The reaction which shows standard heat of formation of water correctly is

Easy

A

$\large {{H}_{2\left( g \right)}}+\frac{1}{2}{{O}_{2\left( g \right)}}\to {{H}_{2}}{{O}_{\left( l \right)}};\,\Delta H\,\,=\,-68.3kcal$
B

${{H}_{2\left( g \right)}}+\frac{1}{2}{{O}_{2\left( g \right)}}\to {{H}_{2}}{{O}_{\left( s \right)}};\,\Delta H\,=\,-68.3$
C

${{H}_{2\left( g \right)}}+\frac{1}{2}{{O}_{2\left( g \right)}}\to {{H}_{2}}{{O}_{\left( g \right)}};\,\Delta H\,=\,-68.3kcal$
D

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l), ∆ H = - 136.6 kcals$

Solution

“The enthalpy change when one mole of a substance is formed in its standard state from its constituent elements in their standard standard states is called standard enthalpy of formation"
Standard states of hydrogen….gas,oxygen….gas and water….liquid;
In 4th option two moles of water is formed ;
So option 1 is correct ;

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