
Among the following gases which gas has the highest value for Henry’s law constant ${K_H}$ at the same temperature
A. ${O_2}$
B. ${N_2}$
C. ${H_2}$
D. $He$
Answer
565.2k+ views
Hint: Henry Law relates partial pressure of gases to the amount of which is dissolved in the solvent. This is also the concentration of the gas in that solvent.
Formula used will be:
$P = {K_H} \times C$
From this formula derives how Partial pressure and Concentration are relation to Henry’s Law constant.
See the changing trend of these factors in the above given options and which one of them will give the highest possible value for the constant for their value of Partial pressure and concentration at same temperature.
Complete step by step answer:
Partial pressure of a gas is the thermodynamic property of a gas. When a gas is dissolved in a solvent, Henry’s law states that, the amount of gas dissolved, or the solubility of this gas will be directly proportional to the partial pressure at a given temperature.
From the given formula,
$P = {K_H} \times C$
By rearranging these terms, we get:
${K_H} = \dfrac{P}{C}$ where, $P$ is the partial pressure, ${K_H}$ is Henry's law constant and $C$is the concentration of gas dissolved.
From this above equation we know that the solubility of the gas is inversely proportional to Henry’s law constant.
${K_H} \propto \,\dfrac{1}{C}$
This means that a gas is easily soluble in water, which would result in more amount of the gas dissolving in water, which would lead to a rise in its concentration and hence the Henry’s law constant should decrease.
$\therefore $ For a gas to have a high value of Henry’s law constant, it should have very less solubility in water.
Hence the Value of Henry’s law constant in decreasing order will be:
$He > {N_2} > {H_2} > {O_2}$
So, Option D is correct.
Note: This assumption is only valid if all the gases are considered to be dissolved in aqueous medium only and the temperature has been maintained constant. If the temperature changes then solubility will increase and if the medium changes then there are chances of association of disassociation also, which will affect its solubility.
Formula used will be:
$P = {K_H} \times C$
From this formula derives how Partial pressure and Concentration are relation to Henry’s Law constant.
See the changing trend of these factors in the above given options and which one of them will give the highest possible value for the constant for their value of Partial pressure and concentration at same temperature.
Complete step by step answer:
Partial pressure of a gas is the thermodynamic property of a gas. When a gas is dissolved in a solvent, Henry’s law states that, the amount of gas dissolved, or the solubility of this gas will be directly proportional to the partial pressure at a given temperature.
From the given formula,
$P = {K_H} \times C$
By rearranging these terms, we get:
${K_H} = \dfrac{P}{C}$ where, $P$ is the partial pressure, ${K_H}$ is Henry's law constant and $C$is the concentration of gas dissolved.
From this above equation we know that the solubility of the gas is inversely proportional to Henry’s law constant.
${K_H} \propto \,\dfrac{1}{C}$
This means that a gas is easily soluble in water, which would result in more amount of the gas dissolving in water, which would lead to a rise in its concentration and hence the Henry’s law constant should decrease.
$\therefore $ For a gas to have a high value of Henry’s law constant, it should have very less solubility in water.
Hence the Value of Henry’s law constant in decreasing order will be:
$He > {N_2} > {H_2} > {O_2}$
So, Option D is correct.
Note: This assumption is only valid if all the gases are considered to be dissolved in aqueous medium only and the temperature has been maintained constant. If the temperature changes then solubility will increase and if the medium changes then there are chances of association of disassociation also, which will affect its solubility.
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