A 1.50 -g sample of quinone $\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)$ is burned in a bomb calorimeter whose total heat capacity is \(8.500 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter increases from 25.00 to \(29.49^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of quinone and per mole of quinone?

To write a balanced chemical equation for the combustion of quinone in a bomb calorimeter, we need to know that quinone reacts with oxygen (O₂) during combustion and produces carbon dioxide (CO₂) and water (H₂O). The unbalanced equation for this reaction is as follows: C₆H₄O₂ + O₂ → CO₂ + H₂O Now we will balance the equation: C₆H₄O₂ + 6O₂ → 6CO₂ + 2H₂O So, the balanced chemical equation for the combustion of quinone is: C₆H₄O₂ + 6O₂ → 6CO₂ + 2H₂O

To calculate the heat of combustion of quinone, we will use the equation: q = C × ΔT Where q is the heat released during combustion, C is the heat capacity of the bomb calorimeter, and ΔT is the temperature change in the bomb calorimeter. Given values are: C = 8.5 kJ/ºC ΔT = 29.49ºC - 25.00ºC = 4.49ºC Now we will calculate the heat released during combustion (q): q = C × ΔT q = 8.5 kJ/ºC × 4.49ºC q = 38.165 kJ Now we will find the heat of combustion per gram and per mole of quinone: 1. Heat of combustion per gram (q/g): Divide the heat released during combustion (q) by the mass of the quinone sample. q/g = q / mass q/g = 38.165 kJ / 1.50 g q/g = 25.443 kJ/g 2. Heat of combustion per mole (q/mol): First, we need to determine the molar mass of quinone. Quinone: C₆H₄O₂ Molar mass: (6 × 12.01) + (4 × 1.01) + (2 × 16.00) = 108.05 g/mol Now we will calculate the heat of combustion per mole by dividing the heat released during combustion (q) by the moles of quinone sample. Moles of quinone = mass / molar mass Moles of quinone = 1.50 g / 108.05 g/mol Moles of quinone = 0.01388 mol q/mol = q / moles of quinone q/mol = 38.165 kJ / 0.01388 mol q/mol = 2748.499 kJ/mol So, the heat of combustion per gram of quinone is 25.443 kJ/g, and per mole of quinone is 2748.499 kJ/mol.