Question

1. when 2.00 g of liquid hexane (c₆h₁₄, mm = 86.18) undergoes combustion in a bomb calorimeter, the temperature rises by 7.12°c. find δeₘ for the reaction in kj/mol hexane. the heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kj/°c.

Explanation:

Step1: Calculate the heat absorbed by the calorimeter

The heat absorbed by the calorimeter \(q = C\times\Delta T\), where \(C = 5.73\space kJ/^{\circ}C\) and \(\Delta T=7.12^{\circ}C\).

$$q = 5.73\times7.12\space kJ$$

$$q = 40.8\space kJ$$

Step2: Calculate the number of moles of hexane

The molar mass of hexane \(MM = 86.18\space g/mol\), and the mass of hexane \(m = 2.00\space g\). The number of moles \(n=\frac{m}{MM}\)

$$n=\frac{2.00\space g}{86.18\space g/mol}=0.0232\space mol$$

Step3: Calculate \(\Delta E_{rxn}\)

Since the heat released by the reaction is equal to the heat absorbed by the calorimeter (\(\Delta E_{rxn}=-\frac{q}{n}\)) (negative because the reaction is exothermic)

$$\Delta E_{rxn}=-\frac{40.8\space kJ}{0.0232\space mol}$$

$$ \Delta E_{rxn}=- 1760\space kJ/mol$$

Answer:

\(-1760\space kJ/mol\)