What is the enthalpy of reaction at #"373 K"# if the heat capacities are the same at #"273 K"# and we know #DeltaH_(rxn)# at #"273 K"#?

2 Answers
SCooke
Nov 5, 2017

A) -20kJ

Explanation:

Enthalpy of a reaction is NOT affected by the temperature. The temperature may affect the reaction rate , but the inherent amount of heat energy produced or consumed by a reaction remains the same no matter what the conditions of the reaction environment are.

Ref:
https://cbc-wb01x.chemistry.ohio-state.edu/~woodward/ch121/ch5_enthalpy.htm

https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Thermodynamics/State_Functions/Enthalpy/Heat_of_Reaction

Truong-Son N.
Nov 5, 2017

Due to the assumption of the question that #C_P# is equal between both the products and reactants at any given temperature, ONLY THEN can we say that #DeltaH_"rxn"# did not change in that temperature range.

Otherwise, there is a small difference. Remember, this is just Hess's Law...

#DeltaH_("rxn","373 K") = DeltaH_("rxn,273 K") + cancel(int_(273)^(373) DeltaC_(P,"rxn")(T)dT)^(~~ 0)#

if #C_(P("Products"))(T) = C_(P("Reactants"))(T)# for both temperatures.

See here for further discussion:
https://chemistry.stackexchange.com/questions/35756/calculating-enthalpy-changes-at-different-temperatures

And see here for further references to back up this answer.

We can write a thermodynamic cycle for that.

  • State 1: Reactants at #"273 K"#
  • State 2: Products at #"273 K"#
  • State 3: Reactants at #"373 K"#
  • State 4: Products at #"373 K"#

#"State 1" stackrel(DeltaH_("rxn","273 K")" ")(->) "State 2"#

#"State 3" stackrel(DeltaH_("rxn","373 K")" ")(->) "State 4"#

#"State 1" stackrel(int_(273)^(373) C_(P("reactants"))(T)dT" ")(->) "State 3"#

#"State 2" stackrel(int_(273)^(373) C_(P("products"))(T)dT" ")(->) "State 4"#

We know the first one, and in principle can determine the last two, but do not know the second one. If we want to go from #3# to #4#, we can pick the path as follows:

#"Reactants at 373 K" -> "Reactants at 273 K" -> "Products at 273 K" -> "Products at 373 K"#

This path is then going to result in calculating #DeltaH_"rxn"# for the reaction at #"373 K"#:

#color(blue)(DeltaH_("rxn","373 K")) = -int_(273)^(373) C_(P("reactants"))(T)dT + DeltaH_("rxn,273 K") + int_(273)^(373) C_(P("products"))(T)dT#

#= color(blue)(barul(|stackrel(" ")(" "DeltaH_("rxn,273 K") + int_(273)^(373) DeltaC_(P,"rxn")(T)dT" ")|))#

See here; this person quotes the same equation I just derived. The astute chemist would recognize that this is just Hess's Law.

In other words, we can calculate #DeltaH_"rxn"# at #"373 K"# by cooling the reactants from #"373 K"# to #"273 K"#, using #DeltaH_"rxn"# at #"273 K"#, and then heating the products back up to #"373 K"#.

NOTE: It is ultimately not entirely clear what the question means by "the heat capacities are the same".

Are we saying

  • #C_(P("Reactants"))(T) ~~ C_(P("Products"))(T)#?
  • #C_(P("Reactants/Products"))("273 K") ~~ C_(P("Reactants/Products"))("373 K")#?

I interpret it is meaning that the products and reactants each have the same heat capacities at a particular temperature, and not that they are constant in the temperature range irrespective of what they are in relation to each other.

In that special case,

#int_(273)^(373) C_(P("products"))(T)dT-int_(273)^(373) C_(P("reactants"))(T)dT#

#= int_(273)^(373) DeltaC_(P,"rxn")dT ~~ 0#

since the integral of #DeltaC_(P,"rxn") ~~ 0# is zero. Only then is #DeltaH_("rxn", "373 K") = DeltaH_("rxn", "273 K")#.

If the question means that #C_(P("Reactants"))("373 K") ~~ C_(P("Reactants"))("273 K")# and that #C_(P("Products"))("373 K") ~~ C_(P("Products"))("273 K")#, but that #C_(P("Products")) ne C_(P("Reactants"))# in general, then we really do NOT have the same #DeltaH_"rxn"# at two different temperatures:

#C_(P("products"))int_(273)^(373) dT - C_(P("reactants"))int_(273)^(373) dT#

#= DeltaC_(P,"rxn") int_(273)^(373) dT#

#= color(red)(DeltaC_(P,"rxn") cdot DeltaT ne 0)# in general

Related questions
Impact of this question
6927 views around the world
You can reuse this answer
Creative Commons License