# How does steam distillation work to extract a substance from a mixture, below its boiling point?

This is a small confusion I have regarding steam distillation. Let us consider a sample of water mixed with an organic compound that is insoluble in water. Let us also consider that the boiling point of this organic compound is at 70 °C at atmospheric pressure and the boiling point of water is 100 °C at atmospheric pressure.

According to what I have learned and what I have researched online, it says the two compounds in the mixture evaporate separately, independent from each other. So, the total pressure of the system at a given time would be the vapor pressure of water and the vapor pressure of organic compound combined.

My confusion is that the articles online says that this combined vapor pressure is higher than the vapor pressure when only one compound is available, so, the mixture reach atmospheric pressure faster so, it boils at lower temperature than 70 °C.

How can this be possible if these two compounds (water and the organic compound) are insoluble in one another, and if these two compounds evaporate independently from each other? I cannot understand it because I have also learned that when the environmental pressure increases it also increases the boiling point of that substance. Since these two compounds are not soluble in each other and evaporate independently, should not the vapor pressure of one compound increase pressure that affects the other compounds boiling point and evaporation rather than the vapor pressures of both these compounds affecting the boiling point and evaporation of whole system of these compounds combined (Because these compounds are separate and does not have intermolecular interactions like a binary mixture). What is the mistake I made in my thought process?

Edit: For example, take benzene and water mixture and their boiling points. Pure water is at 100 °C, benzene at 80 °C, benzene + water at 65 °C. The reason suggested is that, at 65 °C, the vapor pressures are 0.23 atm for water, and 0.77 atm for benzene. The total pressure over this mixture at 65 °C is 0.23 atm + 0.77 atm = 1.00 atm. So any mixture water + benzene boils at 65 °C.

Why does this combined vapor pressure reaching the atmospheric pressure lower the boiling point of both liquids, while these two liquids are in equilibrium with their vapor independent from each other? What kind of interaction does these two liquids has that causes this?

• It is rather water distillation with organic vapor than steam distillation of the organic. As it is supposed the organic substance has (much) higher boiling point than water, typically essential oils from plants. Commented Dec 4, 2023 at 18:57
• For 2 separate liquids, the partial pressure of each is the same regardless of presence of the other. And the total pressure is sum of partial pressures. In reality, there is deviation due limited mutual solubility. Commented Dec 4, 2023 at 19:01
• Your problem looks like the mixture water + benzene, which are immiscible. At atmospheric pressure, water boils at $100$°C, and benzene boils at $80$°C. Whatever its proportions, a mixture water + benzene boils at temperature lower than $80$°C, probably at $65$°C. Why ? The published tables mention that, at $65$°C, the vapor pressures are $0.23$ atm for water, and $0.77$ atm for benzene. The total pressure over this mixture at $65$°C is $0.23$ + $0.77$ = $1.00$ atm. So any mixture water + benzene boils at $65$°C. Commented Dec 4, 2023 at 20:15
• @Maurice yes, What my confusion is why both of these liquids boil at a lower temperature? What interaction does these two liquids have that causes this. Commented Dec 5, 2023 at 5:15
• @Mithoron No, it does not explain why the mixture boils at a lower boiling point. Commented Dec 5, 2023 at 5:18

With comparable boiling points like for water and benzene mixture at constant pressure, neither of liquids boils when sum of partial pressures of their vapors overcomes air pressure. They are just quickly evaporating, as this evaporation is pushing air and own vapors away, not being controlled by diffusion nor convection like ordinary evaporation.

In a way, it is "boiling without boiling", or "boiling from liquid surfaces" what is not real boiling as forming bubbles inside the liquid.

If the mixture is overheated above its pseudo-boiling point, the summary vapor pressure gets higher than environmental pressure and evaporation significantly speeds up, cooling the mixture back to the pseudo-boiling temperature. similarly as if a single phase liquid is overheated and starts sudden violent boiling, cooling itself down.

Such a distillation is usually arranged in the way the vapor of more volatile liquid (usually water, but not in this case) is brought to the heated mixture of liquids, stripping vapor of the other liquid and taking it away to the condenser. The vapor input and mixture heating is being balanced so the net volume change rate of the mixture is about zero. This arrangement covers evaporation loses of water (or more volatile liquid) during distillation, kinetically helps releasing vapors to gaseous phase and prevents mixture overheating, leading potentially to sudden true violent boiling.

One could have two separate liquid reservoirs in the common, heated open system with restricted outlet. Each of liquids would have vapour tension lower than sum of their vapor tensions, so neither of them would boil, but just evaporate. But there would be steady flow of vapors from the outlet, like if they were boiling.

With raising difference in boiling points, the reality converges to boiling of more volatile liquid. As the less volatile liquid contribution to the summary vapor pressure is minimal. The system temperature then frequently randomly change between the temperature for summary vapor pressure being equal to air pressure and the more volatile liquid boiling point. That depends of fine balancing of the evaporation rate and heat transfer rate.

Consider the analogy: Imagine an open system with sensors for partial pressure of nitrogen and oxygen. Sources of these compressed gases would provide respective gas until partial pressures reach 0.21 atm for oxygen and 0.8 atm for nitrogen. If only one of gas sources were active nothing much would happen. But with both sources active, there would be steady flow of gas from the system.

The 2 non miscible liquid system effectively works like if there was a pump, pumping away vapor if total vapor pressure is higher than 1 atm, so the space above liquid is never fully saturated by vapors.

Imagine that as some temperature, the saturated vapor pressures of benzene and water are $$\pu{0.8 atm}$$, resp. $$\pu{0.45 atm}$$, in total $$\pu{1.25 atm}$$. But the pressure balancing with air keeps it at $$\pu{1 atm}$$, so the relative vapor saturation is just 80%. With quite high absolute vapor tensions, evaporation is fast.

• Does this mean the vapor pressure of one component in the mixture is not trying to add extra pressure to other component, but rather reduces the net applied atmospheric pressure to the other component, so both of these liquids can boil (or fast evaporate) at a lower temperature? Am I correct? Commented Dec 6, 2023 at 6:47
• No, they rather cooperate by summing their lower than ambient partial pressures, like any ordinary gases, so the summary vapor pressure equals or get higher than external pressure. Commented Dec 6, 2023 at 6:51