How can we convert DeltaH_f^@ at 298K to DeltaH_f^@ at different temperatures ?? Also tell how to convert bond energy at 298K to bond energy at different temperature

1 Answer
Mar 18, 2017

You'd need to heat the product from standard temperature to nonstandard temperature at the same pressure and add that heating enthalpy.

However, I don't think bond enthalpy changes significantly at a different temperature. We can think of it like an activation energy for that particular molecule's dissociation reaction (in the gas phase to keep it simple):

"AB"(g) -> "A"(g) + "B"(g)

Activation energy is almost completely temperature independent, so we can assume that so is bond enthalpy.


Suppose you have the DeltaH_f^@ for "NH"_3 at "298.15 K", for the reaction

3"H"_2(g) + "N"_2(g) -> 2"NH"_3(g),

and suppose the reactant concentrations were constantly replenished such that the equilibrium is constantly pushed way towards "NH"_3.

The DeltaH_f^@ (=DeltaH_"rxn"^@ since all reactants were in their standard states) for "NH"_3(g) in this reaction is -45.94 pm 0.35 "kJ/mol" from the NIST webbook.

This is the thermodynamic cycle for this process, to get the big picture:

Usually we know DeltaH_"rxn"^@. To fill in DeltaH for heating curves, we have the following formula for heating from T_1 to T_2:

DeltaH_(T_1)^(T_2) = int_(T_1)^(T_2) C_P dT

This requires that you know the constant-pressure heat capacity of
your compound(s) and that you assume that it stays constant in the relevant temperature range.

C_P can be looked up via the NIST webbook. For example, "NH"_3 has:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C7664417&Type=JANAFG&Plot=on

a C_P of ~~ "35.64 J/mol"cdot "K" at "298.15 K" and ~~ "42.01 J/mol"cdot"K" at "500 K".

So, from using the average C_P in between the two temperatures, heating "NH"_3 would approximately contribute a DeltaH of:

DeltaH_(298.15)^(500) = int_(298.15)^(500) (42.01 + 35.64)/2 dT

~~ 38.825T_2 - 38.825T_1

= 38.825DeltaT

= 38.825(500 - 298.15)

= "7836.83 J/mol" -> "7.937 kJ/mol"

(notice the resemblance to "q = mc_sDeltaT".)

The DeltaH_f^@ for "NH"_3(g) was -45.94 pm 0.35 "kJ/mol" from the NIST webbook. So, at "500 K", it would be approximately:

(-45.94 + 7.937) "kJ/mol" ~~ color(blue)(-"38.10 kJ/mol")

It's less negative at a higher temperature, meaning that it is less exothermic to form "NH"_3(g) at higher temperatures.

This makes physical sense because we are just assuming the heat released due to making it at a lower temperature is re-absorbed to heat the surroundings further (thus making the enthalpy released less negative).