For example, carbon monoxide reacts with hydrogen to synthesize methanol in the presence of some catalysts, but the pressure needs to be $\pu{50 atm}$ and the temperature needs to be $\pu{523 K}$.

Why the pressure? What does the pressure adds to the reaction?


3 Answers 3


ChemGuide has a good introductory article here.

The effects of increasing pressure and temperature are, to an extent, equivalent. Increased pressure leads to increased collisions and increased collision strength between molecules, allowing the (usually high) activation energy barrier to be overcome at a noticeable rate; at standard temperatures and pressures, for example, collisions between $\ce{CO}$ and $\ce{H2}$ are far too infrequent and involve far too little energy for the formation of $\ce{CH3OH}$.

Remark. How do we achieve high pressures and temperatures? We could pressurize a steel cylinder and use it as a reaction vessel, for one, but a novel approach involves sonochemistry. Ultrasound waves in solution lead to cavitation, the formation and rapid collapse of bubbles, which are highly pressurized and at high temperatures. The use of ultrasound can even open up new reaction pathways not usually available; for example, radical species are often transient, but the forcing conditions in these bubbles allow them to survive long enough to react.

  • $\begingroup$ I think I got it. so, it's just an optimized way of achieving the reaction. We could achieve the same reaction at 283K and a much higher pressure or at 1 atm and a much higher temperature, is this it? temperature and pressure are somewhat equivalent in a chemical system? what is the ratio T/P that accounts for pressure change equivalence to temperature change $\endgroup$ Commented Jul 21, 2016 at 1:22
  • $\begingroup$ High temperature and high pressure can both serve to accelerate a reaction based on kinetic considerations, yes; in other senses, they may not be similar. I only speak of qualitative considerations of high temperature and high pressure; I'm not sure you can quantitatively describe their relationship. Note also that thermodynamic considerations, as highlighted by sixtytrees, make this more complex. By Le Chatelier's principle, high pressures favor reactions where there are less moles of gas in the product, whereas high temperatures favor endothermic reactions. $\endgroup$ Commented Jul 21, 2016 at 1:34
  • $\begingroup$ Almost, but not quite. Can you achieve the same reaction at room temperature, but higher pressure? Yes, you can. In diamond anvil (used for high pressure experiments) you can achieve the same result. It chemically doable, but expensive. However, just increasing temperature and decreasing pressure will not help. $\endgroup$
    – sixtytrees
    Commented Jul 21, 2016 at 1:35
  • $\begingroup$ Here is why. High temperature will accelerate both direct and back reaction. But low pressure would shift equilibrium to dissociation. So at high T and low P you will very fast arrive at 1% MeOH, 99% of 2H2+CO. $\endgroup$
    – sixtytrees
    Commented Jul 21, 2016 at 1:36
  • $\begingroup$ Engineers keep 1500 C temperature when making steel. They would have increased temperature of MeOH synthesis if it was helpful. Increasing pressure in large volumes is hard. That is why they don't go above 50 atm. $\endgroup$
    – sixtytrees
    Commented Jul 21, 2016 at 1:39

The reaction you are talking about is an equilibrium. Pressure is added to shift equilibrium towards methanol.

CO+2H2 <-> MeOH

Consider an equilibrium pressure. In one second number of MeOH molecules dissociate = number of molecules formed. MeOH doesn't need a partner to dissociate. But CO heeds two partners (2 H2 molecules) to turn into MeOH. If you decrease the pressure MeOH will dissociate at the same rate. But collisions of CO with H2 will be less frequent. So less MeOH will form.

Here [Y] is the concentration of Y (in moles/L)

Let system be at equilibrium at given pressure:

K$_{eq}$ = [CO]*[H$_2$]$^2$/[MeOH]

Increase pressure by a factor of two. What happens to concentration of MeOH?

K$_{eq}$ = (2*[CO])(2[H$_2$])$^2$/(x*[MeOH])

Left parts are equal, so right parts are equal too.

[CO][H$_2$]$^2$/[MeOH] = (2[CO])(2[H$_2$])$^2$/(x*[MeOH])

After dividing by [CO], [H2], [MeOH] we have 1=2*2$^2$/x

X=8. We have 8 times more MeOH after increasing the pressure by a factor of 2. If there was no reaction the amount of MeOH would just increase by 2. So we have 4 times more MeOH produced by just doubling the pressure.

This was mostly the thermodynamics aspect (how much MeOH we have when the reaction finishes). There is also kinetics (how fast reaction goes initially). Kinetics increases as pressure$^3$. Temperature also increases the speed of the reaction.

Pressure is needed to use the Le Chatelier's principle. This is one of a dozen fundamental principles of chemistry that you should look into. The higher the concentration of the reagent the more product will be produced. If the pressure is low formation of gasses that will spread around is preferred. At higher pressures formation of gas doesn't increase entropy (measure of disorder) by much and enthalphy (measure of heat) starts playing a role.

Note that a lot of processes depend on pressure. Water can vaporize by decreasing pressure or condensed by increasing it. Human's breathing requires high (1 atmosphere or so) pressure.

For MeOH synthesis you have 4 limiting cases:

Low P, low T: no reaction and if you wait forever you have little product at equilibrium time

Low P, high T: reaction goes fast, but against you MeOH falls apart rapidly

High P, low T: very high yield of MeOH if you wait forever, but reaction is somewhat slow

High P, high T: reaction is fast, but how much product you get depends on ratio of P and T. P should grow faster than T to keep equilibrium favorable.

Equilibrium is dictated by Gibbs energy G. Gibbs energy is just G = H - TS here H is measure of heat produced/consumed (enthalpy) T is temperature and S is measure of chaos/disorder produced.

Crystal has high order, so it has low S (enthropy), but formation of crystals is exothermic (favorable enthalpy). Gas is different. It has a lot of entropy (S is big), but little enthalpy of formation (it takes energy to turn crystal into gas). If you have high T (lots of energy around) then S becomes important. If you have low T, then H becomes important. This is why at high temperature you have gas and at low temperature you have crystals.

Now, enthropy of gas increases as you increase the volume it can occupy (gas feels it has less freedom when it is locked in a tight cell, but more freedom in bigger volume. So, if pressure is low 1 molecule MeOH dissociates three molecules form, they fly away from each other oad hardly ever meet. S increased a lot.

At high pressure they have little space and don't gain much enthropy. At the same time coming together is allows then to release some heat H (enthalphy).

Pressure doesn't add enthalpy H. It takes away entropic benefit S for MeOH to fall apart. Pressure says "dear gases, I know you love freedom and don't want to marry each other. But I will push you so close to each other that you will hit each other with elbows all the time. So yo uhave no freedom to lose when you form MeOH".

  • $\begingroup$ I did not fully understand your reply, specially this statement: At higher pressures formation of gas doesn't increase entropy (measure of disorder) by much and enthalphy (measure of heat) starts playing a role. $\endgroup$ Commented Jul 21, 2016 at 1:19
  • $\begingroup$ so, basically pressure is needed to increase the yield of the reaction without the need to increase the temperature, is that it? So if pressure is so much more efficient than temperature at adding the needed enthalpy of formation, what dictates the point where pressure is preferable over temperature for a chemical compound systesis in lab? $\endgroup$ Commented Jul 21, 2016 at 1:29
  • $\begingroup$ Since when do you measure pressure in mol/L? $\endgroup$ Commented Jul 21, 2016 at 3:48
  • $\begingroup$ @Martin I measure concentration in mol/L. At a given temperature pressure is defined by concentration of ideal gas. Don't put your words in my mouth. $\endgroup$
    – sixtytrees
    Commented Jul 21, 2016 at 12:12
  • 1
    $\begingroup$ You should really read up on the be nice policy. Calling me a demagogue is not nice. Apart from that I am not trying to appeal to anyone. I think your answer is bad as it contains too many words, and not enough explanation. This is my personal opinion and I have downvoted such answers before this one and I will continue to do so. I am sorry if you take that personal, it is not. The content you are writing is ambiguous and hence I downvoted it. On another note, you might want to have a look at how to use MathJax , so that you posts don't fall apart on different displays. $\endgroup$ Commented Jul 25, 2016 at 8:07

This is old but I wanted to add a quantitative / physical perspective here.

As you know, increasing pressure has a similar effect on reactions as increasing pressure. Recalling the ideal gas law, $$pV = nRT$$

We have that $p$ is proportional to $T$. In other words, if you hold all other variables constant, increase in $p$ leads to increase in $T$, and vice-versa.

Pressure can be expressed as force per unit area, or energy per unit volume:

$$ p = \frac{\vec{F}}{A} = \frac{E}{V}$$

So it follows that increasing pressure increases force on molecules. Force effects the acceleration of particles by $\vec{a} = \frac{\vec{F}}{m}$, and velocity follows: $\vec{v} = \int \vec{a} dt$.

Temperature can be expressed using the velocities of particles (by a classical Boltzmann distribution), $$ T = \frac{\sum_i^N m_i\vec{v_{i \alpha}}^2}{k_b N}$$

(Note the numerator is kinetic energy). So you can see that increasing $p$ and $T$ have the same physical result: faster motion of particles. Faster particles means more collisions, and more collisions means faster reactions, or occurrence of a reaction where it could not happen otherwise, because an energy-barrier is overcome ($p \propto E$ and $T \propto E_k$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.