Henry's Law — Gas solubility
Solutions
10
JEE Qs
8%
Hard
60
min
Master the formula p_gas = K_H * x_gas, pay close attention to units, and understand the impact of temperature and the significance of K_H value on gas solubility.
🧮 Key Formulas
✅ Key Points for JEE
- 1Henry's Law states that the partial pressure of a gas in vapor phase (p_gas) is proportional to the mole fraction of the gas (x_gas) in the solution.
- 2The Henry's Law constant (K_H) is specific for each gas-solvent pair and its value changes with temperature. Higher K_H indicates lower solubility of the gas for a given partial pressure.
- 3Solubility of a gas in a liquid decreases with increasing temperature, as dissolution of gases is typically an exothermic process (Le Chatelier's principle).
- 4Henry's Law is applicable for gases at low pressures and moderate temperatures, and when the gas does not undergo chemical reaction with the solvent (e.g., HCl in water does not follow Henry's law due to ionization).
⚠️ Common Mistakes
- ✕Using total pressure instead of the partial pressure of the specific gas for p_gas in the formula.
- ✕Confusing the units of K_H and partial pressure; they must be consistent (e.g., both in bar or atm).
- ✕Not understanding the inverse relationship between K_H and gas solubility (higher K_H means lower solubility).
- ✕Applying Henry's Law to gases that react chemically with the solvent or at very high pressures/low temperatures where ideal behavior deviates significantly.
📝 Practice Questions
See allQ73.Consider the following cases of standard enthalpy of reaction (ΔH∘r in kJmol−1) 7 C2H6( g) + O2( g) →2CO2( g) + 3H2O(l)ΔH∘1 = −1550 2 C (graphite) + O2( g) →CO2( g) ΔH∘2 = −393.5 The magnitude of ΔH∘fC2H6( g) is _______ 1 H2( g) + O2( g) →H2O(l) ΔH∘3 = −286 2 kJmol−1 (Nearest integer).
Q70.When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ? (1) 0.8 (2) 0.4 (3) 0.2 (4) 0.6 Q71.0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9 gH2O. Molar mass of (X) is _____ gmol−1 .
Q69.Arrange the following solutions in order of their increasing boiling points. (i) 10−4M NaCl (ii) 10−4M Urea (iii) 10−3M NaCl (iv) 10−2M NaCl (1) (i) < (ii) < (iii) < (iv) (2) (iv) < (iii) < (i) < (ii) (3) (ii) < (i ) ≡(iii) < (iv) (4) (ii) < (i) < (iii) < (iv)
Q52.Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is : (1) 1.79 m (2) 2.79 m (3) 2 m (4) 3 m
Q64.Consider a binary solution of two volatile liquid components 1 and 2. x1 and y1 are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of 1 vs 1 x1 y1 are given respectively as: P02−P01 (1) P0 P02−P01 (2) P0 2 , 1 , P01 P02 P02 P02 P01−P02 (3) P0 P01−P02 (4) P0 1 , 2 , P02 P02 P01 P02
Q69.Consider the given plots of vapour pressure (VP) vs temperature(T/K). Which amongst the following options is correct graphical representation showing ΔTf , depression in the freezing point of a solvent in a solution? (1) (2) (3) (4)
NCERT Chapters
- Class 12 Chemistry Ch 2: Solutions