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Why does changing the volume of the reaction vessel have no effect on this equilibrium?
CO(g) + Fe3O4(s) <===> CO2(g) + 3FeO(s)

  • Chemistry -

    The only factors that effect the rate of reaction are the temperature used for heating, concentration of your reactants , physical state of your reactants, pressure and the catalyst used.
    However, reducing the volume of the reaction vessel increases the number of collisions so the reaction rate increases as pressure increases.

  • Chemistry -

    Although there is nothing wrong with what Veronica wrote, the response doesn't answer the question. The response talks in terms of reaction rates while the question asks for shift in equilibrium; i.e., Le Chatelier's Principle. The reason there is no shift in the equilibrium is because there is no difference in number of moles on each side. Note there is one mole gas on the right and one mole gas on the left; changing the volume changes the concentration and it changes the pressure BUT since the number of moles are the same there is no shift. Under an increase in pressure, the reaction would shift to the side with the smaller number of moles. Same number of moles, no shift.

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