Find ΔErxn for the reaction in kJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment is 5.79 kJ/∘C ?

Jun 26, 2017

I got $- 3.99 \times {10}^{3}$ $\text{kJ/mol}$.

Bomb calorimeters are constant-volume calorimeters, which implies that the heat flow $q$ is equal to the change in internal energy, $\Delta E$, i.e.

${q}_{V} = \Delta E$

We know that the heat flow is given by:

${q}_{V} = {C}_{V} \Delta T$

where you were given that the constant-volume heat capacity is ${C}_{V} = \text{5.79 kJ/"^@ "C}$ (rather than the specific heat capacity in $\text{kJ/g"^@ "C}$).

So, we can first calculate ${q}_{V}$ to be:

q_V = "5.79 kJ/"^@ "C" cdot (38.22^@ "C" - 25.74^@ "C")

$=$ ${72.2}_{592}$ $\text{kJ}$

We then use the definition of ${q}_{V} = \Delta E$, dividing by the mols of limiting reagent (hexane) to get:

${q}_{V} / {n}_{L R} = \Delta {\overline{E}}_{s o l n}$,

where ${n}_{L R}$ is the mols of limiting reagent and $\Delta {\overline{E}}_{s o l n} \equiv \frac{\Delta {E}_{s o l n}}{n} _ \left(L R\right)$ is the change in molar internal energy of the solution.

The heat ${q}_{V}$ transferred out from the product into the calorimeter is about $\text{72.26 kJ}$. With respect to the system, ${q}_{V} < 0$.

From here, we just need the mols of hexane:

1.560 cancel"g hexane" xx ("1 mol hexane")/(6cdot12.011 + 14cdot1.0079 cancel"g")

$= {0.0181}_{02}$ $\text{mols hexane combusted}$

Therefore, the change in molar internal energy for the reaction, which is EXOTHERMIC with respect to the system, is:

$\textcolor{b l u e}{\Delta {\overline{E}}_{r x n}} = - \Delta {\overline{E}}_{s o l n} = \textcolor{red}{-} \frac{{q}_{V}}{n} _ \left(L R\right) = \textcolor{red}{-} \text{72.2592 kJ"/"0.018102 mols hexane}$

$=$ $\textcolor{b l u e}{- 3.99 \times {10}^{3}}$ $\textcolor{b l u e}{\text{kJ/mol hexane}}$

to three sig figs.