Question

Air is a mixture of gases that is about 78.0% n2 by volume. when air is at standard pressure and 25.0 ∘c, the n2 component will dissolve in water with a solubility of 4.88×10−4 m. what is the value of henry's law constant for n2 under these conditions? express your answer with the appropriate units. enter the unit m using the compound form mol/l.

Answer 1

Step first:

Since, 78.0 percent  $N_{2}$ by volume is present in the air which implies there are 78 moles of $N_{2}$ present in 100 moles of air.

Now,

Mole fraction of $N_{2}$ = $\frac{Moles of N_{2}}{Total moles of gases in the mixture}$

Mole fraction  =$\frac{78}{100} mol$

= 0.78

Partial pressure is equal to the multiplication of total pressure and mole fraction.

Partial pressure = $0.78\times 1 atm$ (as 1 atm is atmospheric pressure)

= 0.78 atm.

Step second:

Henry's law constant is calculated by:

$S_{g}=k\times P_{g}$

where,

$S_{g}$ is solubility of gas

$P_{g}$ is partial pressure of gas

$k$ is henry's law constant

Substitute the value of solubility and partial pressure to find the value of Henry's law constant in above formula:

$4.88\times 10^{-4} M=k\times 0.78 atm$

k = $\frac{4.88\times 10^{-4}M}{0.78 atm}$

= $6.25\times 10^{-4} M atm^{-1}$

Thus, Henry's Law constant is $6.25\times 10^{-4} mol L^{-1} atm^{-1}$.

Answer 2

The value of Henry's law constant for N₂ under these conditions : kH = 6,256 .10⁻⁴ mol/L.atm

Further explanation

In the transfer from the gas phase to the liquid phase, diffusion process occurs.

The solubility of the gas itself decreases with increasing temperature

The partial pressure of gas itself is influenced by the concentration of the gas and temperature

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

Can be formulated

$\large{\boxed{\bold{S~=~kH~\times~P}}}$

S = gas solubility, mol / L

kH = Henry constant, mol / L.atm

P = gas partial pressure

The gas partial pressure is proportional to the percentage of the gas volume in the mixture

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

$\frac{V_N}{V_{tot}} = \frac{P_N}{P_{tot}}$

$\frac{78}{100}  = \frac{P_N}{1}$

P = 0.78 atm

Nitrogen gas solubility itself 4.88 × 10⁻⁴ M (mol/L)

so the constant H:

kH = S / P

kH = 4.88 × 10⁻⁴M / 0.78

kH = 6,256 .10⁻⁴

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Keywords: Henry's Law, partial pressure, gas solubility