A student performs a neutralization reaction involving an acid and a base in an open polystyrene coffee-cup calorimeter. How would the calculated value of #DeltaH# differ from the actual value if there was significant heat loss to the surroundings?
A coffee-cup calorimeter is a constant-pressure system, seeing as how it is open to the air, and thus, nothing is restricting the produced gases from expanding all the way.
#DeltaH = q_P#,
#q_"cal"#is the heat absorbed or released by the calorimeter's solution. #q_"rxn"#is the heat generated or absorbed by the reaction occurring in the calorimeter
As a result,
That means we now have to define a realistic
#-q_"cal ideal" = q_("rxn")#
#-(q_"cal realistic" + q_("cal", "lost")) = q_("rxn ideal")#
#-q_"cal realistic" = q_("rxn ideal") + q_("cal lost")#
#-q_"cal realistic" = q_("rxn ideal") - q_("rxn lost")#
#-q_"cal realistic" = q_("rxn realistic")#
This means that
#DeltabarH_"rxn" = q_"rxn realistic"/(n_"product") < q_"rxn ideal"/(n_"product")#
and the enthalpy of reaction would be less than expected, because the temperature of the solution changed by less, and thus