Question

The standard enthalpies of formation of $C O _{2}(g), H _{2} O (l)$ 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

• A. + 2900 kJ
• B. - 2900 kJ
• C. - 16.11 kJ
• D. + 16.11 kJ

Answer 1

Answer: (C)

PLAN $\Delta_{C} H^{\circ}$ (Standard heat of combustion) is the standard enthalpy change when one mole of the substance is completely oxidised.
Also standard heat of formation $\left(\Delta_{f} H^{\circ}\right)$ can be taken as the standard of that substance.
$H_{ CO _{2}}^{\circ}=\Delta_{f} H^{\circ}\left( CO _{2}\right)=-400\, kJ \, mol ^{-1}$
$H_{ H _{2} O }^{\circ}=\Delta_{f} H^{\circ}\left( H _{2} O \right)=-300\, kJ\, mol ^{-1}$
$H_{\text {glucose }}^{\circ}=\Delta_{f} H^{\circ}($ glucose $)=-1300\, kJ \, mol ^{-1}$
$H_{ O _{2}}^{\circ}=\Delta_{f} H^{\circ}\left( O _{2}\right)=0.00$
$C _{6} H _{12} O _{6}(s)+6 O _{2}(g) \longrightarrow 6 CO _{2}(g)+6 H _{2} O (l)$
$\Delta_{c} H^{\circ}($ glucose $)=6\left[\Delta_{f} H^{\circ}\left( CO _{2}\right)+\Delta_{f} H^{\circ}\left( H _{2} O \right)\right]-\left[\Delta_{f} H^{\circ}\left( C _{6} H _{12} O _{6}\right)+6 \Delta_{f} H^{\circ}\left( O _{2}\right)\right]$
$=6[-400-300]-[-1300+6 \times 0]$
$=-2900 \, kJ\, mol ^{-1}$
Molar mass of $C _{6} H _{12} O _{6}=180 \, g\, mol ^{-1}$
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 $R$ $\left(8.314 \, J\, K ^{-1} \, mol ^{-1}\right)$ in $JK ^{-1} \, mol ^{-1}$ must be converted into $kJ$ by dividing the unit by $1000$ .