The value of Henry’s law constant for some gases at 293 K is given below. Arrange the gases in the increasing order of their solubility. He: 144.97 kbar, H2: 69.16 kbar, N2: 76.48 kbar, O2: 34.86 kbar

Question

The value of Henry’s law constant for some gases at 293 K is given below. Arrange the gases in the increasing order of their solubility. He: 144.97 kbar, H2: 69.16 kbar, N2: 76.48 kbar, O2: 34.86 kbar (A) He < N2 < H2 < O2 (B) O2 < H2 < N2 < He (C) H2 < N2 < O2 < He (D) He < O2 < N2 < H2

Tardigrade Admin · Accepted Answer

K H =( p / x ) At constant pressure, K H is inversely proportional to x. Therefore, the higher the K H lower the solubility. The increasing solubility order is: He < N 2< H 2< O 2

Q. The value of Henry’s law constant for some gases at $293 K$ is given below. Arrange the gases in the increasing order of their solubility.
$He : 144.97 kba r$ , $H_{2} : 69.16 kba r$ ,
$N_{2} : 76.48 kba r$ , $O_{2} : 34.86 kba r$

$He < N_{2} < H_{2} < O_{2}$

$O_{2} < H_{2} < N_{2} < He$

$H_{2} < N_{2} < O_{2} < He$

$He < O_{2} < N_{2} < H_{2}$

$< b r / > K_{H} = \frac{p}{x} < b r / >$
At constant pressure, $K_{H}$ is inversely proportional to $x$ . Therefore, the higher the $K_{H}$ lower the solubility.
The increasing solubility order is:
$He < N_{2} < H_{2} < O_{2}$

Q. The value of Henry’s law constant for some gases at 293K is given below. Arrange the gases in the increasing order of their solubility. He:144.97kbar, H2​:69.16kbar, N2​:76.48kbar, O2​:34.86kbar

He<N2​<H2​<O2​

O2​<H2​<N2​<He

H2​<N2​<O2​<He

He<O2​<N2​<H2​