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:
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.
