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

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$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(g)}; Δ H = - 68.3 kcal$

$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(s)}; Δ H = - 68.3 kcal$

$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}$ ; $Δ H = - 68.3 kcal$

$2 H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to 2 H_{2} O; Δ H = - 136.6 kcal$

answer is A.

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Detailed Solution

Water having chemical formula $H_{2} O$ is usually found in liquid, solid and gaseous form. And its liquid form is known to be the most stable and standard form.

And so the reactants for full-filling the condition of standard heat of formation must also be in its elemental state.

$∆ {H_{f}}^{0}$ generally denotes the standard enthalpy of formation. It states that during the formation of 1 mole of substance when its constituent elements are in standard states, the enthalpy change is represented by the standard enthalpy of formation.

Hence, among the following reactions,

$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O; Δ H = - 68.3 kcal$