The standard enthalpies of formation of CO_{2 | vedclass

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(C) The combustion reaction for glucose is: $C_6H_{12}O_{6(s)} + 6O_{2(g)} \longrightarrow 6CO_{2(g)} + 6H_2O_{(\ell)}$The standard enthalpy of combus

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$-16.11 \ kJ/g$

Vedclass · Answer

(C) The combustion reaction for glucose is: $C_6H_{12}O_{6(s)} + 6O_{2(g)} \longrightarrow 6CO_{2(g)} + 6H_2O_{(\ell)}$The standard enthalpy of combustion $(\Delta_{c}H^{\circ})$ is calculated as:$\Delta_{c}H^{\circ} = [6 	imes \Delta_{f}H^{\circ}(CO_2) + 6 	imes \Delta_{f}H^{\circ}(H_2O)] - [\Delta_{f}H^{\circ}(C_6H_{12}O_6) + 6 	imes \Delta_{f}H^{\circ}(O_2)]$Substituting the given values:$\Delta_{c}H^{\circ} = [6 	imes (-400) + 6 	imes (-300)] - [-1300 + 6 	imes 0]$$\Delta_{c}H^{\circ} = [-2400 - 1800] + 1300 = -4200 + 1300 = -2900 \ kJ/mol$The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$.Enthalpy of combustion per gram $= \frac{-2900 \ kJ/mol}{180 \ g/mol} = -16.11 \ kJ/g$.

The standard enthalpies of formation of $CO_{2(g)}$,$H_2O_{(\ell)}$ and glucose$_{(s)}$ 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

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