Question
Consider the following reactions,
$$\left( {\text{i}} \right){H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to $$ $${H_2}O\left( l \right), - {x_1}\,kJ\,mo{l^{ - 1}}$$
$$\left( {{\text{ii}}} \right)\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to $$ $${H_2}O\left( l \right),{x_2}\,kJ\,mo{l^{ - 1}}$$
$$\left( {{\text{iii}}} \right)C{O_2}\left( g \right) + {H_2}\left( g \right) \to $$ $$CO\left( g \right) + {H_2}O\left( l \right), - {x_3}\,kJ\,mo{l^{ - 1}}$$
$$\left( {{\text{iv}}} \right){C_2}{H_2}\left( g \right) + \frac{5}{2}{O_2}\left( g \right) \to $$ $$2\,C{O_2}\left( g \right) + {H_2}O\left( l \right), + {x_4}\,kJ\,mo{l^{ - 1}}$$
Enthalpy of formation of $${H_2}O\left( l \right)$$ is
A.
$$ - {x_2}\,kJ\,mo{l^{ - 1}}$$
B.
$$ + {x_3}\,kJ\,mo{l^{ - 1}}$$
C.
$$ - {x_4}\,kJ\,mo{l^{ - 1}}$$
D.
$$ + {x_1}\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - {x_2}\,kJ\,mo{l^{ - 1}}$$
Solution :
Enthalpy of formation : The amount of heat evolved or absorbed during the formation of $$1\,mole$$ of a compound from its constituent elements is known as heat of formation. So, the correct answer is
$$\eqalign{
& {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right), \cr
& \Delta H = - {x_2}\,kJ\,mo{l^{ - 1}} \cr} $$