The combustion of glucose (C₆H₁₂O₆) at 1 bar and 298 K releases 2802.5 kJ mol^-1 of heat.
C₆H₁₂O₆ (s) + 6 O₂ (g) $\rightarrow$ 6 CO₂ (g) + 6 H₂O (l) Calculate the standard enthalpy change of formation of glucose (in kJ mol^-1) . Use $\Delta_{\mathrm{f}} H_{208}^{\circ}$ data in Appendix 7.

Question

The combustion of glucose (C₆H₁₂O₆) at 1 bar and 298 K releases 2802.5 kJ mol^-1 of heat.
C₆H₁₂O₆ (s) + 6 O₂ (g)

\rightarrow

6 CO₂ (g) + 6 H₂O (l) Calculate the standard enthalpy change of formation of glucose (in kJ mol^-1) . Use

\Delta_{\mathrm{f}} H_{208}^{\circ}

data in Appendix 7.

Quizplus · Accepted Answer

The standard enthalpy change of formation of glucose can be calculated using the enthalpy change of the combustion reaction and the standard enthalpies of formation of the products and reactants. The enthalpy change of formation of glucose is negative because it is an exothermic process.

The Combustion of Glucose (C6H12O6) at 1 Bar and 298 →\rightarrow→

The Combustion of Glucose (C₆H₁₂O₆) at 1 Bar and 298 $\rightarrow$