At NTP, the solubility of natural gas in water is 0.8 mole of gas/kg of water. What is the Henry’s law constant for natural gas?

Options

A. 8 kN/`m^2`
B. 7.90 x `10^-3 `Pa
C. 71.36 bar
D. 105 mmHg

Answer

To find the Henry's law constant for natural gas, we can follow these steps:

### Step 1: Understand the given data
We know that the solubility of natural gas in water at NTP (Normal Temperature and Pressure) is 0.8 moles of gas per kg of water.

### Step 2: Calculate the number of moles of water
To find the number of moles of water in 1 kg (1000 g), we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of water (H₂O) is approximately 18 g/mol. Thus,
\[ N_w = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

### Step 3: Calculate the mole fraction of natural gas
The mole fraction (x) of natural gas in the solution can be calculated using the formula:
\[ x_{ng} = \frac{n_{ng}}{n_{ng} + n_w} \]
Where \( n_{ng} \) is the number of moles of natural gas (0.8 mol) and \( n_w \) is the number of moles of water (55.56 mol).
\[ x_{ng} = \frac{0.8}{0.8 + 55.56} \approx \frac{0.8}{56.36} \approx 0.0142 \]

### Step 4: Apply Henry's Law
Henry's Law states that:
\[ P_{ng} = K_H \cdot x_{ng} \]
Where \( P_{ng} \) is the partial pressure of the natural gas, and \( K_H \) is the Henry's law constant. At NTP, the pressure of natural gas (\( P_{ng} \)) is 1.01325 bar.

### Step 5: Rearrange to find Henry's law constant
Rearranging the equation gives:
\[ K_H = \frac{P_{ng}}{x_{ng}} \]
Substituting the values we have:
\[ K_H = \frac{1.01325 \, \text{bar}}{0.0142} \approx 71.36 \, \text{bar} \]

### Conclusion
The Henry's law constant for natural gas is approximately **71.36 bar**.

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