# When "32.4 mL" of liquid benzene (C_6H_6, rho = "0.879 g/mL") reacts with "81.6 L" of oxygen gas, measured at "1.00 atm" and "25°C", 1.19xx10^3 "kJ" of heat is released at const. pressure. What is DeltaH^@ for the following reaction?

## ($R = \text{0.0821 L"cdot"atm/K"cdot"mol}$) $2 \text{C"_6"H"_6(l) + 15"O"_2(g) -> 12"CO"_2(g) + 6"H"_2"O} \left(l\right)$

Apr 11, 2018

$\Delta {H}_{r x n}^{\circ} = - 3.264 \times {10}^{3} \text{kJ/mol}$

Why is it important (i.e. useful!) to use $\text{kJ/mol}$ instead of $\text{kJ}$?

At constant pressure, the heat released IS the $\Delta H$ of the reaction. The combusted quantity is

$32.4 \cancel{\text{mL" xx "0.879 g"/cancel"mL" = "28.48 g benzene}}$

$28.48 \cancel{\text{g benzene" xx ("1 mol benzene")/(78.1134 cancel"g benzene") = "0.3646 mols benzene}}$
vs. ${\text{2.735 mols O}}_{2}$ needed

and we can probably tell that the ${\text{O}}_{2}$ is in excess... assuming ideal gases,

$n = \frac{P V}{R T} = \left(\text{1.00 atm" cdot "81.6 L")/("0.08206 L"cdot"atm/mol"cdot"K" cdot "298.15 K}\right)$

$= {\text{3.335 mols O}}_{2}$

Indeed, it is in excess by $\text{0.6006 mols}$. We base the mols off of the limiting reactant benzene.

The reaction apparently releases $\text{1190 kJ}$ of heat at constant pressure, so

$\textcolor{b l u e}{\Delta {H}_{r x n}^{\circ}} = - \text{1190 kJ"/"0.3646 mols benzene}$

$= \textcolor{b l u e}{- 3.264 \times {10}^{3} \text{kJ/mol}}$