Question

What partial pressure of C2H4 gas (in mm Hg) is required to maintain a solubility of 4.92×10-2 g/L in water at 25 °C? kH for C2H4 at 25 °C is 4.78×10-3 mol/L·atm.

Answer 1

Answer: The partial pressure of $C_2H_4$ is 281 mmHg

Explanation:

We are given:

Solubility of ethene gas = $4.92\times 10^{-2}g/L$

To convert this solubility into mol/L, we divide the given solubility by the molar mass of the gas, which is 28 g/mol

$\text{Solubility (in mol/L)}=\frac{\text{Solubility (in g/L)}}{\text{Molar mass}}$

$\text{Solubility (in mol/L)}=\frac{4.92\times 10^{-2}g/L}{28g/mol}=0.176\times 10^{-2}mol/L$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{C_2H_4}=K_H\times p_{C_2H_4}$

where,

$K_H$ = Henry's constant = $4.78\times 10^{-3}mol/L.atm$

$C_{C_2H_4}$ = molar solubility of ethene gas = $0.176\times 10^{-2}mol/L$

Putting values in above equation, we get:

$0.176\times 10^{-2}mol/L=4.78\times 10^{-3}mol/L.atm\times p_{C_2H_4}\\\\p_{C_2H_4}=\frac{0.176\times 10^{-2}mol/L}{4.78\times 10^{-3}mol/L.atm}=0.370atm$

Converting this into mmHg, we use the conversion factor:

1 atm = 760 mmHg

So, $0.370atm\times \frac{760mmHg}{1atm}=281mmHg$

Hence, the partial pressure of $C_2H_4$ is 281 mmHg