Negative Delta H, Temperature Increase
Date: Winter 2010-11
I am sure there is a simple response to this question but
I have begun chasing my tail on this concept. How can delta H be
negative (system's point of view) and temperature increase? For
example, the combustion of methanol.
It is just a matter of convention. Delta H (enthalpy change) is
simply DEFINED as the heat energy added to a system in a
process. If delta H is negative, that just means that the system
Richard E. Barrans Jr., Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming
Mathematically, dH = H(final) - H(initial), which means that if dH
is negative that H(final) is less than H(initial).
Since enthalpy change is another way of saying transfer of heat when
no PV work is involved (otherwise it would be internal energy), this
means that when dH is negative, heat is released from the system.
This would also account for the first statement above. The only way
H(final) is less than H(initial) is if heat was released from the
If heat is released from the reaction system, heat must enter the
environment. Since we can only measure the environment temperature -
we cannot stick a thermometer into the reaction itself (since the
"reaction system" are individual molecules reacting, we cannot
capture by thermometer the temperature change of the individual
molecules), we note that the temperature of the environment
increases, conclude that the environment gained heat, and infer (by
First and Third Law of Thermodynamics) that the reaction system must
have released heat, H(final) < H(initial), and dH is negative.
Greg (Roberto Gregorius)
The temperature increase that you record for an exothermic reaction
is in the surroundings. So, if you burn methanol in a spirit burner,
the system is the methanol in the wick and the oxygen it meets, this
releases heat and the hot gases rise to increase the temperature of,
for example, a beaker of water above the flame. It also radiates
heat to your hand if you hold it by the flame and this increases the
temperature of your skin. In all cases heat is lost from the system
(the reaction) to the surroundings. The system loses enthalpy
(stored heat energy) so delta H is negative and the temperature of
the surroundings increases.
You may not be happy with this explanation, although it is the usual
one given, as it depends on how we define the system and
surroundings. So here is another way to look at it:
All of these types of enthalpy changes involve the transformation of
energy from potential energy (from opposite charges held at a
distance from each other) to random kinetic energy (heat) or vice
versa. The higher the potential energy the higher the enthalpy of
the species, the higher the random kinetic energy the higher the
heat (and average kinetic energy per particle is proportional to the
temperature). In an exothermic change you move from higher potential
energy (higher enthalpy) in the reactants (corresponds to weaker or
longer bonds) to lower potential energy (lower enthalpy) in the
products (corresponds to stronger or shorter bonds). This "enthalpy"
released manifests itself as heat energy (random kinetic energy)
i.e. particles are "jiggling" about more. Initially, I suppose, the
particles that took part in the reaction are jiggling more, but this
is soon transferred to other particles and, eventually our
thermometer in the vicinity of the reaction. So we see a temperature increase.
In my experience, this is one of the most problematic concepts for
students of Chemistry and I believe if comes back to how we define
"enthalpy". It is not an easy concept for students to grasp and it
does not help that we cannot measure it directly and that it only,
in these types of systems, manifests itself when it is converted into heat.
Your question is common. The question results from some confusion regarding
the definition of the terms involved, so let us just "walk through the
definitions and issues".
1. The heat of reaction (technically, called the "enthalpy" of reaction) is
the amount of heat (in calories or Joules) either absorbed or given off
under a specified set of conditions of temperature and pressure. By
agreement these conditions are 25 C. (298.15 kelvins) and 1 atmosphere
pressure if all reactants and products are solids and/or liquids. While
certain "corrections" have to be applied if these conditions of temperature
and pressure are not met, for the most part these "corrections" are small
and only become important if you are analyzing thermochemical data, rather
than just looking for some measure of how much heat is absorbed or given
2. The next question is what is the change of the heat (enthalpy) of
reaction when the reaction (in your example the combustion of methanol
(CH3OH + 3/2 O2 ---> CO2 + 2 H2O) if the temperature is changed from 25 C.
to some other, but constant, temperature. Even at this "new" temperature
corrections have to be applied to whether the various reactants and products
are gases, liquids or solids, but these "details", while quantitatively
important, do not affect the concepts. The reactants collectively have a
certain heat capacity (The amount of heat lost by their disappearance in the
reaction, and the amount of heat gained by the products by their appearance
in the reaction). This difference dependences upon the chemical substances,
and the temperature difference between 25 C. and the "new" temperature. In
quantitative measurements factors in addition to the change in temperature,
( like a phase change from liquid to gas, for example) need to be applied.
3. So see, your question is not so simple when you want a complete answer.
For most cases, just knowing the delta H at the standard temperature 25 C.
and 1 atmosphere pressure is sufficient to show the trends. You are looking
at the heat absorbed or given off by a reaction and how that quantity
changes over fairly small changes in temperature doesn't make a large
4. All of the above is more complicated when solutions (that depend upon the
concentration of the reactants and products). In the case of solutions the
effects of solvents, especially water, can play a dominant role, but that is
too complicated to address here.
In Thermodynamics, by definition, combustion reactions are generally
strongly exothermic and therefore the change in Enthalpy (delta H) is
generally strongly negative; Meaning heat is generated and is
Consider the definition of enthalpy,
where H = U + pV
H is the enthalpy of the system,
U is the internal energy of the system,
p is the pressure at the boundary of the system and its environment, and
V is the volume of the system.
Hence, dH = dU + p dV + dp V, where dV and dp are both zero because in
a closed system, pressure and volume are constant. Therefore a change
in enthalpy is due to the change in the internal energy of a system.
So it follows, in an exothermic reaction, combustion for example, dU
is negative, meaning the internal energy of the system is decreasing
by way of releasing heat, and therefore, outside the system, the
The opposite case would be an endothermic reaction, that is, delta H
is positive, and therefore the reaction is an absorption of heat.
Hope that helps.
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Update: June 2012