Henry's Law (original) (raw)

Last Updated : 20 Dec, 2023

The behavior of a gas is described by Henry's law. The amount of dissolved gas in a liquid is proportional to the partial pressure of the gas above the liquid, according to this law. In addition, Henry's law constant refers to this proportionality factor. The law was created by English scientist William Henry. In the year 1803, he proposed this law. This law is used to calculate the number of gases dissolved in water by experts. Let's take a closer look at the formula for Henry's law.

Henry's Law

When the temperature is held constant, Henry's law states that the amount of gas dissolved in a liquid is precisely proportional to its partial pressure above the liquid. Henry's law constant (usually represented by 'kH') is the constant of proportionality for this connection. The dissolution of oxygen and nitrogen in the blood of divers who are diving underwater is a nice example of Henry's law in operation. As a result, when decompressing underwater, this disintegration changes, causing decompression sickness. Another notable example is the taste of carbonated soft drinks.

The gas above the drink is practically pure carbon dioxide before the bottle is opened, and the pressure is higher than atmospheric pressure. This gas is released from the bottle after it is opened. As a result, the partial pressure of CO2 above the liquid falls dramatically. As dissolved carbon dioxide gas escapes, the gases are degassed.

Henry's Law Formula

Henry's law is expressed as a mathematical formula:

**P α C

**P = K H × C

Where,

Sample Questions

**Question 1: What are the prerequisites for employing Henry's Law?

**Answer:

When the molecules are in equilibrium, Henry's law holds true. The rule of Henry does not apply to gases at high pressures (for example, when N2 (g) is put into the bloodstream at high pressure, it becomes exceedingly soluble and hazardous).

**Question 2: What are Henry's Law's Limitations?

**Answer:

**Limitations of Henry's Law,

**Question 3: Calculate the solubility of gaseous oxygen in water at a temperature of 293 K and a partial pressure of 4 bar for gaseous oxygen. (K H for O 2 **is 34840 bar.L.mol -1 )

**Solution:

Given: P = 4 bar, KH = 34840 bar.L.mol-1

Since,

P = KH × C

∴ C = P / KH

∴ C = 4/34840

**∴ C = 0.00011 mol/L

**Question 4: At a temperature of 293 K, the value of K H for carbon dioxide is 2.1 × 10 2 atm.L.mol -1 . The gas's solubility (in water) would be 3 × 10 -3 M at what partial pressure?

**Solution:

Given: KH = 2.1 × 102 atm.L.mol-1, C = 3 × 10-3 M

Since,

P = KH × C

∴ P = 2.1 × 102 × 3 × 10-3

**∴ P = 6.3 × 10 -1 atm.

**Question 5: Find Henry's constant for helium dissolved in water at 0.7 atm pressure and a neon concentration of 2.4 M.

**Solution:

Given: P = 0.7 atm, C = 2.4 M

Since,

P = KH × C

∴ KH = P/C

∴ KH = 0.7 / 2.4

**∴ K H = 0.2916 atm.L.mol -1

**Question 6: If Henry’s law constant of the gas is 1.238 atm.L.mol -1 and the dissolved gas's concentration is 1.2 M, then calculate the partial pressure of the gas in the atmosphere.

**Solution:

Given: KH = 1.238 atm.L.mol-1, C = 1.2 M

Since,

P = KH × C

∴ P = 1.238 × 1.2

**∴ P = 1.48 atm.