Question

Find the value of KH for N2 gas in water at 303 K in k bar. The value of KH for N2 gas in water at 298 K is 86.76 k bar.​

Answer 1

Answer:

Total pressure is 5 atm. The mole fraction of nitrogen is 0.8.

Hence,the partial pressure of nitrogen=P

T

×X=0.8×5=4 atm

According to Henry's law, P=KX

x is the mole fraction

K is the henry's law constant

P is the pressure in atm.

Substitute values in the above expression.

X=

1×10

5

atm

4 atm

=4×10

−5

=

10

n

n=4.0×10

−4

mol

Answer 2

Given:

Value of $\[{K_H}\]$ for $\[{N_2}\]$ gas in water at 298 K $\[ = 86.76\,k\,bar\]$

To Find:  Value of $\[{K_H}\]$ for $\[{N_2}\]$ gas in water at 303 K in k bar

Solution:

According to Henry's law,

$\[p = {K_H}x\]$

Where, $p=$ partial pressure, $x=$ mole fraction, $\[{{K_H}}\]=$ Henry's constant

Therefore,

$\[x = \frac{p}{{{K_H}}}\]$

As pressure is directly proportional to temperature, replace pressure with temperature.

At temperature 298 K,

$\[{x_} = \frac{{298\,K}}{{86.76\,k\,bar}}\]$               ... (i)

At temperature 303 K,

$\[{x_} = \frac{{303\,K}}{{z\,k\,bar}}\]$                   ... (ii)

Equating equations (i) and (ii), we have

$\[\frac{{303\,K}}{z} = \frac{{298\,K}}{{86.76\,k\,bar}}\]$

$\[z = \frac{{303\,K \times 86.76\,k\,bar}}{{298\,K}}\]$

$\[z = 88.21\,k\,bar\]$

Hence, the value of $\[{K_H}\]$ for $\[{N_2}\]$ gas in water at 303 K in k bar is $\[88.21\,k\,bar\]$.