Question

The standard heat of formation of $CH_4, CO_2$ and $H_2O (I)$ are $ -76.2, -394.8$ and $-285.82\, kJ\, mol^{-1}$, respectively. Heat of vaporization of water is $44\, kJ\, mol^{-1}$. Calculate the amount of heat eveolved when $22.4\, L$ of $CH _4$, kept under normal conditions, is oxidized into its gaseous products

• A. 802 kJ
• B. 878.4 kJ
• C. 702 kJ
• D. 788.4 kJ
• E. 500 kJ

Answer 1

Answer: (A)

Given : $C +2 H _{2} \longrightarrow CH _{4}$

$\Delta_{f} H^{\circ}=-76.2\,kJ\,mol ^{-1}\,.....(i)$

$C + O _{2} \longrightarrow CO _{2}$

$\Delta_{f} H^{\circ}=-349.8\,kJ\,mol ^{-1}\,.....(ii)$

$H _{2}+\frac{1}{2} O _{2} \longrightarrow H _{2} O (l)$

$\Delta_{f} H^{\circ}=-285.82\,kJ\,mol ^{-1}\,.....(iii)$

$H _{2} O (l) \longrightarrow H _{2} O (g)$

$\Delta_{v} H^{\circ}=44 kJ mol ^{-1}\,.....(iv)$

Oxidation of $CH _{4}$ into its gaseous products is given as

$CH _{4}+2 O _{2} \longrightarrow CO _{2}+2 H _{2} O \Delta H=?$

To get the required equation, rewrite the above equation.

$CH _{4} \longrightarrow C +2 H _{2}$

$\Delta_{f} H^{\circ}=+76.2\, kJ\, mol ^{-1} \,.....(i)$

$C + O _{2} \longrightarrow CO _{2} $

$\Delta_{f} H^{\circ}=-394.8\,kJ\, mol ^{-1} \,...(ii)$

$2 H _{2}+ O _{2} \longrightarrow 2 H _{2} O (l)$

$\Delta_{f} H^{\circ}=2 \times-285.82\,kJ\,mol ^{-1} \,....(iii)$

$2 H _{2} O (l) \longrightarrow H _{2} O (g) $

$\Delta_{v} H^{\circ}=2 \times 44\, kJ\, mol ^{-1} .....(iv)$

give $CH _{4}(g)+2 O _{2}(g) \longrightarrow CO _{2}+2 H _{2} O (g) $

$\Delta H^{\circ}=76.2-394.8-2 \times 285.82+88 $

$ \Delta H=-80224$

Thus, heat evolved when $22.4 \,L$ ( or 1 mole of $CH _{4}$ ) is oxidised into its gaseous products is $802.64 \,kJ$.