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

Answer 1

Answer: (A)

802 kJ

Answer 2

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 $802.64 \,kJ$.