Match List I with List II:
|
List I \(K_H \text{ values }/ K\text{ bar}\) |
List II Gas |
| (A) 145 | (I) \(CO_2\) |
| (B) 89 | (II) \(He\) |
| (C) 76.5 | (III) \(N_2\) (at 293 K) |
| (D) 1.67 | (IV) \(N_2\) (at 303 K) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is option 1. (A)-(II), (B)-(IV), (C)-(III) D-(I).
|
List I \(K_H \text{ values }/ K\text{ bar}\) |
List II Gas |
| (A) 145 | (II) \(He\) |
| (B) 89 | (IV) \(N_2\) (at 303 K) |
| (C) 76.5 | (III) \(N_2\) (at 293 K) |
| (D) 1.67 | (I) \(CO_2\) |
Henry's law constant $K_H$ is higher for gases that are less soluble in a liquid.
- He (helium) is least soluble $\Rightarrow$ highest $K_H \Rightarrow$ matches 145 $\rightarrow$ (A)-(II)
- For the same gas, increase in temperature decreases solubility, so $K_H$ increases
$\Rightarrow K_H$ of $N_2$ at $303\text{ K} > 293\text{ K}$
$\Rightarrow 89 \rightarrow N_2 (303\text{ K}) \Rightarrow$ (B)-(IV)
$\Rightarrow 76.5 \rightarrow N_2 (293\text{ K}) \Rightarrow$ (C)-(III)
- $CO_2$ is most soluble $\Rightarrow$ lowest $K_H$ $\Rightarrow 1.67 \rightarrow$ (D)-(I)
Thus, the correct matching is: (A)-(II), (B)-(IV), (C)-(III), (D)-(I)