# A student conducting a calorimetry investigation determines a negative DeltaH. What does the negative value indicate about the reaction?

Jun 14, 2017

We don't know. Not unless it is specified what kind of $q$, and thus $\Delta H$ was obtained...

• If ${q}_{s o l n}$ is negative, i.e. the heat flow and enthalpy for the solution was exothermic, $\Delta {H}_{r x n} > 0$, so the reaction was endothermic and heat was released from the solution ($\Delta T < 0$, solution gets colder).
• If ${q}_{s o l n}$ is positive, i.e. the heat flow and enthalpy for the solution was endothermic, $\Delta {H}_{r x n} < 0$, so the reaction was exothermic and heat was absorbed into the solution ($\Delta T > 0$, solution heats up).

A calorimeter experiment in general is when one measures the change in temperature of a system due to a chemical process occurring in (usually) a solution.

I assume you mean a constant-pressure scenario, i.e. a coffee-cup calorimeter open to the atmosphere, as heat flow $q = n \Delta H$, the molar enthalpy of some arbitrary change-of-state process, ONLY at constant pressure. Upon determining the change in temperature, and knowing the specific heat capacity of the solvent, one could then calculate the heat involved in the process (with respect to the solution):

${q}_{s o l n} = m {C}_{P} \Delta T$,

where:

• $m$ is the mass of the solution in $\text{g}$. One can measure the initial solution volume (before the process occurs) and assume a density equal to that of the pure solvent, whatever it may be. That allows for the determination of this mass.
• ${C}_{P}$ is the specific heat capacity at CONSTANT PRESSURE. For water at ordinary temperatures, it is about $\text{4.184 J/g"^@ "C}$.
• $\Delta T = {T}_{2} - {T}_{1}$ is the change in temperature in $\text{^@ "C}$ (because ${C}_{P}$ used $\text{^@ "C}$), sign included.

Note that ${q}_{s o l n}$, is with respect to the solution, specifically, i.e. it is the heat transferred between the system (the solution) and its surroundings (the object in the solution and everything else).

If you wish to know the molar enthalpy of reaction, $\Delta {\overline{H}}_{r x n}$ (where the system is the object in the solution and the immediate surroundings are the solution), you first need to determine ${q}_{s o l n}$, and then assume conservation of energy, i.e.

${q}_{r x n} + {q}_{s o l n} = 0$

Thus, ${q}_{r x n} = - {q}_{s o l n}$. And finally, use the definition of enthalpy to obtain:

$\textcolor{b l u e}{\Delta {\overline{H}}_{r x n} = - {q}_{s o l n} / \left({n}_{\text{object}}\right)}$

where ${n}_{\text{object}}$ is the mols of whatever you put into the solution, whether it was a solute, or the limiting reactant of a chemical reaction.

Thus...

• If ${q}_{s o l n}$ is negative, i.e. the heat flow and enthalpy for the solution was exothermic, $\Delta {H}_{r x n} > 0$, so the reaction was endothermic and heat was released from the solution ($\Delta T < 0$, solution gets colder).
• If ${q}_{s o l n}$ is positive, i.e. the heat flow and enthalpy for the solution was endothermic, $\Delta {H}_{r x n} < 0$, so the reaction was exothermic and heat was absorbed into the solution ($\Delta T > 0$, solution heats up).
Jun 14, 2017

That the reaction was $\text{exothermic..........}$, and that the enthalpy of the reactants was GREATER than the enthalpy of the products.

#### Explanation:

By convention we report an enthalpy change that RELEASES energy as negative. To give an actual example, we know that when we burn methane we produce heat, which we use to cook our breakfast or warm our homes........i.e.

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) + \Delta$

We can certainly measure and report the $\text{enthalpy change}$, and here $\Delta \equiv \Delta {H}_{\text{combustion}} = - 890 \cdot k J \cdot m o {l}^{-} 1$, and when we write $\text{per mole}$ we mean $\text{per mole of reaction as written.}$