Question

Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8)(C10H8). When 1.025 gg of naphthalene is burned in a bomb calorimeter, the temperature rises from 24.25 ∘C∘C to 32.33 ∘C Find ΔErxn for the combustion of naphthalene. The heat capacity of the calorimeter, determined in separate experiment, is 5.11kJ/∘C .

Answer 1

Answer: The enthalpy of the reaction is -5167.71 kJ

Explanation:

To calculate the heat absorbed by the calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 5.11 kJ/°C

$\Delta T$ = change in temperature = $T_2-T_1=(32.33-24.25)^oC=8.08^oC$

Putting values in above equation, we get:

$q=5.11kJ/^oC\times 8.08^oC=41.29kJ$

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of naphthalene = 1.025 g

Molar mass of naphthalene = 128.2 g/mol

Putting values in above equation, we get:

$\text{Moles of naphthalene}=\frac{1.025g}{128.2g/mol}=0.00799mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -41.29 kJ

n = number of moles of naphthalene = 0.00799 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-41.29kJ}{0.00799mol}=-5167.71kJ/mol$

Hence, the enthalpy of the reaction is -5167.71 kJ