relative enthalpy of alkanes and alcohols.

As alkanes release more energy this would suggest the bonds within alkanes are less stable than the bonds within alcohols.  The more stable the less enthalpy.

Happy to hear others points of view also.

Hey,
There's this one question ive come across:
If the heat of combustion if alkanes is greater than their corresponding alcohols, what does this suggest about the bonding between alkanes and their corresponding alcohols?
Cheers in advance

[zvezda]

Interesting question: This is what i would say.

The products of both combustions are Carbon Dioxide and Water.  this allows you to compare the relative enthalpy of alkanes and alcohols.

As alkanes release more energy this would suggest the bonds within alkanes are less stable than the bonds within alcohols.  The more stable the less enthalpy.

Happy to hear others points of view also.

Ahh yeah. Good point.
However one thing, why is it that less enthalpy means more stable?
Thanks

[lzxnl]

More negative delta H => stronger bonds formed or weaker bonds broken (think about it)
The products are the same => same bonds formed, so weaker bonds broken if the delta H is more negative.
Lower enthalpy means less chemical energy (not quite, but you don't need to worry about that) so it means more stability. Think about it. H-H and O-O bonds are quite weak, but O-H bonds and the hydrogen bonding in water are quite strong, so the combustion of hydrogen gas is very exothermic given the size of hydrogen.

[zvezda]

More negative delta H => stronger bonds formed or weaker bonds broken (think about it)
The products are the same => same bonds formed, so weaker bonds broken if the delta H is more negative.
Lower enthalpy means less chemical energy (not quite, but you don't need to worry about that) so it means more stability. Think about it. H-H and O-O bonds are quite weak, but O-H bonds and the hydrogen bonding in water are quite strong, so the combustion of hydrogen gas is very exothermic given the size of hydrogen.

Yeah yeah that makes sense. Thanks guys

[Mao]

Hey,
There's this one question ive come across:
If the heat of combustion if alkanes is greater than their corresponding alcohols, what does this suggest about the bonding between alkanes and their corresponding alcohols?
Cheers in advance

That is an interesting question. I would like to point out that generally, these kind of direct comparisons can only be made for isomers (e.g. cyclohexane vs hexene), but not for molecules with differing chemical formulas. In this particular case though, we can make a direct comparison of "bond strength" (i.e. enthalpy of formation), because:

1. The other reactant in combustion is O2, an elementary molecule, and by definition has an enthalpy of formation of zero.

2. The products for the combustion of alkane vs alkanol are exactly the same.

The only difference between the two combustion reactions is the difference of half of an oxygen molecule on the reactant side, which has an enthalpy of formation of zero, so does not affect the calculation of the enthalpy of other species. This is why we can directly compare alkane and alkanol, even though they are not isomers.

Consider another question, that asks you to interpret the heat of combustion of an alkane and an alkene. Since the products are not exactly the same (they differ by one water molecule), you cannot meaningfully compare the formation enthalpies UNLESS you know the enthalpy of formation of water.

Just something to keep in mind.

[zvezda]

That is an interesting question. I would like to point out that generally, these kind of direct comparisons can only be made for isomers (e.g. cyclohexane vs hexene), but not for molecules with differing chemical formulas. In this particular case though, we can make a direct comparison of "bond strength" (i.e. enthalpy of formation), because:

1. The other reactant in combustion is O2, an elementary molecule, and by definition has an enthalpy of formation of zero.

2. The products for the combustion of alkane vs alkanol are exactly the same.

The only difference between the two combustion reactions is the difference of half of an oxygen molecule on the reactant side, which has an enthalpy of formation of zero, so does not affect the calculation of the enthalpy of other species. This is why we can directly compare alkane and alkanol, even though they are not isomers.

Consider another question, that asks you to interpret the heat of combustion of an alkane and an alkene. Since the products are not exactly the same (they differ by one water molecule), you cannot meaningfully compare the formation enthalpies UNLESS you know the enthalpy of formation of water.

Just something to keep in mind.

What do you mean by enthalpy of formation?

I was also wondering, how can redox reactions be either exo or endothermic of they can be at all?
Thanks

[lzxnl]

The enthalpy of formation is the enthalpy change when one mole of the substance is formed from its elements at standard states, the state that the elements would be in at 298 K and 101.3 kPa, and the most stable form is chosen (so diatomic nitrogen as opposed to monoatomic nitrogen, graphite for carbon etc).
As an example, the enthalpy of formation of diatomic oxygen gas is zero, because it IS the elemental form of oxygen. For ozone, however, its enthalpy of formation is the enthalpy difference between 3/2 oxygen molecules and 1 ozone molecule.
For something more complex like C6H6, benzene, the enthalpy of formation of this is the enthalpy change of the reaction 6C(graphite)+3H2(g) => C6H6(l)

Redox reactions can be either exothermic or endothermic. What's wrong with that? Perhaps heat transfer occurs when the electrons are transferred too.

[zvezda]

The enthalpy of formation is the enthalpy change when one mole of the substance is formed from its elements at standard states, the state that the elements would be in at 298 K and 101.3 kPa, and the most stable form is chosen (so diatomic nitrogen as opposed to monoatomic nitrogen, graphite for carbon etc).
As an example, the enthalpy of formation of diatomic oxygen gas is zero, because it IS the elemental form of oxygen. For ozone, however, its enthalpy of formation is the enthalpy difference between 3/2 oxygen molecules and 1 ozone molecule.
For something more complex like C6H6, benzene, the enthalpy of formation of this is the enthalpy change of the reaction 6C(graphite)+3H2(g) => C6H6(l)

Redox reactions can be either exothermic or endothermic. What's wrong with that? Perhaps heat transfer occurs when the electrons are transferred too.

Ah ok. But then touching on mao's point: why would he mention the presence of elementary oxygen as being necessary to figure the original question out? Cant you easily have another molecule as thw reactant and still make some sort of conclusion about the bonding strengths of alkanes and their corresponding alcohols?

Also, i came across a question on the AN guide which mentioned that absorption at 1600cm^-1 on the IR spectrum suggests the presence of a carbonyl group and not a C=C??

[lzxnl]

Ah ok. But then touching on mao's point: why would he mention the presence of elementary oxygen as being necessary to figure the original question out? Cant you easily have another molecule as thw reactant and still make some sort of conclusion about the bonding strengths of alkanes and their corresponding alcohols?

Also, i came across a question on the AN guide which mentioned that absorption at 1600cm^-1 on the IR spectrum suggests the presence of a carbonyl group and not a C=C??

Mao's point is pretty clever.
I'll illustrate with an example.

Let's say we want to compare ethane and ethanol, or C2H6 and C2H5OH
The reactions are:
C2H6 + 3.5 O2 => 3H2O + 2CO2     Reaction 1
C2H5OH + 3O2 => 3H2O + 2CO2    Reaction 2

The products are exactly the same.
So if we want to see which is more stable, we consider the reaction C2H6 + 0.5 O2 => C2H5OH
By Hess's law, the enthalpy change of this reaction is equal to the enthalpy of formation of the right hand side, minus the enthalpy of formation of the left hand side. As the enthalpy of formation of oxygen gas is zilch, the enthalpy change of this reaction allows us to compare the enthalpies of ethane and ethanol.
This would be done by, saying, flipping reaction 2 around and adding that to reaction 1, so that the enthalpy change of ethane to ethanol is simply the enthalpy of combustion of ethane minus the enthalpy of combustion of ethanol.