The combustion of glucose in oxygen at 298 K is given by the following thermochemical equation:
C₆H₁₂O₆ (s) + 6O₂ → 6CO₂ (g) + 6H₂O (l) $\Delta_{T} H_{298}^{\theta}$ = -2808 kJ mol^-1. How much heat energy (in kJ) is supplied when 275 g of glucose burn in a plentiful supply of air so that combustion is complete?

Question

The combustion of glucose in oxygen at 298 K is given by the following thermochemical equation:
C₆H₁₂O₆ (s) + 6O₂ → 6CO₂ (g) + 6H₂O (l)

\Delta_{T} H_{298}^{\theta}

= -2808 kJ mol^-1. How much heat energy (in kJ) is supplied when 275 g of glucose burn in a plentiful supply of air so that combustion is complete?

Quizplus · Accepted Answer

The molar mass of glucose (C₆H₁₂O₆) is 180 g/mol. Therefore, 275 g of glucose is 275/180 = 1.5278 mol. The heat energy released is 1.5278 mol × (-2808 kJ/mol) = -4286.19 kJ.

The Combustion of Glucose in Oxygen at 298 K Is ΔTH298θ\Delta_{T} H_{298}^{\theta}ΔT​H298θ​

The Combustion of Glucose in Oxygen at 298 K Is $\Delta_{T} H_{298}^{\theta}$