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2 years ago

If air contains 78% Nitrogen by volume, what is the solubility of nitrogen in water at 25 degree celisus

Chemistry

1 answer:

Svet_ta [14]2 years ago

7 0

The solubility of nitrogen in water at 25 °C= 4.88 x 10⁻⁴ mol/L

<h3>Further explanation</h3>

Given

78% Nitrogen by volume

Required

The solubility of nitrogen in water

Solution

Henry's Law states that the solubility of a gas is proportional to its partial pressure

Can be formulated

S = kH. P.

S = gas solubility, mol / L

kH = Henry constant, mol / L.atm

P = partial gas pressure

In the standard 25 C state, the air pressure is considered to be 1 atm, so the partial pressure of N₂ -nitrogen becomes:

Vn / Vtot = Pn / Ptot

78/100 = Pn / 1

Pn = 0.78 atm

Henry constant for N₂ at 25 °c = 1600 atm/mol.L=6.25.10⁻⁴ mol/L.atm

The solubility :

$\tt S=6.25.10^{-4}\times 0.78\\\\S=4.88\times 10^{-4}~mol/L$

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Step 1. Write the equations for the two half-reactions

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Step 2. Identify the cathode and the anode

The half-cell with the more negative E° (Zn) is the anode.

Step 3. Calculate E°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

E° = +0.76 V

<h2>Part (b). Calculate K for the reaction </h2>

The relation between E° and K is

E° = (RT)/(nF)lnK

where

R = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

n = the moles of electrons transferred

F = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]lnK

0.76 = 0.012 85 lnK

lnK = 0.76/0.012 85 = 59.16

K =e^59.16 = 5.0 × 10^25

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3 years ago

If air contains 78% Nitrogen by volume, what is the solubility of nitrogen in water at 25 degree celisus​

If air contains 78% Nitrogen by volume, what is the solubility of nitrogen in water at 25 degree celisus