Thermodynamics And Thermochemistry Result 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 $ $kJ$
(b) $-2900 $ $kJ$
(c) $-16.11 $ $kJ$
(d) $+16.11 $ $kJ$
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Answer:
Correct Answer: 6. ( c )
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.
$\begin{aligned} H_{\mathrm{CO}_2}^{\circ} & =\Delta_f H^{\circ}\left(\mathrm{CO}2\right)=-400 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ H{\mathrm{H}2 \mathrm{O}}^{\circ} & =\Delta_f H^{\circ}\left(\mathrm{H}2 \mathrm{O}\right)=-300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ H{\text {glucose }}^{\circ} & =\Delta_f H^{\circ}(\text { glucose })=-1300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ H{\mathrm{O}_2}^{\circ} & =\Delta_f H^{\circ}\left(\mathrm{O}_2\right)=0.00 \\ \mathrm{C}6 \mathrm{H}{12} \mathrm{O}_6(s) & +6 \mathrm{O}_2(g) \longrightarrow 6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l) \end{aligned}$
$\begin{aligned} \Delta_c H^{\circ}(\text { glucose })= & 6\left[\Delta_f H^{\circ}\left(\mathrm{CO}_2\right)+\Delta_f H^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)\right] \\ & \quad-\left[\Delta_f H^{\circ}\left(\mathrm{C}6 \mathrm{H}{12} \mathrm{O}_6\right)+6 \Delta_f H^{\circ}\left(\mathrm{O}_2\right)\right] \\ = & 6[-400-300]-[-1300+6 \times 0] \\ = & -2900 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$
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 $R$ ( $8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ) in $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ must be converted into kJ by dividing the unit by $1000$ .
BETA
