Discussão:Metal alcalino

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Addition to the article "Metal alcalino"

The sentence "2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H₂ ↑" describes an irreversible process, that is:

2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H₂ ↑ $~\Leftrightarrow$ 2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H_{2 (g)} $~\land \neg$( 2 LiOH _(aq) + H_{2 (g)} → 2 Li _(s) + 2 H₂O _(l) )

N.B.

The sentence "NaCl _(aq) + AgNO_{3 (aq)} → NaNO_{3 (aq)} + AgCl ↓" is another example of the sentences that describe the irreversible chemical processes.

NaCl _(aq) + AgNO_{3 (aq)} → NaNO_{3 (aq)} + AgCl ↓ $~\Leftrightarrow$ NaCl _(aq) + AgNO_{3 (aq)} → NaNO_{3 (aq)} + AgCl _(s) $~\land \neg$( NaNO_{3 (aq)} + AgCl _(s) → NaCl _(aq) + AgNO_{3 (aq)} )

The sentence "2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H₂ ↑" describes a redox process, that is:

2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H₂ ↑ $~\Rightarrow$ 2 Li⁰ _(s) + 2 H⁺¹₂O _(l) → 2 Li⁺¹OH _(aq) + H⁰_{2 (g)}

The sentence "2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H_{2 (g)}" may be inferred from the following sentences:

Li _(s) → Li⁺_(aq) + e⁻

2 H⁺_(aq) + 2 e⁻ → H_{2 (g)}

H₂O _(l) = H⁺_(aq) + OH⁻_(aq)

LiOH _(aq) = Li⁺_(aq) + OH⁻_(aq)

Inference

$~\triangleleft$ Li _(s) → Li⁺_(aq) + e⁻ $~\land$ 2 H⁺_(aq) + 2 e⁻ → H_{2 (g)} $~\land$ H₂O _(l) = H⁺_(aq) + OH⁻_(aq) $~\land$ LiOH _(aq) = Li⁺_(aq) + OH⁻_(aq) $~\Rightarrow$

2 Li _(s) → 2 Li⁺_(aq) + 2 e⁻ $~\land$ 2 H⁺_(aq) + 2 e⁻ → H_{2 (g)} $~\land$ H₂O _(l) = H⁺_(aq) + OH⁻_(aq) $~\land$ LiOH _(aq) = Li⁺_(aq) + OH⁻_(aq) $~\Rightarrow$

2 Li _(s) + 2 H⁺_(aq) + 2 e⁻ → 2 Li⁺_(aq) + 2 e⁻ + H_{2 (g)} $~\land$ 2 H₂O _(l) = 2 H⁺_(aq) + 2 OH⁻_(aq) $~\land$ 2 LiOH _(aq) = 2 Li⁺_(aq) + 2 OH⁻_(aq) $~\Rightarrow$

2 Li _(s) + 2 H⁺_(aq) → 2 Li⁺_(aq) + H_{2 (g)} $~\land$ 2 H⁺_(aq) = 2 H₂O _(l) − 2 OH⁻_(aq) $~\land$ 2 Li⁺_(aq) = 2 LiOH _(aq) − 2 OH⁻_(aq) $~\Rightarrow$

2 Li _(s) + 2 H₂O _(l) − 2 OH⁻_(aq) → 2 LiOH _(aq) − 2 OH⁻_(aq) + H_{2 (g)} $~\Rightarrow$

2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H_{2 (g)} $~\triangleright$

The sentence "2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H₂ ↑" may be inferred from the following sentences:

Li _(s) → Li⁺_(aq) + e⁻ _{& Eox = −3.045 V}

2 H⁺_(aq) + 2 e⁻ → H_{2 (g) & Ered = 0 V}

H₂O _(l) = H⁺_(aq) + OH⁻_(aq)

LiOH _(aq) = Li⁺_(aq) + OH⁻_(aq)

Inference

$~\triangleleft$ Li _(s) → Li⁺_(aq) + e⁻_{& Eox = −3.045 V} $~\land$ 2 H⁺_(aq) + 2 e⁻ → H_{2 (g) & Ered = 0 V} $~\land$ H₂O _(l) = H⁺_(aq) + OH⁻_(aq) $~\land$ LiOH _(aq) = Li⁺_(aq) +

+ OH⁻_(aq) $~\Rightarrow$

2 Li _(s) + 2 H⁺_(aq) + 2 e⁻ → 2 Li⁺_(aq) + 2 e⁻ + H_{2 (g) & Eredox = Ered − Eox = (0 − −3.045) V = +3.045 V} $~\land$ 2 H₂O _(l) = 2 H⁺_(aq) + 2 OH⁻_(aq)

$~\land$ 2 LiOH _(aq) = 2 Li⁺_(aq) + 2 OH⁻_(aq) $~\Rightarrow$

2 Li _(s) + 2 H⁺_(aq) → 2 Li⁺_(aq) + H_{2 (g) & Eredox = 3.045 V} $~\land$ 2 H⁺_(aq) = 2 H₂O _(l) − 2 OH⁻_(aq) $~\land$ 2 Li⁺_(aq) = 2 LiOH _(aq) − 2 OH⁻_(aq) $~\Rightarrow$

2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H_{2 (g) & Eredox > 0 V} $~\Leftrightarrow$

2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H₂ ↑ $~\triangleright$

The sentence "2 Li _(s) + 2 H₂O _(l) → _{T = 298.15 K} 2 LiOH _(aq) + H_{2 (g) & Eredox = 3.045 V}" contains the adverbial clause "T = 298.15 K" and may be inferred from the following sentences:

T = 298.15 K $~\to$ Li _(s) → Li⁺_(aq) + e⁻ _{& Eox = −3.045 V}

T = 298.15 K $~\to$ 2 H⁺_(aq) + 2 e⁻ → H_{2 (g) & Ered = 0 V}

T = 298.15 K $~\to$ H₂O _(l) = H⁺_(aq) + OH⁻

T = 298.15 K $~\to$ LiOH _(aq) = Li⁺_(aq) + OH⁻_(aq)

Inference

$~\triangleleft$ ( T = 298.15 K $~\to$ Li _(s) → Li⁺_(aq) + e⁻ _{& Eox = −3.045 V} ) $~\land$ ( T = 298.15 K $~\to$ 2 H⁺_(aq) + 2 e⁻ → H_{2 (g) & Ered = 0 V} )

$~\land$ ( T = 298.15 K $~\to$ H₂O _(l) = H⁺_(aq) + OH⁻_(aq) ) $~\land$ ( T = 298.15 K $~\to$ LiOH _(aq) = Li⁺_(aq) + OH⁻_(aq) ) $~\Rightarrow$

T = 298.15 K $~\to$ ( Li _(s) → Li⁺_(aq) + e⁻ _{& Eox = −3.045 V} $~\land$ 2 H⁺_(aq) + 2 e⁻ → H_{2 (g) & Ered = 0 V} $~\land$ H₂O _(l) = H⁺_(aq) + OH⁻_(aq)

$~\land$ LiOH _(aq) = Li⁺_(aq) + OH⁻_(aq) ) $~\Rightarrow$

T = 298.15 K $~\to$ 2 Li _(s) + 2 H₂O _(l) → 2 LiOH _(aq) + H_{2 (g) & Eredox = 3.045 V} $~\Leftrightarrow$

2 Li _(s) + 2 H₂O _(l) → _{T = 298.15 K} 2 LiOH _(aq) + H_{2 (g) & Eredox = 3.045 V} $~\triangleright$