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OpenStudy (anonymous):

why would the position of equilibrium in a reaction with equal numbers of gas on either side of equation be unaffected by pressure?

OpenStudy (jfraser):

equal moles of gas create a 1:1 or 2:2 ratio within the Keq expression. if you double or triple the pressures, you're multiplying both by the same factor, so there's no net change in the position of the equilibrium

OpenStudy (anonymous):

Thaaanks:D

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