The value of Henry's law constant for few gases at 298 K is given below. Arrange these gases in the increasing order of their solubility .
`N_2` : 76.48 kbar. `O_2` : 34.86 kbar, `H_2`: 69.16 kbar

Options

A. `O_2 lt N_2 lt H_2`
B. `O_2 lt H_2 lt N_2`
C. `H_2 lt N_2 lt O_2`
D. `N_2 lt H_2 lt O_2`

Answer

To solve the problem of arranging the gases in increasing order of their solubility based on Henry's law constant, we can follow these steps:

### Step 1: Understand Henry's Law
Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The relationship can be expressed as:
\[ S = k_H \cdot P \]
where:
- \( S \) is the solubility of the gas,
- \( k_H \) is Henry's law constant,
- \( P \) is the partial pressure of the gas.

### Step 2: Analyze the Given Data
We have the following Henry's law constants for the gases at 298 K:
- \( N_2 \): 76.48 kbar
- \( O_2 \): 34.86 kbar
- \( H_2 \): 69.16 kbar

### Step 3: Determine the Relationship Between Henry's Law Constant and Solubility
From Henry's law, we know that a higher value of \( k_H \) indicates lower solubility. Therefore, we can say:
- The gas with the highest \( k_H \) will have the lowest solubility.
- The gas with the lowest \( k_H \) will have the highest solubility.

### Step 4: Compare the Values
Now, let's compare the values of \( k_H \):
- \( N_2 \): 76.48 kbar (highest)
- \( H_2 \): 69.16 kbar
- \( O_2 \): 34.86 kbar (lowest)

### Step 5: Arrange the Gases in Increasing Order of Solubility
Based on the values:
- **Lowest Solubility**: \( N_2 \) (76.48 kbar)
- **Medium Solubility**: \( H_2 \) (69.16 kbar)
- **Highest Solubility**: \( O_2 \) (34.86 kbar)

Thus, the increasing order of solubility is:
\[ N_2 < H_2 < O_2 \]

### Final Answer
The gases arranged in increasing order of their solubility are:
1. \( N_2 \)
2. \( H_2 \)
3. \( O_2 \)