Combustion of 2.14 g of liquid benzene (C₆H₆) causes a temperature rise of 16.2 °C in a constant-pressure calorimeter that has a heat capacity of 5.53 kJ/°C. What is ?H for the following reaction? C₆H₆(l) + $\frac { 15 } { 2 }$ O₂(g) ? 6 CO₂(g) + 3 H₂O( $\ell$ )

Question

Quizplus · Accepted Answer

The heat absorbed by the calorimeter is calculated as the product of the heat capacity and the temperature change: $5.53 \, \text{kJ/°C} \times 16.2 \, \text{°C} = 89.526 \, \text{kJ}$. Since this is the heat released by the combustion of 2.14 g of benzene, we convert the mass of benzene to moles (2.14 g / 78.11 g/mol = 0.0274 mol). The enthalpy change per mole of benzene is then $-89.526 \, \text{kJ} / 0.0274 \, \text{mol} = -3.27 \times 10^3 \, \text{kJ/mol}$.

Combustion of 2 152\frac { 15 } { 2 }215​ O2(g) ? 6 CO2(g) + 3 H2O

Combustion of 2 $\frac { 15 } { 2 }$ O₂(g) ? 6 CO₂(g) + 3 H₂O