Thermodynamics and Thermochemistry 2 Question 6
6. The standard enthalpies of formation of $\mathrm{CO}{2}(g), \mathrm{H}{2} \mathrm{O}(l)$ and glucose $(s)$ at $25^{\circ} \mathrm{C}$ are $-400 \mathrm{~kJ} / \mathrm{mol},-300 \mathrm{~kJ} / \mathrm{mol}$ and $-1300 \mathrm{~kJ} / \mathrm{mol}$, respectively. The standard enthalpy of combustion per gram of glucose at $25^{\circ} \mathrm{C}$ is
(2013 Adv.)
(a) $+2900 \mathrm{~kJ}$
(b) $-2900 \mathrm{~kJ}$
(c) $-16.11 \mathrm{~kJ}$
(d) $+16.11 \mathrm{~kJ}$
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Solution:
- 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.
$$ \begin{gathered} H_{\mathrm{CO}{2}}^{\circ}=\Delta{f} H^{\circ}\left(\mathrm{CO}{2}\right)=-400 \mathrm{~kJ} \mathrm{~mol}^{-1} \ H{\mathrm{H}{2} \mathrm{O}}^{\circ}=\Delta{f} H^{\circ}\left(\mathrm{H}{2} \mathrm{O}\right)=-300 \mathrm{~kJ} \mathrm{~mol}^{-1} \ H{\text {glucose }}^{\circ}=\Delta_{f} H^{\circ}(\text { glucose })=-1300 \mathrm{~kJ} \mathrm{~mol}^{-1} \ H_{\mathrm{O}{2}}^{\circ}=\Delta{f} H^{\circ}\left(\mathrm{O}{2}\right)=0.00 \ \mathrm{C}{6} \mathrm{H}{12} \mathrm{O}{6}(s)+6 \mathrm{O}{2}(g) \longrightarrow 6 \mathrm{CO}{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \end{gathered} $$
$\Delta_{c} H^{\circ}$ (glucose) $=6\left[\Delta_{f} H^{\circ}\left(\mathrm{CO}{2}\right)+\Delta{f} H^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]$
$$ \begin{aligned} & -\left[\Delta_{f} H^{\circ}\left(\mathrm{C}{6} \mathrm{H}{12} \mathrm{O}{6}\right)+6 \Delta{f} H^{\circ}\left(\mathrm{O}_{2}\right)\right] \ = & 6[-400-300]-[-1300+6 \times 0] \ = & -2900 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
$$ \text { Molar mass of } \mathrm{C}{6} \mathrm{H}{12} \mathrm{O}_{6}=180 \mathrm{~g} \mathrm{~mol}^{-1} $$
Thus, standard heat of combustion of glucose per gram
$$ =\frac{-2900}{180}=-16.11 \mathrm{~kJ} \mathrm{~g}^{-1} $$
To solve such problem, students are advised to keep much importance in unit conversion. As here, value of $R$ $\left(8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right.$ ) in $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ must be converted into $\mathrm{kJ}$ by dividing the unit by 1000 .
