Carole Yarbrough

2022-01-09

Mothballs are composed primarily of the hydrocarbon naphth alene $\left({C}_{10}{H}_{8}\right)$. When 1.025 g of naphthalene burns in a bomb calorimeter, the temperature rises from ${24.25}^{\circ }C$ to ${32.33}^{\circ }C$. Find $\mathrm{\Delta }{E}_{rxn}$ for the combustion of naphthalene. The heat capacity of the bomb calorimeter, determined in a separate experiment , is $5.11kJ{/}^{\circ }C$.

jean2098

Step 1
The temperature change $\left(\mathrm{\Delta }T\right)$ in a bomb calorimeter is related to the heat absorbed by the entire calorimeter assembly (qcal).
${q}_{cal}={C}_{cal}×\mathrm{\Delta }T$
Where ${C}_{cal}$ is the heat capacity of the bomb calorimeter.
Since no heat escapes the bomb calorimeter.
${q}_{rxn}=-{q}_{cal}$
Combining the two equations
${q}_{rxn}=-{C}_{cal}×\mathrm{\Delta }T$
Step 2
Given that 1.025 g of naphthalene $\left({C}_{10}{H}_{8}\right)$ burned in a bomb calorimeter raises the temperature rises from ${24.25}^{\left\{}\circ C$ to ${32.33}^{\circ }C$. Determine the $\mathrm{\Delta }{E}_{rxn}\left(kJ/mol\right)$ for the combustion of naphthalene if the heat capacity of the bomb calorimeter is $5.11kJ{/}^{\circ }C$.
Recall that
$\mathrm{\Delta }{E}_{rxn}=\frac{{q}_{rxn}}{n}$
Step 3
Solving for ${q}_{rxn}$.
${q}_{rxn}=-5.11\frac{k}{{}^{\left\{\circ \right\}}C}×\left({32.33}^{\circ }C-{24.25}^{\circ }C\right)=-41.2888kJ$
Step 4
Solving for nn naphthalene from g naphthalene using molecular weight $\left(MW=128.1705g/mol\right)$.

Step 5
Solving for $\mathrm{\Delta }{E}_{rxn}$

Fasaniu