Henry's law constant for CO2 in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
It is given that,
KH = 1.67 × 108 Pa
PCO2 = 2.5 atm = 2.5 × 1.01325 × 105 Pa
= 2.533125 × 105 Pa
According to Henry's law
PCO2 = KHx
⇒ x = \(\frac{p_{CO_2}}{K_H}\)
= \(\frac{2.533125\times 10^5}{1.67 \times 10^8}\)
= 0.00152
We can write,
x = \(\frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}}\) ≈ \(\frac{n_{CO_2}}{n_{H_2O}}\)
[Since, nCO2 is negligible as compared to nHO2]
In 500 mL of soda water, the volume of water = 500 mL
[Neglecting the amount of soda present]
We can write,
500 mL of water = 500 g of water
= \(\frac{500}{18}\) mol of water
= 27.78 mol of water
Now, \(\frac{n_{CO_2}}{n_{H_2O}}\) = x
\(\frac{n_{CO_2}}{27.78}\) = 0.00152
nCO2 = 0.042 mol
Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g
= 1.848 g