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?

Question

Natalie Anton · Accepted Answer

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.

By definition:

#DeltaH = q_P#,

where #q_P# is heat flow #q# at constant pressure. There are two types of #q# to discuss.

As a result, #q_"cal" = -q_"rxn"#. The question is asking what would you get if some #q_"cal"# was lost to the surroundings, since clearly, #q_"rxn"# was released (if exothermic) or absorbed (if endothermic) by the reaction, and is the same for the same reaction.

That means we now have to define a realistic #q_"cal"# value which accounts for the heat loss:

#-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")#

or that #color(blue)(q_("rxn realistic") = q_("rxn ideal") - q_("rxn lost"))#.

This means that #q_"rxn realistic"#, which includes the heat loss, is now smaller than the ideal #q_"rxn"#. That means:

#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 #|q_"cal realistic"| < |q_"cal ideal"|# and #|q_"rxn realistic"| < |q_"rxn ideal"|#.

Axel Alvarado · Answer

undefined The calculated ΔH would be lower than the actual value due to the loss of heat to the surroundings.