At $293 \mathrm{~K}$, the Henry law constant in water for $\mathrm{N}_2$ and $\mathrm{O}_2$ are $76.48…
At $293 \mathrm{~K}$, the Henry law constant in water for $\mathrm{N}_2$ and $\mathrm{O}_2$ are $76.48 \mathrm{k}$ bar and $34.86 \mathrm{k}$ bar respectively. What is the ratio of mole fractions of $\mathrm{N}_2$ and $\mathrm{O}_2$ in water?
(Assume partial pressures of $\mathrm{N}_2$ and $\mathrm{O}_2$ same at $293 \mathrm{~K}$ )
2.19
0.95
0.60
0.45
Solution
According to Henry's law :-
$
\mathrm{P}=\mathrm{xK}_{\mathrm{H}}
$
For $\mathrm{N}_2$ :- $\mathrm{P}_{\mathrm{N}_2}=\mathrm{x}_{\mathrm{N}_2} \mathrm{~K}_{\mathrm{H}}\left(\mathrm{N}_2\right)$
For $\mathrm{O}_2$ :- $\mathrm{P}_{\mathrm{O}_2}=\mathrm{x}_{\mathrm{O}_2} \mathrm{~K}_{\mathrm{H}}\left(\mathrm{O}_2\right)$
We have, $\mathrm{P}_{\mathrm{N}_2}=\mathrm{P}_{\mathrm{O}_2}, \mathrm{~K}_{\mathrm{H}}\left(\mathrm{N}_2\right)=76.48 \mathrm{k}$ bar, $\mathrm{K}_{\mathrm{H}}\left(\mathrm{O}_2\right)=3486 \mathrm{k}$ bar
$\begin{aligned} & \Rightarrow \frac{\mathrm{x}_{\mathrm{N}_2}}{\mathrm{x}_{\mathrm{O}_2}}=\frac{\left(\frac{\mathrm{P}_{\mathrm{N}_2}}{\mathrm{~K}_{\mathrm{H}}\left(\mathrm{N}_2\right)}\right)}{\left(\frac{\mathrm{P}_{\mathrm{O}_2}}{\mathrm{~K}_{\mathrm{H}}\left(\mathrm{O}_2\right)}\right)}=\frac{\frac{\mathrm{P}}{\mathrm{N}_2}}{\mathrm{~K}_{\mathrm{H}}\left(\mathrm{N}_2\right)} \times \frac{\mathrm{K}_{\mathrm{H}}\left(\mathrm{O}_2\right)}{\frac{\mathrm{P}}{\mathrm{O}_2}} \\ & =\frac{\mathrm{K}_{\mathrm{H}}\left(\mathrm{O}_2\right)}{\mathrm{K}_{\mathrm{H}}\left(\mathrm{N}_2\right)}=\frac{34.86}{76.48}=0.45\end{aligned}$