True or False: at equilibrium, the concentration of reactants and products does not change with time

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True or False: at equilibrium, the concentration of reactants and products does not change with time

Chemistry
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true
what about "when the reactants and products in a chemical equation are tripled, the value of the new equilibrium constant is the initial equilibrium constant raised to the third power
false. the concentrations of the reactants and products may change, but the ratio that defines K must remain constant.

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okay and then "a large equilibrium constant indicates that the reaction takes place at a very fast rate" that's true isn't it
also false. the size of K says nothing about how quickly equilibrium is achieved, just that the ratio favors the concentration of products.
yeah but the rate law says that rate =k(concentration)
so higher k makes the rate law quicker
different K's
whats the rate law constant?
the rate law constant, k, describes the speed with which reactants are used up, the equilibrium constant, KEQ, defines the ratio of concentrations present in a system at equilibrium. Rate law problems are rarely equilibrium problems
okay so why doesn't a large EQ constant indicate that the reaction takes place very fast
what does a large EQ constant mean?
Here's the equation for the equilibrium constant:\[K{_E}{_Q} = \frac{[products]}{[reactants]}\] A large value of K means that the concentrations of the products are much much larger than the concentrations of the reactants. It may take a long time to get there, but the only thing the value of KEQ does is give us what the ratio is, nothing about how quickly the reaction gets there
Be careful. You have to keep in mind every chemical system capable of equilibrium supports at least TWO reactions: the second being the reverse of the first. So, yes, if you triple the concentrations of the reactants and products, the rate of the individual reactions (forward and revesre) will significantly increase, in agreement with the general principles of chemical kinetics. However, since the rate of BOTH reactions (forward and reverse) will increase, what happens to the equilibrium concentrations is less clear than you may at first think, and, in fact, the new concentrations at equilibrium will end up being those that keep the value of K constant. Think of K as being related to the required ratio of the rate of forward and reverse reaction at equilibrium. Even if the rates increase or decrease, because you have added more or taken away concentration, the ratio of the rates (which determines the concentrations) will not change. It's important in this context to keep in mind that chemical equilibrium is dynamic equilibrium: at equilibrium, reactants are still being rapidly converted to products -- but products are just as rapidly being converted back to reactants, so while there is much furious reacting still going on, the net concentrations don't change.

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