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Correct me If I am wrong, given a reaction at equilibrium, if we say add more concentration of products to the reaction, the rate going backwards would increase to try to get back to equilibrium.

I was wondering if the rate forward stays the same or slows down?

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In simple cases, the rate is equal to the rate constant "k" times the concentration. The forward rate constant and the backward rate constant don't change, but the rate of change (rate constant times concentration) will change as the concentration of a reactant or product changes.

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  • $\begingroup$ I see, so there is a distinction between rate constant denoted by a K and rate. So in my above example the Rate backward is slowing down after being initially sped up, and the rate forward is being increased because reactants are being formed from products? $\endgroup$ – Jeff Jan 29 '14 at 2:19
  • $\begingroup$ yes, also a capital "K" typically refers to an equilibrium constant and lower case "k" refers to a rate constant. Keq = kforward/kbackward $\endgroup$ – ron Jan 29 '14 at 2:35
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Yes, if we add products, then the reverse reaction will try to make more reactants. If we take away products, then the forward reaction will try to make more products. Not on topic: If you increase pressure, the side with less moles will be favoured. And more things. This is all Le Chatelier's Principle.

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