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).