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Thermodynamics

Section: Chemistry

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Question : 3 of 34

Marks: +1, -0

Enthalpy of formation of

\mathrm{CO}_2(g), \mathrm{H}_2\mathrm{O}(l)

\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(s)

-393,-286

-1170\ \mathrm{kJ}\ \mathrm{mol}^{-1}

respectively. The quantity of heat liberated when 18 g of

\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6(s)

is burnt completely in oxygen is

[AP EAPCET 23 May 2025 S2]

$520\ \mathrm{kJ}$
$145\ \mathrm{kJ}$
$290\ \mathrm{kJ}$
$420\ \mathrm{kJ}$

Solution:

The complete combustion reaction is

\; \mathrm{C_6H_{12}O_6}(s)+6\,\mathrm{O_2}(g) \rightarrow 6\,\mathrm{CO_2}(g)+6\,\mathrm{H_2 0}(n)

\; \Delta H^{\circ}_{\text{comb}} = \sum \Delta H^{\circ}_{\text{froduct}} - \sum \Delta H^{\circ}_{\text{frectant}}

\; = [6 \times (-393) + 6 \times (-286)] - [1 \times (-1170) + 6 \times 0]

\; = -2904\ \mathrm{kJ}

\; \text{ Moles of glucose } \;\; = \frac{\; \text{ Mass } \;}{\; \text{ Molar mass } \;} = \frac{18\ \mathrm{g}}{180}

\; = 0.1\ \mathrm{mol}

\; \text{ Heat liberated } \;\; = \text{ Moles } \times \Delta H^{\circ}_{\text{combustion}}

\; = 0.1 \times (-2904)

\;\; \approx -290\ \mathrm{kJ}

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