Enthalpy changes.

[Gareth56]

I'm brushing up on my chemistry and looking at Enthalpy and it says in my book 'Hill & Holman' that:

 

Standard Conditions are 1 atmosphere pressure, Temperature of 298K and substances being in their normal physical states.

 

Now H&H states that the definition of the Standard enthalpy change of formation of a substances is the heat evolved or absorbed when one mole of the substance is formed from its elements under standard conditions.

 

So in the reaction:

 

Mg(s) + 1/2O2 ----> MgO(s) DeltaH(standard formation) = -602kJ/mol

 

What I don't quite understand, probably because I've forgotten, is that if you want the above reaction to go then you have to burn the Mg in oxygen to get it to react to form MgO so how can the above reaction be under standard conditions if being under standard conditions involves carrying out the reaction at 298K. Surely burning the Mg is carrying out the reaction at temperature higher than 298K.

 

In other words a lump of Mg placed in contact with O2 at 298K will not react but the above reaction states that the value obtained is under standard conditions of 298K.

Edited July 17, 2008 by Gareth56

[insane_alien]

you can work out the enthalpy of reaction at any temperature. even if the reaction cannot take place at that temperature.

 

for the sake of consistency the standard was taken to be at STP.

 

The enthalpy of reaction is calculated by taking the enthalpies of formation of the products and subtracting the nethalpies of formation of the reactants. this will give you the enthalpy of reaction.

[Gareth56]

How do you do that then? Also putting the standard symbol in the Delta H isn't necessarily correct then?

[insane_alien]

ijust told you how to calculate the standard enthalpy of reaction. the whole bit about the difference in enthalpy between products and reactants.

 

the standard symbol is correct. if it is not at STP then putting it in is inncorrect and a note of the temperature must be made.

[Gareth56]

If you could possibly indulge me a tad further I'd be grateful.

 

The values of enthalpies of formation in tables are given under STP conditions (they are in my books anyway) which with regard to temperature is 298K. What I don't understand and forgive me for not understanding you concise explanation is how can they quote a value of -602kJ/mol for the formation of MgO at 298K when the reaction doesn't even go at that temperature.

[insane_alien]

well it does go as the magnesium oxidises albeit slowly.

 

The reaction doesn't need to be able to happen for you to get a reaction enthalpy as energy is a conserved quantity. For example, you could heat the magnesium up to 1000*C and let the reaction proceed there and then cool the products back down to STP and you'd have an enthalpy change of -602kJ/mol.

 

The only reasons standard reaction enthalpies are given at STP is for consistency. If they were all at temperature and pressure regimes where the reaction was possible then it would make multi reaction calculations more complex than they need to be.

[Gareth56]

For example, you could heat the magnesium up to 1000*C and let the reaction proceed there and then cool the products back down to STP and you'd have an enthalpy change of -602kJ/mol.

 

Thank you. For completeness of the answer sought I don't suppose you know the method of the above technique? How do you work out that you have an enthalpy change of -602kJ/mol after heating the Mg up to 1000*C then allowing it to cool?

[insane_alien]

well generally you don't.

 

you measure the formation enthalpies of the reactants and products then take the difference of them.

 

alternatively, if you have no choice you can use a bomb calorimeter.

[Gareth56]

well generally you don't.

 

you measure the formation enthalpies of the reactants and products then take the difference of them.

 

alternatively, if you have no choice you can use a bomb calorimeter.

 

 

OK, thanks anyway.

[CaptainPanic]

Enthalpy of formation (or reaction) is used to calculate the temperature change of a substance or mixture due to a reaction taking place. It is define so that all compounds involved are at 298K at the start of the reaction. The value given then says how much energy is available (or required). With the specific heat you can then calculate the temperature change.

 

So,

 

What happens in real:

You take a strip of magnesium, place it in air, ignite it, form MgO and a lot of heat. How much heat? Well, -602 kJ/mol ... (so, in this case the heat of combustion is the same as the heat of formation, but that's not always the case). So, that means that this heats up... and we knew this already, because this burns with a pretty beautiful and very hot flame

 

 

Example of the calculation:

So, say you want to do the combustion of magnesium (forming 1 mol MgO, or 24.31+16.00 = 40.31 g, or 0.04031 kg), and your strip of magnesium is nicely at 298K. And the air (oxygen) around you is also 298K.

 

It generates -602 kJ/mol... or -602 kJ/0.04031 kg = 14934 kJ/kg.

Then you look up the specific heat of MgO (877 J/kgK, or 0.877 kJ/kgK).

 

This enables you to calculate how hot the MgO will become (assuming it is not losing heat to the surrounding air):

14934 kJ/kg / 0.877 = 17028 K

 

That hot (the answer surprised me too)? Well… there is the fact that with every mole of O2, about 4 moles of N2 enter the system, and they too need to be heated up… so the actual temperature will be much less.

 

I’m too lazy to calculate that too… but I found that experiments have shown that magnesium in air (so, heating up all the nitrogen too) gives 2800 deg C temperatures! (They read the color of the flame to estimate the temperature, no thermometer will survive the heat).

[Gareth56]

Thank you very much for such a comprehensive explanation.

 

So basically what you're saying is that all the figures for standard enthalpy changes found in tables tell you that the reactions begin at 298K but the reactants may require an input of energy such as a flame as in the case off the Mg + O2 example to get them to go and the reactions themselves are NOT taking place at 298K.

 

Those 'O' levels were a long time ago

[hermanntrude]

gareth, it's also worth noting that enthalpy has only a partial effect on whether a reaction is spontaneous. This is why we see reactions which require energy and reactions which give out energy, both spontaneous. For more details on what determines whether a reaction "goes" at a specific set of conditions, read up on Gibb's free energy

[CaptainPanic]

Indeed... and as I like examples myself:

A common spontaneous reaction that actually requires extra energy: the dissolution of salt (NaCl).

 

When you dissolve a lot of salt into water, the solution will cool down... but it's still spontaneous... just to show that the two are not necessarily linked...

[Gareth56]

Presumably this is why a petrol/oxygen mixture requires a spark to ignite it? On it's own at 298K the petrol/oxygen mixture is stable with respect to the products of the combustion.

[CaptainPanic]

If you are interested to learn more about spontaneous reactions, and non-spontaneous reactions, I suggest that you study the topic of "Gibbs free energy" a bit, and then come back to this forum if you have any questions. To explain the whole theory goes perhaps a bit far at this moment... besides, the wiki page is not too bad. It would help if you understand "entropy" too.