Question

How many millimoles of ${\mathrm{N}}_2$ gas would dissolve in 1 liter of water if ${\mathrm{N}}_2$ gas is bubbled through water at $293\mathrm{K},$ and exerts a partial pressure of$0.987\mathrm{bar}.$Given that Henry’s law constant for ${\mathrm{N}}_2$ at $293\mathrm{K}\mathrm{is}76.48\mathrm{kbar}$?

Answer 1

Step 1: Given data:

Given;$\mathrm{P}=0.987\mathrm{bar},\mathrm{K}=76.48\mathrm{Kbar}=76480\mathrm{bar}$

Step 2: Henry’s law :

According to Henry's law, at a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

Thus, there is an increase of partial pressure of the gas with an increase in the solubility of the gas which depends on the mole fraction in the liquid mixture.
The solubility of a gas dissolved in the aqueous solution is dependent on its mole fraction in the aqueous solution.

This mole fraction is determined with the help of Henry's law.
Mathematically, Henry’s law can be written as:
$p=K_{H}x$
Where $p=$ partial pressure of the gas dissolved in the solvent
$K_H=$ Henry’s constant
$x=$ mole fraction of the dissolved gas

Step 3: Applying Henry’s law for calculating the number of millimoles of N₂:

Now, in the case of nitrogen gas dissolved in the water, the mole fraction can be determined as:
$x_{N_2}={\displaystyle \frac{p_{N_2}}{K_H}}$

Step 4: Calculating the mole fraction of N₂:
Substituting the given values in the above equation we can get:
$x_{N_2}={\displaystyle \frac{0.987}{76480}}=1.29\times {10}^{-5}$
Also, 1 liter of water contains = $55.5moles$
Let the number of moles of nitrogen in the aqueous solution be $a$ .
Thus,

$x_{N_2}={\displaystyle \frac{a}{a+55.5}}$
The number of moles of nitrogen is very less in comparison to that of the moles of water.
Thus,

${\mathrm{x}}_{{\mathrm{N}}_2}=\frac{\mathrm{a}}{\mathrm{a}+55.5}\approx \frac{\mathrm{a}}{55.5}$

Step 5: Calculating the number of millimoles of N₂:
We have already calculated the value of the mole fraction. Thus, substituting the value, we have:
$$\begin{gathered} 1.29\times {10}^{-5}=\frac{a}{55.5} \\ \Rightarrow a=1.29\times {10}^{-5}\times 55.5 \end{gathered}$$
Thus, the number of moles of nitrogen will be equal to:
$$\begin{gathered} a=7.16\times {10}^{-4}mol \\ =7.16\times {10}^{-4}mol\times \frac{1000mmol}{1mol} \end{gathered}$$
Hence, the amount of nitrogen in millimoles will be:
$a=0.716mmol$
Hence, for the given condition, the millimoles of N₂ gas would dissolve in 1 litre of water is given as:$0.716\mathrm{mmol}$