Question

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?

Answer 1

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.

• `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, `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"|`.