Pogovor:Alkalijska kovina

Addition to the article "Alkalijska kovina"

0. The sentence "2Na _(s) + 2H₂O _(l) → 2NaOH _(aq) + H₂ ↑" consists of the subject ( 2Na _(s) + 2H₂O _(l) ), the predicate ( → ), and the object

( 2 NaOH _(aq) + H₂ ↑ ).

N.B.

NaOH _(aq) = NaOH _{(dissolved in H2O (l))}

1. The sentence "2Na _(s) + 2H₂O _(l) → 2NaOH _(aq) + H₂ ↑" describes an irreversible process, that is:

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

N.B.

$~\land$ is [logical] and.

$~\neg$ is [logical] not.

$~\neg$( 2NaOH _(aq) + H_{2 (g)} → 2Na _(s) + 2H₂O _(l) ) $~\Leftrightarrow$ It is false that 2NaOH _(aq) + H_{2 (g)} → 2Na _(s) + 2H₂O _(l).

Other examples of the sentences that describe irreversible processes

2Na _(s) + 2HCl _(aq) → 2NaCl _(aq) + H₂ ↑

$~\Leftrightarrow$ 2Na _(s) + 2HCl _(aq) → 2NaCl _(aq) + H_{2 (g)} $~\land \neg$( 2NaCl _(aq) + H_{2 (g)} → 2Na _(s) + 2HCl _(aq) )

2Na _(s) + 2NH_{3 (l)} → 2NaNH_{2 (dissolved in NH3 (l))} + H₂ ↑

$~\Leftrightarrow$ 2Na _(s) + 2NH_{3 (l)} → 2NaNH_{2 (dissolved in NH3 (l))} + H_{2 (g)} $~\land \neg$( 2NaNH_{2 (dissolved in NH3 (l))} + H_{2 (g)} → 2Na _(s) + 2NH_{3 (l)} )

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

2Na _(s) + 2H₂O _(l) → 2NaOH _(aq) + H₂ ↑ $~\Rightarrow$ 2Na⁰ _(s) + 2H⁺¹₂O _(l) → 2Na⁺¹OH _(aq) + H⁰_{2 (g)}

N.B.

2Na⁰ _(s) + 2H⁺¹₂O _(l) → 2Na⁺¹OH _(aq) + H⁰_{2 (g)} $~\Rightarrow$ Oxl( Na in Na _(s) ) < Oxl( Na in NaOH _(aq) ) $~\land$ Oxl( H in H₂O _(l) ) > Oxl( H in H_{2 (g)} )

Other examples of the sentences that describe redox processes

2Na _(s) + 2HCl _(aq) → 2NaCl _(aq) + H₂ ↑ $~\Rightarrow$ 2Na⁰ _(s) + 2H⁺¹Cl _(aq) → 2Na⁺¹Cl _(aq) + H⁰_{2 (g)}

2Na _(s) + 2NH_{3 (l)} → 2NaNH_{2 (dissolved in NH3 (l))} + H₂ ↑ $~\Rightarrow$ 2Na⁰ _(s) + 2NH⁺¹_{3 (l)} → 2Na⁺¹NH_{2 (dissolved in NH3 (l))} + H⁰_{2 (g)}

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

Na _(s) → Na⁺_(aq) + e⁻,

2H⁺_(aq) + 2e⁻ → H_{2 (g)},

H₂O _(l) = H⁺_(aq) + OH⁻_(aq),

NaOH _(aq) = Na⁺_(aq) + OH⁻_(aq).

Inference

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

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

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

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

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

N.B.

The sentence "Na _(s) → Na⁺_(aq) + e⁻" describes an oxidative process.

The sentence "2H⁺_(aq) + 2e⁻ → H_{2 (g)}" describes a reductive process

The sentence "2Na _(s) + 2HCl _(aq) → 2NaCl _(aq) + H_{2 (g)}" may be inferred from the following compound sentence "Na _(s) → Na⁺_(aq) + e⁻

$~\land$ 2H⁺_(aq) + 2e⁻ → H_{2 (g)} $~\land$ HCl _(aq) = H⁺_(aq) + Cl⁻_(aq) $~\land$ NaCl _(aq) = Na⁺_(aq) + Cl⁻_(aq)".

The sentence "2Na _(s) + 2NH_{3 (l)} → 2NaNH_{2 (dissolved in NH3 (l))} + H_{2 (g)}" may be inferred from the following compound sentence

"Na _(s) → Na⁺_{(dissolved in NH3 (l))} + e⁻ $~\land$ 2H⁺_{(dissolved in NH3 (l))} + 2e⁻ → H_{2 (g)} $~\land$ NH_{3 (l)} = H⁺_{(dissloved in NH3 (l))} + NH₂⁻_{(dissolved in NH3 (l))}

$~\land$ NaNH_{2 (dissolved in NH3 (l))} = Na⁺_{(dissolved in NH3 (l))} + NH₂⁻_{(dissolved in NH3 (l))}".

4. The sentence "2Na _(s) + 2H₂O _(l) → 2NaOH _(aq) + H₂ ↑" may be inferred from the following sentences:

Na _(s) → Na⁺_(aq) + e⁻ _{& Eox = −2.71 V},

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

H₂O _(l) = H⁺_(aq) + OH⁻_(aq),

NaOH _(aq) = Na⁺_(aq) + OH⁻_(aq).

Inference

$~\triangleleft$ Na _(s) → Na⁺_(aq) + e⁻ _{& Eox = −2.71 V} $~\land$ 2H⁺_(aq) + 2e⁻ → H_{2 (g) & Ered = 0 V} $~\land$ H₂O _(l) = H⁺_(aq) + OH⁻_(aq)

$~\land$ NaOH _(aq) = Na⁺_(aq) + OH⁻_(aq) $~\Rightarrow$

2Na _(s) → 2Na⁺_(aq) + e⁻ _{& Eox = −2.71 V} $~\land$ 2H⁺_(aq) + 2e⁻ → H_{2 (g) & Ered = 0 V} $~\land$ H⁺_(aq) = H₂O _(l) − OH⁻_(aq)

$~\land$ Na⁺_(aq) = NaOH _(aq) − OH⁻_(aq) $~\Rightarrow$

2Na _(s) + 2H⁺_(aq) + 2e⁻ → 2Na⁺_(aq) + 2e⁻ + H_{2 (g) & Eredox = Ered − Eox = (0 − −2.71) V = +2.71 V} $~\land$ 2H⁺_(aq) = 2H₂O _(l) − 2OH⁻_(aq)

$~\land$ 2Na⁺_(aq) = 2NaOH _(aq) − 2OH⁻_(aq) $~\Rightarrow$

2Na _(s) + 2H₂O _(l) − 2OH⁻_(aq) → 2NaOH _(aq) − 2OH⁻_(aq) + H_{2 (g) & Eredox = 2.71 V} $~\Rightarrow$

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

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

N.B.

The sentence "Na _(s) → Na^{+_(aq) + e− _{& Eox = −2.71 V}" contains the adverbial clause "E_ox = −2.71 V".}

The sentence "2H⁺_(aq) + 2e⁻ → H_{2 (g) & Ered = 0 V}" contains the adverbial clause "E_red = 0 V".

5. The sentence "2Na _(s) + 2H₂O _(l) → _{T = 298.15 K} 2NaOH _(aq) + H_{2 (g)} ↑" contains the conditional clause "T = 298.15 K" and may be inferred

from the following complex sentences:

T = 298.15 K $~\to$ Na _(s) → Na⁺_(aq) + e⁻ _{& Eox = −2.71 V},

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

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

T = 298.15 K $~\to$ NaOH _(aq) = Na⁺_(aq) + OH⁻_(aq),

where A $~\to$ B $~\Leftrightarrow$ "If A, then B." (see the article "Material condition").

Inference

$~\triangleleft$ ( T = 298.15 K $~\to$ Na _(s) → Na⁺_(aq) + e⁻ _{& Eox = −2.71 V} ) $~\land$ ( T = 298.15 K $~\to$ 2H⁺_(aq) + 2e⁻ → H_{2 (g) & Ered = 0 V} )

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

T = 298.15 K $~\to$ ( Na _(s) → Na⁺_(aq) + e⁻ _{& Eox = −2.71 V} $~\land$ 2H⁺_(aq) + 2e⁻ → H_{2 (g) & Ered = 0 V} $~\land$ H₂O _(l) = H⁺_(aq) + OH⁻_(aq)

$~\land$ NaOH _(aq) = Na⁺_(aq) + OH⁻_(aq) ) $~\Rightarrow$

T = 298.15 K $~\to$ 2Na _(s) + 2H₂O _(l) → 2NaOH _(aq) + H_{2 & Eredox = 2.71 V} $~\Rightarrow$

T = 298.15 K $~\to$ 2Na _(s) + 2H₂O _(l) → 2NaOH _(aq) + H_{2 & Eredox > 0 V} $~\Leftrightarrow$

T = 298.15 K $~\to$ 2Na _(s) + 2H₂O _(l) → 2 NaOH _(aq) + H₂ ↑ $~\Leftrightarrow$

2Na _(s) + 2H₂O _(l) → _{T = 298.15 K} 2NaOH _(aq) + H₂ ↑ $~\triangleright$

6. Any chemical process takes place in some system.

6.1 The complex sentence

"m(NaOH _(aq)) >> 2 mol • M(NaOH) $~\to$ 2Na _(s) + 2H₂O _(l) → _{T = 298.15 K} 2NaOH _(aq) + H_{2 (g) & PΔV ≈ RT = 2.479 kJ / mol}"

describes the chemical process in a closed system (= the system which does not exchange matter with its surroundings).

That sentence consists of:

a) the subordinate clause "m(NaOH _(aq)) >> 2 mol • M(NaOH)"

b) the main clause "2Na _(s) + 2 H₂O _(l) → _{T = 298.15 K} 2NaOH _(aq) + H_{2 (g) & PΔV ≈ RT = 2.479 kJ/mol}"

That main clause contains the conditional clause "T = 298.15 K" and the other adverbial clause "PΔV ≈ RT = 2.479 kJ/mol".

N.B.

m(NaOH _(aq)) is the mass of the solution of sodium hydroxide in water.

M(Na) is the mass of one mole of sodium. M(Na) = 23 g/mol

Δm(Na) = −2 mol • M(Na) = −46 g

M(H₂O) is the mass of one mole of water. M(H₂O) = 18 g/mol

Δm(H₂O) = −2 mol • M(H₂O) = −36 g

M(NaOH) is the mass of one mole of sodium hydroxide. M(NaOH) = 40 g/mol

Δm(NaOH) = +2 mol • M(NaOH) = 80 g

M(H₂) is the mass of one mole of diatomic hydrogen. M(H₂) = 2 g/mol

Δm(H₂) = +1 mol • M(H₂) = 2 g

ρ(Na _(s)) = 0.97 g/cm³

T = 298.15 K $~\land$ C_oncentration(NaOH _(aq)) = 1 mol/L $~\to$ ρ(NaOH _(aq)) ≈ 1.04 g/cm³

T = 298.15 K $~\to$ ρ(H_{2 (g)}) = 8.18 • 10⁻⁵ g/cm³

R = 8.314 J • mol⁻¹ • K⁻¹

PΔV = P • ( Δm(Na) / ρ(Na _(s)) + Δm(H₂O) / ρ(NaOH _(aq)) + Δm(NaOH) / ρ(NaOH _(aq)) + Δm(H₂) / ρ (H_{2 (g)}) ) • 10⁻⁶ =

= P • ( −46 / 0.97 + −36 / 1.04 + 80 / 1.04 + 2 / (8.18 • 10⁻⁵) ) • 10⁻⁶ ≈

≈ P • ( −46 + −36 + 80 + 2•10^{5 / 8.18 ) • 10−6 =}

= P • ( −2 + 2•10⁵ / 8.18 ) • 10⁻⁶ ≈

≈ P • (2 / (8.18 • 10⁻⁵)) • 10⁻⁶ =

= P • (1 mol • M(H₂) / ρ(H_{2 (g)})) • 10⁻⁶ = R • T = 8.314 • 298.15 ≈ 2479 J/mol = 2.479 kJ /mol.

6.2 The sentence

m_s(NaOH _(aq)) > 116 g • (1 − 10⁻² • 2C_s(NaOH _(aq)))⁻¹ $~\to$ 2Na _(s) + 2H₂O _(l) → _{T = 293.15 K} 2NaOH _(aq) + 2H_{2 (g) & Cf(NaOH (aq)) =}

_{= (Cs(NaOH (aq)) • 10⁻² • ms(NaOH (aq)) + Δm(NaOH)) • 10² • (ms(NaOH (aq)) + Δm(H2O) + Δm(NaOH))⁻¹}

describes the chemical process in a closed system too.

That sentence consists of:

a) the subordinate clause "m_s(NaOH _(aq)) > 116 g / (1 − 2C_s(NaOH _(aq)) • 10⁻²),

b) the main clause "2Na _(s) + 2H₂O _(l) → _{T = 293.15 K} 2NaOH _(aq) + 2 _{2 (g) & Cf(NaOH (aq)) =}

_{= (Cs(NaOH (aq)) • 10⁻² • ms(NaOH (aq)) + Δm(NaOH)) • 10² / (ms(NaOH (aq) + Δm(H2O) + Δm(NaOH))}

The main clause contains the conditional clause "T = 293.15 K" and the other adverbial clause "C_f(NaOH _(aq)) = (C_s(NaOH _(aq)) • 10⁻² • m_s(NaOH _(aq)) + Δm(NaOH)) • 10² / (m_s(NaOH _(aq) + Δm(H₂O) + Δm(NaOH))".

N.B.

m_s(NaOH _(aq)) is the mass of the solution of sodium hydroxide in water before the beginning of the chemical process

C_s(NaOH _(aq)) is the concentration of the solution of sodium hydroxide in water before the beginning of the chemical process.

C_s(NaOH _(aq)) is measured in mass percents. C_s(NaOH _(aq)) ≥ 0%

C_f(NaOH _(aq)) is the concetration of the solution of sodium hydrooxide in water after the end of the chemical process.

C_f(NaOH _(aq)) is measured in mass percents. C_f(NaOH _(aq)) ≤ C_lim(NaOH _(aq)) ≈ 52%

Δm(NaOH) = +2 mol • M(NaOH) = 80 g

Δm(H₂O) = −2 mol • M(H₂O) = −36 g

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