Equilibrium law
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If the reversible reaction:
2H2O(g) --> 2H2(g) + O2(g) occurs at 1 atm pressure and 20% of the steam is converted to hydrogen and oxygen, how do we find the equilibrium partial pressures (in atm) of the components of the equilibrium?
140408
Corrie writes ...
Let's assume you start with 2 mole of water vapour at 1 atm pressure (and no hydrogen or oxygen).
Then at equilibrium at 1 atm, there will 2 - 0.4 = 1.6 moles of H2O, 0.4 moles of H2 and 0.2 moles of O2: a total of 2.2 moles of gas.
The partial pressure of water vapour at equilbrium is thus (1.6/2.2) x 1 atm.
And so on, for the other two gases.
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updated: 14 April 2008
