# 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?

Nov 6, 2016

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:

$\Delta H = {q}_{P}$,

where ${q}_{P}$ is heat flow $q$ at constant pressure. There are two types of $q$ to discuss.

• ${q}_{\text{cal}}$ is the heat absorbed or released by the calorimeter's solution.
• ${q}_{\text{rxn}}$ is the heat generated or absorbed by the reaction occurring in the calorimeter

As a result, ${q}_{\text{cal" = -q_"rxn}}$. The question is asking what would you get if some ${q}_{\text{cal}}$ was lost to the surroundings, since clearly, ${q}_{\text{rxn}}$ 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 ${q}_{\text{cal}}$ value which accounts for the heat loss:

-q_"cal ideal" = q_("rxn")

$- \left({q}_{\text{cal realistic" + q_("cal", "lost")) = q_("rxn ideal}}\right)$

-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 $\textcolor{b l u e}{{q}_{\text{rxn realistic") = q_("rxn ideal") - q_("rxn lost}}}$.

This means that ${q}_{\text{rxn realistic}}$, which includes the heat loss, is now smaller than the ideal ${q}_{\text{rxn}}$. 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 $| {q}_{\text{cal realistic"| < |q_"cal ideal}} |$ and $| {q}_{\text{rxn realistic"| < |q_"rxn ideal}} |$.