Question

The standard heat of formation of $C{H}_4,C{O}_2$ and $H_{2}O(I)$ are $-76.2,-394.8$ and $-285.82kJmo{l}^{-1}$, respectively. Heat of vaporization of water is $44kJmo{l}^{-1}$. Calculate the amount of heat eveolved when $22.4L$ of $C{H}_4$, 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 : $C+2H_2⟶C{H}_4$

${\Delta }_f{H}^{\circ }=-76.2kJmo{l}^{-1}.....(i)$

$C+O_2⟶C{O}_2$

${\Delta }_f{H}^{\circ }=-349.8kJmo{l}^{-1}.....(ii)$

$H_2+\frac{1}{2}{O}_2⟶H_{2}O(l)$

${\Delta }_f{H}^{\circ }=-285.82kJmo{l}^{-1}.....(iii)$

$H_{2}O(l)⟶H_{2}O(g)$

${\Delta }_v{H}^{\circ }=44kJmo{l}^{-1}.....(iv)$

Oxidation of $C{H}_4$ into its gaseous products is given as

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O\Delta H=?$

To get the required equation, rewrite the above equation.

$C{H}_4⟶C+2H_2$

${\Delta }_f{H}^{\circ }=+76.2kJmo{l}^{-1}.....(i)$

$C+O_2⟶C{O}_2$

${\Delta }_f{H}^{\circ }=-394.8kJmo{l}^{-1}...(ii)$

$2H_2+O_2⟶2H_{2}O(l)$

${\Delta }_f{H}^{\circ }=2\times -285.82kJmo{l}^{-1}....(iii)$

$2H_{2}O(l)⟶H_{2}O(g)$

${\Delta }_v{H}^{\circ }=2\times 44kJmo{l}^{-1}.....(iv)$

give $C{H}_4(g)+2O_2(g)⟶C{O}_2+2H_{2}O(g)$

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

$\Delta H=-80224$

Thus, heat evolved when $22.4L$ ( or 1 mole of $C{H}_4$ ) is oxidised into its gaseous products is $802.64kJ$.