A nano-tube packed fuel tank is full of Hydrogen, and there is a steady pressure of Hydrogen in the tank. While hydrogen gas is being removed from the fuel tank to power the car, the pressure in the fuel tank drops very little for some time. Explain this observation using Le Chatelier's principle.

My attempt at a solution: We look at the equilibrium $\ce{H2 (\text{absorbed by car}) <=> H2 (\text{in tank})}$. Initially, this equilibrium lies to the right, so it moves to the left. But this just means that the hydrogen in the fuel tank is being used up, and so the pressure should drop.

  • $\begingroup$ I'm not sure but perhaps the equilibrium of liquid hydrogen in the tank and hydrogen vapours in the tank should be considered. $\endgroup$ May 5 '17 at 15:02
  • $\begingroup$ The official answer considers the equilibrium H2 ( absorbed ) <---> H2 ( gaseous ) . One way I can think of to make this work is that H2 (absorbed ) is actually the hydrogen absorbed by the nano-tube. Initially, the two are in equilibrium, but as H2 (gaseous ) is used up, the equilibrium shifts to the right, and thus the pressure changes only a little. Two questions come to mind : What exactly is meant by the 'hydrogen absorbed' by the nano-tube? Why does the pressure change at all? Shouldn't there be absolutely no change? $\endgroup$
    – John
    May 5 '17 at 15:36
  • 2
    $\begingroup$ I think that your explanation just above is essentially correct. The adsorbed H2 can be considered as equivalent to liquid H2 (should such a thing exist at normal temperatures). Then the H2 gas pressure is equivalent to the vapour pressure and is constant as long as some 'liquid' still remains in the tank. The equilibrium is between adsorbed (liquid) H2 and H2 gas. $\endgroup$
    – porphyrin
    May 5 '17 at 17:52
  • $\begingroup$ Why would some hydrogen be adsorbed though? And shouldn't there be no change at all in the pressure? $\endgroup$
    – John
    May 5 '17 at 18:54
  • 1
    $\begingroup$ For practical reasons for hydrogen to be stored in large quantities it is best to do so adsorbed onto a substrate rather than have to cool it and to use enormous high pressures. $\endgroup$
    – porphyrin
    May 12 '17 at 15:19

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