Question

The standard enthalpies of formation of $CO_2(g), H_2O(l)$ and glucose(.v) at 25$^{\circ}$C- are 400 kJ/ mol. -300 kJ/mol and - 1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25$^{\circ}$C is

• A. #ERROR!
• B. - 2900 kJ
• C. - 16.11 kJ
• D. #ERROR!

Answer 1

The Correct Option is C

PLAN A CH $^{\circ}$ (Standard heat of combustion) is the standard enthalpy change when one mole of the substance is completely oxidised. Also standard heat of formation (A fH$^{\circ}$) can be taken as the standard of that substance $ \, \, \, \, \, H^{\circ} _{CO_2} = \Delta _f H^{\circ} (CO_2) = -400 kJ moI^{-1}$ $ \, \, \, \, \, H^{\circ}_{H_2O} = \Delta _f H^{\circ} (H_2O) = -300 kJ moI^{-1}$ $ \, \, \, \, \, H^{\circ} _{glucose} = \Delta _f H^{\circ} (glucose) = -1300 kJ moI^{-1}$ $ \, \, \, \, \, H^{\circ}_{O_2} = \Delta _f H^{\circ} (O_2) = 0.00$ $C_6H_12O_6(s) + 6O_2(g) \longrightarrow 6CO_2(g) + 6H_2O(l)$ $\Delta _cH^{\circ} (glucose) = 6[\Delta _fH^{\circ}(CO_2) + \Delta _fH^{\circ}(H_2O)]$ $ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -[\Delta _f H^{\circ} (C_6H_12O_6) + 6 \Delta _f H^{\circ} (O_2)]$ $ \, \, \, \, \, \, \, \, \, \, \, \, = 6[- 400 -300] - [-1300 + 6 \times 0]$ $ \, \, \, \, \, \, \, \, \, \, \, \, \, = -2900 kJ \, moI^{-1}$ Molar mass of $ C_6H_{12}O_6 = 180g mol^{-1} $ Thus, standard heat of combustion of glucose per gram $ \, \, \, \, \, \, \, \, =\frac{-2900}{180} = -16.11 kJ g^{-1}$ To solve such problem, students are advised to keep much importance in unit conversion. As here, value of R $(8.314JK^{-1} mol^{-1})$ in $JK^{-1} mol^{-1})$ must be converted into kJ by dividing the unit by 1000.