Solutions
4.0 Henry's law
4.0 Henry's law
According to Henry's law, the partial pressure of a gas over the solution is directly proportional to mole fraction of gas in solution.
Partial pressure is defined as the pressure exerted by an individual gas in a mixture of gases occupying the same volume.
So, if four gases $A$, $B$, $C$, $D$ are present in a mixture of gases then the partial pressure exerted by gas $A$ is the product of total pressure and mole fraction of gas $A$ in the mixture.
Mathematically, $$\begin{equation} \begin{aligned} {P_A} \propto {x_A} \\ {P_A} = {K_H}{x_A} \\\end{aligned} \end{equation} $$
where,
$P_A$=partial pressure exerted by gas $A$,
$x_A$=mole fraction of gas $A$ in the mixture,
${K_H}$=Henry's law constant.
Henry's law constant depends on following factors:
1. Temperature
2. Nature of gas
Temperature
Usually, ${K_H}$ increases with increase in temperature.
It means solubility decreases with increase in temperature.
Relation between Henry's law constant $({K_H})$ and temperature $(T)$ is given by Van't Hoff equation.
$$\ln \frac{{{K_{{H_2}}}}}{{{K_{{H_1}}}}} = - \frac{{\Delta H^\circ }}{R}.\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right)$$
where,
${K_H}$: Henry law constant at any temperature $T$.
$K_H^\theta $:Henry law constant at $298K$.
${\Delta H^\circ }$: Enthalpy of solution
$R$: Universal gas constant
Nature of gas
It is different for different gases.
$$P = {K_H}x$$$$ \Rightarrow x \propto 1/{K_H}$$
It means, higher the ${K_H}$ lower is the solubility of gas.
$ \to $ For a gas at given temperature, ${K_H}$ remains constant.
So, for two different pressure ${P_1}$ and ${P_2}$ let the solubility of gas is ${x_1}$ and ${x_2}$ respectively, then
$${P_1} = {K_H}{x_1}$$
and $${P_2} = {K_H}{x_2}$$
$$ \Rightarrow{P_1}/{P_2} = {x_1}/{x_2}$$
As, ${x_1} = {n_1}/total\ moles$
and ${x_2} = {n_2}/total\ moles$
$$ \Rightarrow{P_1}/{P_2} = {n_1}/{n_2}$$
where, $n$ represent moles, $x$ represents mole fraction.
$ \to $ We know, $$P = {K_H}x$$
If solution is made up of one gas and solvent then,
$$P = {K_H}({n_g}/{n_g} + {n_{solvent}})$$
$${n_g} + {n_{solvent}}$$ \cong $${n_{solvent}}$$
$$P = {K_H}({n_g}/{n_{solvent})}$$
$ \to $ The gases which disassociate in the given liquid are more soluble in that liquid.
For example: $C{O_2}$, $HCHO$, vinyl chloride are more soluble in water, as on mixing these gases disassociates.
It implies that these gases have low henry law constant in water.
$ \to $ Solubility of oxygen is high in cold water. That's why aquatic species prefers cold water more.
Applications of Henry law
1) Soft drinks are sealed under high pressure to increase solubility of $C{O_2}$ in drinks.
2) Due to high pressure under water solubility of gases in scuba divers blood increases. When they come out of water solubility of gases decreases as pressure decreases, hence the blood releases dissolved gases which leads to formation of bubbles of Nitrogen in blood which in turn blocks capillaries and leads to medical condition called bends, which can lead to death.To reduce this, Helium is used in tanks of divers.
3) At high altitudes due to low partial pressure of oxygen, there is low concentration of oxygen in blood of climbers and people living there. So, they feel weakness and inability to think clearly which are symptoms of medical condition called Anoxia.
Limitation of Henry law constant
1) It can be applied to only sufficiently dilute solution.
2) The component should not react to each other.
3) Gas and its solutions are essentially ideal.
4) Pressure of gas is not too high.
5) Temperature is not too low.
{while using Henry law take care of units and chemically reacting systems}
Question 5. If Nitrogen gas is bubbled through water at $293$ $K$, how many millimole of Nitrogen gas would dissolve in $1$ litre of water. Assume that Nitrogen exerts a partial pressure of $0.987\ bar$. Given that Henry’s law constant for Nitrogen at $293\ K$ is $76.48\ kbar$.
Solution: The solubility of gas is related to the mole fraction in aqueous solution.
The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus:
$${x_{{N_2}}} = \frac{{{p_{_{{N_2}}}}}}{{{K_H}}} = 1.29\times 10–5$$
As $1 litre$ of water contains $55.5 mol$ of it, therefore if $n$ represents number of moles of ${N_2}$ in solution,
$${x_{{N_2}}} = \frac{n}{{n + 55.5}} = \frac{n}{{55.5}} = 1.29 \times {10^{ - 5}}$$
($n$ in denominator is neglected as it is $< < 55.5$)
Thus, $$n = 0.716\ mmol$$
$ \to $ Effect of temperature: Dissolution of gas in a liquid is exothermic process. So, on increasing the temperature solubility of gases decreases by le-chatelier's principle.
Liquid solution
A Solution is called liquid solution if solvent is liquid and solute can be liquid, solid or gas. In liquid solutions, generally liquid solvent is volatile in nature. The solute can be volatile or Non volatile.
