($K_H \text{ for } \text{N}_2 \text{ is } 6.5 \times 10^7 \text{ mm Hg}$)
Solving for Nitrogen Mole Fraction in Water
This solution calculates the mole fraction of nitrogen ($\text{N}_2$) dissolved in water using Henry's Law, based on the air composition, total pressure, and the provided Henry's Law constant ($K_H$).
Key Information Provided
- Temperature: 298 K
- Total Pressure ($P_{total}$): 10 atm
- Mole percentage of $\text{N}_2$ in air: 80%
- Henry's Law Constant ($K_H$) for $\text{N}_2$: $6.5 \times 10^7$ mm Hg
Step 1: Determine Mole Fraction of N₂ in Air
The mole percentage of $\text{N}_2$ in air is given as 80%. Convert this percentage to a mole fraction ($x_{\text{N}_2, \text{air}}$):
$x_{\text{N}_2, \text{air}} = \frac{80}{100} = 0.80$
Step 2: Calculate Partial Pressure of N₂
Convert the total pressure from atm to mm Hg to match the units of the Henry's Law constant.
$P_{total} = 10 \text{ atm} \times 760 \frac{\text{mm Hg}}{\text{atm}} = 7600 \text{ mm Hg}$
Calculate the partial pressure of $\text{N}_2$ ($P_{\text{N}_2}$) above the water using its mole fraction in air and the total pressure:
$P_{\text{N}_2} = x_{\text{N}_2, \text{air}} \times P_{total}$
$P_{\text{N}_2} = 0.80 \times 7600 \text{ mm Hg} = 6080 \text{ mm Hg}$
Step 3: Apply Henry's Law to Find Mole Fraction in Water
Henry's Law is expressed as $P_{gas} = K_H \times x_{gas}$, where $x_{gas}$ is the mole fraction of the gas in the liquid. Rearrange the formula to solve for the mole fraction of $\text{N}_2$ in water ($x_{\text{N}_2, \text{water}}$):
$x_{\text{N}_2, \text{water}} = \frac{P_{\text{N}_2}}{K_H}$
Substitute the calculated partial pressure and the given Henry's Law constant:
$x_{\text{N}_2, \text{water}} = \frac{6080 \text{ mm Hg}}{6.5 \times 10^7 \text{ mm Hg}}$
$x_{\text{N}_2, \text{water}} \approx 0.93538 \times 10^{-4}$
$x_{\text{N}_2, \text{water}} \approx 9.35 \times 10^{-5}$
Final Answer
The mole fraction of $\text{N}_2\text{(g)}$ in water at 298 K is determined to be approximately $9.35 \times 10^{-5}$.
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