Question

28,033 Learners found this answer helpful

question 4

a 21.8 g sample of ethanol (\\(\text{c}_2\text{h}_5\text{oh}\\)) is burned in a bomb calorimeter, according to the following reaction. if the temperature rises from 25.0 to 62.3°c, determine the heat capacity of the calorimeter. the molar mass of ethanol is 46.07 g/mol.

\\\text{c}_2\text{h}_5\text{oh}(l) + 3\\,\text{o}_2(g) \
ightarrow 2\\,\text{co}_2(g) + 3\\,\text{h}_2\text{o}(g) \quad \delta h^\circ_{\text{rxn}} = -1235\text{ kj}\\

• 4.99 kj/°c
• 5.65 kj/°c
• 63.7 kj/°c
• 33.1 kj/°c
• 15.7 kj/°c

Explanation:

Calculate moles of ethanol burned

Using the Stoichiometry knowledge point

$$ n = \frac{21.8\text{ g}}{46.07\text{ g/mol}} = 0.47319\text{ mol} $$

Determine heat released by the reaction

Using the Thermochemical Equations and Enthalpy of Reaction knowledge points

$$ q_{\text{rxn}} = n \times \Delta H^{\circ}_{\text{rxn}} = 0.47319\text{ mol} \times (-1235\text{ kJ/mol}) = -584.39\text{ kJ} $$

Calculate temperature change of calorimeter

The temperature change \(\Delta T\) is:

$$ \Delta T = T_{\text{final}} - T_{\text{initial}} = 62.3^{\circ}\text{C} - 25.0^{\circ}\text{C} = 37.3^{\circ}\text{C} $$

Solve for calorimeter heat capacity

The heat absorbed by the calorimeter is equal in magnitude to the heat released by the reaction:

$$ q_{\text{cal}} = -q_{\text{rxn}} = 584.39\text{ kJ} $$

$$ C_{\text{cal}} = \frac{q_{\text{cal}}}{\Delta T} = \frac{584.39\text{ kJ}}{37.3^{\circ}\text{C}} \approx 15.7\text{ kJ/}^{\circ}\text{C} $$

Answer:

• (A) 4.99 kJ/°C
• (B) 5.65 kJ/°C
• (C) 63.7 kJ/°C
• (D) 33.1 kJ/°C
• (E) 15.7 kJ/°C (Correct answer)