- #1
DocZaius
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I am curious about a statement that is usually tacked onto the end of a derivation showing enthalpy with constant pressure being equal to the heat into the system.
First Law of Thermodynamics
[itex]
\Delta U = Q - W_{by}
[/itex]
Define Enthalpy and look at its change:
[itex]
H=U+PV \\
\Delta H = \Delta U + \Delta (PV)
[/itex]
Plug in First Law:
[itex]
\Delta H = Q - W_{by} + \Delta (PV)
[/itex]
Under constant pressure:
[itex]
\Delta H = Q - P\Delta V + P\Delta V \\
\Delta H = Q
[/itex]
And here is the part that I've heard and read more than once (Khan uses it in his enthalpy video for example). It is said that considering enthalpy at a constant pressure is useful for us, since that is the condition under which most of our chemical experiments occur (e.g. those done inside test tubes, etc.). But I always wonder why we can say that. My intuition would be that when a chemical reaction occurs, particularly a violent one, the pressure conditions at the center of the reaction would not necessarily adhere to those of its environment. I understand that an open test tube would be connected to the atmospheric pressure of the room, but surely for a short period of time and under very local conditions (e.g. at the center of the reaction) there must be a significant pressure gradient around the reaction? I would think that force per area in surfaces considered around the reaction would be relatively high. And is it not during that short period of time and at that specific location that we are considering quantities such as enthalpy?
My intuition appears to be wrong in this case, since clearly people have been using enthalpy under constant pressure as a useful quantity for a long time. But I am curious why it is that before any equilibrium is reached, we can safely assume constant pressures in energetic chemical reactions.
First Law of Thermodynamics
[itex]
\Delta U = Q - W_{by}
[/itex]
Define Enthalpy and look at its change:
[itex]
H=U+PV \\
\Delta H = \Delta U + \Delta (PV)
[/itex]
Plug in First Law:
[itex]
\Delta H = Q - W_{by} + \Delta (PV)
[/itex]
Under constant pressure:
[itex]
\Delta H = Q - P\Delta V + P\Delta V \\
\Delta H = Q
[/itex]
And here is the part that I've heard and read more than once (Khan uses it in his enthalpy video for example). It is said that considering enthalpy at a constant pressure is useful for us, since that is the condition under which most of our chemical experiments occur (e.g. those done inside test tubes, etc.). But I always wonder why we can say that. My intuition would be that when a chemical reaction occurs, particularly a violent one, the pressure conditions at the center of the reaction would not necessarily adhere to those of its environment. I understand that an open test tube would be connected to the atmospheric pressure of the room, but surely for a short period of time and under very local conditions (e.g. at the center of the reaction) there must be a significant pressure gradient around the reaction? I would think that force per area in surfaces considered around the reaction would be relatively high. And is it not during that short period of time and at that specific location that we are considering quantities such as enthalpy?
My intuition appears to be wrong in this case, since clearly people have been using enthalpy under constant pressure as a useful quantity for a long time. But I am curious why it is that before any equilibrium is reached, we can safely assume constant pressures in energetic chemical reactions.
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