Sciencemadness Discussion Board

Enthalpy - Clearing a confussion

Sniffity - 26-8-2015 at 14:37

Hey!

I've had a sort of blackout, and can't figure this out as easy as it is probably is. >_>

So; I know that when analyzing a system under CONSTANT PRESSURE, Enthalpy = q(p), meaning heat released or absorbed by the system in a thermodynamic process.

Thus; if a reaction under CONSTANT pressure is exothermic, it'll have a negative deltaH, and if it's endothermic, it'll have a positive deltaH.

So far, no doubts.

But; two doubts:

When someone says: The Heat released in a reaction is 50 kJ. Give me the change in enthalpy.

In theory, being picky about it, saying -50kJ would be wrong, because we don't know the reaction was carried out under constant pressure. Correct?

If the reaction is not carried out under constant pressure: What can we say about Enthalpy? Does it become meaningless? Can we still say that (delta)H is negative for exothermic and positive for endothermic process?

Another non-related question that's been bugging me: We say that a mole of a given gas occupies a volume of 24L at STP, aprox. So, if I was to put one mole of a given gas into a 25 L container at STP... wouldn't it expand to fill the container? My reasoning is that it would expand, but this would come along with a change in its temperature (an increase) thus it would no longer be at STP. Is this correct?

[Edited on 26-8-2015 by Sniffity]

blogfast25 - 26-8-2015 at 15:21

When someone says "the heat released by the reaction is 50 kJ" they mean just that. It means that in the specific conditions of reaction (which may well be non-STP) the change in Enthalpy was - 50 kJ.

A mol of gas taking up 25 L instead of 24 L would be (slightly) non-STP. It could very well be at standard temperature, in which case its pressure will be slightly below STP.

gsd - 26-8-2015 at 16:02

Say you have a completely empty container having volume 22.4 lit. Now add 1 gm mole of gas into it. Maintain the gas temperature at 273.15 K, then you will have the pressure inside the container as 1 Bar.

The ideal gas equation is: (P1*V1)/T1 = (P2*V2)/T2

Now if your container has a volume of 25 lit, then the inside pressure will be:

(22.4/25) = 0.896 Bar.

Alternatively if you want internal pressure as 1 Bar then you will have to maintain the temperature at:

(25/22.4)*273.15 = 304.85 K (31.7 Deg C)

Gsd

Sniffity - 26-8-2015 at 16:27

Quote: Originally posted by blogfast25  
When someone says "the heat released by the reaction is 50 kJ" they mean just that. It means that in the specific conditions of reaction (which may well be non-STP) the change in Enthalpy was - 50 kJ.


But to talk about enthalpy, the conditions of the reaction, whatever they are, should include constant pressure, correct? Otherwise we would not calculate the pressure-volume work by simply applying P(delta)V in the enthalpy equation.