Calculate the standard enthalpy of formation of acetylene $(C_2{H}_2)$ from its elements:

$2C(graphite)+H_2(g)C_2{H}_2(g)$

The equations for each step and the corresponding enthalpy changes are:

$C($ graphite $)+O_2(g)C{O}_2(g),H_0=-393\mathrm{.}5k\frac{J}{m}ol$

$H_2(g)+\frac{1}{2}{O}_2(g)H_{2}O(l),H=-285\mathrm{.}8k\frac{J}{m}ol$

$2C_2{H}_2(g)+5O_2(g)4C{O}_2(g)+2H_{2}O(l),H^O=-2598\mathrm{.}8k\frac{J}{m}ol$

$H_O=-393\mathrm{.}5k\frac{J}{m}ol$

$H^O=-285\mathrm{.}8k\frac{J}{m}ol$

$H^O=-2598\mathrm{.}8k\frac{J}{m}ol$

$+226\mathrm{.}6k\frac{J}{m}ol$

$-226\mathrm{.}6k\frac{J}{m}ol$

$-2\mathrm{.}266k\frac{J}{m}ol$

$2\mathrm{.}266k\frac{J}{m}ol$