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Thermodynamics — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics2 Marks
Question
The standard heat of formation of $\mathrm{CH}_4(\mathrm{~g}), \mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ are $-76.2,-394.8$ and -241.6 kJ mol ${ }^{-1}$ respectively. Calculate the amount of heat evolved by burning $1 \mathrm{~m}^3$ of methane measured at NTP.
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Answer

The burning of methane may be expressed as,$\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ } \ \text{CO}_2(\text{g})+2\text{H}_2\text{O(g)}$
$\Delta_\text{f}\text{H}^\circ=[\Delta_\text{f}\text{H}^\circ(\text{CO}_2)+2\Delta_\text{f}\text{H}^\circ(\text{H}_2\text{O})]\\-[\Delta_\text{f}\text{H}^\circ(\text{CH}_4)+2\Delta_\text{f}\text{H}^\circ(\text{O}_2)]$
$=[-394.8+2\times(-241.6)]-[-76.2+2\times0]$
$=-801.8\text{kJ}$
1 mole or 22.4 L of $\mathrm{CH}_4$ evolve heat $=801.8 \mathrm{~kJ} 1 \mathrm{~m}^3$ or 1000 L of $\mathrm{CH}_4$ evolve heat $=\frac{801.8 \times 1000}{22.4}=35794.6 \mathrm{~kJ}$

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