Question

The value of Henry’s law constant for some gases at $\text{ 293 K }$ is given below. Arrange the gases in the increasing order of their solubility.
$\text{ He = 144}\text{.97 kbar }$ , ${\text{H}}_{\text{2}}\text{ = 69}\text{.16 kbar }$ ,${\text{N}}_{\text{2}}\text{ = 76}\text{.48 kbar }$ ,${\text{O}}_{\text{2}}\text{ = 34}\text{.86 kbar }$.
A) $\text{ He }<{\text{N}}_{\text{2}}<{\text{H}}_{\text{2}}<{\text{O}}_{\text{2}}$
B) ${\text{O}}_{\text{2}}<{\text{H}}_{\text{2}}<{\text{N}}_{\text{2}}<\text{He }$
C) ${\text{H}}_{\text{2}}<{\text{N}}_{\text{2}}<{\text{O}}_{\text{2}}<\text{ He }$
D) $\text{ He }<{\text{O}}_{\text{2}}<{\text{N}}_{\text{2}}<{\text{H}}_{\text{2}}$

Answer 1

Hint: Mass of a gas dissolved per unit volume of a solvent is proportional to the pressure of the gas in equilibrium with the solution at a constant temperature. If X is the mole fraction of gas is dissolved per unit volume of a solvent and p is the partial pressure of the gas in equilibrium with the solution, then, at a constant temperature,
$\text{ p = }{\text{K}}_{\text{H}}\text{ X }$
Where ${\text{K}}_{\text{H}}$ is the proportionality constant. It is called the henry's law constant. The amount of gas dissolved in the liquid is inversely related to the henry law constant${\text{K}}_{\text{H}}$ .

Complete step by step solution:
Henry law is stated in terms of pressure and mole fraction. According to which the pressure of the gas over a solution in which gas is dissolved is proportional to the mole fraction of gas dissolved in the solution. That is,
$\text{ p = }{\text{K}}_{\text{H}}\text{X }$
On further simplifying we have,
${\text{K}}_{\text{H}}\text{=}{\displaystyle \frac{\text{p}}{\text{X}}}$
From the stated relation, we can conclude that the proportionality constant ${\text{K}}_{\text{H}}$ depend on on the pressure of the gas and inversely related to the mole fraction of gas.
We are given the Henry's law constant k for various gases at a constant temperature as follows
$\begin{array}{rl}& \text{ He = 144}\text{.97 kbar }\\ & {\text{H}}_{\text{2}}\text{ = 69}\text{.16 kbar }\\ & {\text{N}}_{\text{2}}\text{ = 76}\text{.48 kbar }\\ & {\text{O}}_{\text{2}}\text{ = 34}\text{.86 kbar }\end{array}$
Let's consider that the pressure for all gases is constant. Then henry law constant would be written as,
${\text{K}}_{\text{H}}\text{=}{\displaystyle \frac{{\text{P}}_{\text{constant}}}{\text{X}}}$
That Henry law constant is now dependent only on the mole fraction of gas. If gas which has high solubility would have a higher mole fraction of gas in the liquid and one which has low solubility will have a low value of mole fraction of gas dissolved in liquid.
Here we know that Henry's constant and mole fraction are inversely related. Thus gas which has a higher value of ${\text{K}}_{\text{H}}$ have lower solubility and which has a low value of ${\text{K}}_{\text{H}}$ having high solubility in liquid. Therefore we can rearrange the given henry law constant according to the increasing solubility order as shown below,
$\text{ He }<{\text{N}}_{\text{2}}<{\text{H}}_{\text{2}}<{\text{O}}_{\text{2}}$

Hence, (A) is the correct option.

Note: Note that, henry law is observed in various biological processes. Respiration and oxygenation of the blood are one of the most common examples of Henry's law. Inhalation of air increases the partial pressure of oxygen in the lung and thus oxygen gas dissolved in the blood which is then further circulated in the body.in the lungs partial pressure of oxygen is high and thus it is less dissolved in lung thus flows in the body, however, the partial pressure of carbon dioxide is low thus it readily dissolves in lungs and exhaled from the body. Henry law is also widely applied in pressure-filled carbonated drinks.