Henry's law constant for solubility of methane in benzene is $4.2 \times 10^{-5}$ mm Hg at a particular constant temperature. At this temperature, calculate the solubility of methane at (i) 750 mm Hg (ii) 840 mm Hg.

Henry's law constant for solubility of methane in benzene is 4.2 × 10⁻⁵ mm Hg at a particular constant tempera

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4.2 \times 10^{-5}

(i) 750 mm Hg

(ii) 840 mm Hg.

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Concept understanding — Henrys Law

Henry's Law: The Physics of "Fizz"

Imagine you open a cold bottle of soda. You hear that familiar psshhht sound. Bubbles rush out. Now think: why were those bubbles inside the bottle in the first place? The liquid wasn't boiling. The answer is Henry's Law.

The Intuition: Gas Wants to Dissolve

Gases are just molecules flying around. When a gas touches a liquid, some of those molecules get "trapped" inside the liquid — they dissolve. But here's the key: the more you push on the gas, the more of it gets forced into the liquid.

Think of a crowded bus. If you push more people toward the door (higher pressure), more people get squeezed inside. If you let the pressure off (open the bottle), people rush out. That's exactly what happens with gas and liquid.

In the soda bottle, carbon dioxide gas is pumped in at high pressure. That pressure forces a huge amount of CO₂ to dissolve into the liquid. When you open the bottle, the pressure above the liquid drops to normal air pressure. Suddenly, the liquid can't hold all that CO₂ anymore — so it escapes as bubbles. That's the fizz.

The Precise Statement

Henry's Law says:

C = k_H \cdot P

Where:

$C$ = concentration of the dissolved gas in the liquid (usually mol/L or g/L)
$P$ = partial pressure of that gas above the liquid (usually atm or kPa)
$k_H$ = Henry's law constant — a number that depends on the specific gas, the liquid, and the temperature

In words: At a constant temperature, the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of that gas above the liquid.

What the Constant $k_H$ Tells You

$k_H$ is not universal. It's different for every gas-liquid pair. For example:

CO₂ in water has a certain $k_H$
O₂ in water has a different $k_H$ (smaller — oxygen doesn't dissolve as easily)

Temperature matters too. Higher temperature means lower $k_H$ — gases become less soluble in hot liquids. That's why a warm soda goes flat faster than a cold one.

Watch out

Henry's Law works only for dilute solutions and non-reacting gases. If the gas reacts chemically with the liquid (like HCl gas dissolving in water to form hydrochloric acid), Henry's Law does not apply — the concentration will be much higher than predicted.

Real-Life Examples

Situation	What Henry's Law explains
Soda fizz	High pressure forces CO₂ in; releasing pressure lets it out

Why this formula?

Henry's Law: Why the Formula Holds

Henry's Law describes the solubility of a gas in a liquid at a constant temperature. The key formula is:

P = k_H \cdot x

Where:

$P$ = partial pressure of the gas above the liquid
$x$ = mole fraction of the gas dissolved in the liquid
$k_H$ = Henry's constant (depends on gas, liquid, and temperature)

Why This Linear Relationship Exists

1. Dynamic Equilibrium at the Interface

Imagine a gas above a liquid. At the molecular level:

Gas molecules constantly strike the liquid surface and dissolve
Dissolved molecules constantly escape back into the gas phase

At equilibrium, the rate of dissolution equals the rate of escape. This is a dynamic balance, not a static one.

2. The Driving Force for Dissolution

The rate at which gas molecules enter the liquid depends on:

How many gas molecules hit the surface — this is proportional to the partial pressure $P$ of the gas
How easily they dissolve — this is captured by $k_H$

So:

\text{Rate}_{\text{dissolve}} \propto P

3. The Driving Force for Escape

The rate at which dissolved molecules leave the liquid depends on:

How many dissolved molecules are near the surface — this is proportional to the mole fraction $x$ of the gas in the liquid
How easily they escape — also captured by $k_H$

So:

\text{Rate}_{\text{escape}} \propto x

4. Equating the Two Rates

At equilibrium:

\text{Rate}_{\text{dissolve}} = \text{Rate}_{\text{escape}}

Therefore:

P \propto x

Introducing the proportionality constant $k_H$ :

P = k_H \cdot x

Why It's Linear (Not Exponential or Logarithmic)

The linearity arises because:

No saturation effects at low concentrations — the molecules don't "crowd" each other
Ideal behavior is assumed — gas molecules don't interact strongly with each other or with the solvent
Temperature is constant — $k_H$ doesn't change …

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Step 1. Henry's law states $x_{solute}=\dfrac{p_{solute}}{K_H}$ , so for the same gas-solvent pair at a fixed temperature, solubility (mole fraction dissolved) is directly proportional to the partial pressure above the solution.

Step 2. Using the constant exactly as printed, $K_H=4.2\times10^{-5}$ mm Hg: at 750 mm Hg, $x_1=\dfrac{750}{4.2\times10^{-5}}$ ; at 840 mm Hg, $x_2=\dfrac{840}{4.2\times10^{-5}}$ .

Step 3. Both of these come out far greater than 1, which is not physically possible for a mole fraction (mole fraction must lie between 0 and 1) -- this signals that the printed constant's exponent is very likely a misprint in this edition (a Henry's law constant for a sparingly soluble gas like methane in a solvent is normally of order $10^{4}$ - $10^{6}$ mm Hg, i.e. a POSITIVE exponent, not $10^{-5}$ ).

Step 4. Whatever the true magnitude of $K_H$ , the RATIO between the two solubilities is unaffected by this, since it cancels out: $\dfrac{x_2}{x_1}=\dfrac{p_2/K_H}{p_1/K_H}=\dfrac{p_2}{p_1}=\dfrac{840}{750}=1.12$ . …

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Q.Henry's law constant for solubility of methane in benzene is 4.2×10−54.2 \times 10^{-5}4.2×10−5 mm Hg at a particular constant temperature. At this temperature, calculate the solubility of methane at

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Q.Henry's law constant for solubility of methane in benzene is $4.2 \times 10^{-5}$ mm Hg at a particular constant temperature. At this temperature, calculate the solubility of methane at

Unlock everything free for 14 days