In which of the following chemical equation, the abbreviations represent the correct states of the reactants and products involved at reaction temperature?

2 H_{2} (l) + O_{2} (l) \to 2 H_{2} O (g)

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2 H_{2} (g) + O_{2} (l) \to 2 H_{2} O (l)

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2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)

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2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)

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Solution

The correct option is C $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$
$2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l)$ is the correction reaction as during formation of water both hydrogen and oxygen are in gaseous form and produces water in liquid form.

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Similar questions

2 H_{2} (l) + O_{2} (l) \to 2 H_{2} O(g)

2 H_{2} (g) + O_{2} (l) \to 2 H_{2} O(l)

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O(l)

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O(g)

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (g)

Δ H

Δ U

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ 2 H_{2} O (l); Δ H = - 86 k J

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l) . . . . . . . . . k J (\pm ?)

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ 2 H_{2} O (l); Δ H = - 286 k J

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l) . . . . . . . . . k J (\pm ?)

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