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Why, when chemical equilibrium is defined as when two opposing reactions are equal in rate, can the products or reactants be favoured?

I understand that temperature can cause either the forward or reverse reaction to be favoured. But then doesn't that mean that the rate of which the reactants or products are being formed differ from each other? So, what does it mean when they define equilibrium as I have said above.

Please keep answers simple and clear.

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    $\begingroup$ The rate must be equal at equilibrium, the products are favoured means that the concentration of products is much greater than the concentrations of the reactants at equilibrium $\endgroup$ – Aditya Anand Feb 16 '16 at 12:49
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A chemical reaction can be reversible, and a reversible reaction can be at equilibrium, but they are not the same thing.

Equilibrium is the state where the forward and reverse reactions are the same rate. This is clearly not true at the start of a reaction because there are no products yet. As the reaction progresses the forward reaction rate decreases and the reverse reaction rate increases until they are equal. At that point the concentrations of reactants and products no longer change.

The position of the equilibrium can be far to the left, so that hardly any product is formed before it reaches equilibrium. This would be because the forward reaction is much slower than the reverse reaction. If the position of the equilibrium is far to the right, almost all of the reactants have been used up because the forward reaction is much faster than the reverse reaction.

If you do something to change the conditions, like increase the temperature or add more reactants, you break the equilibrium and the reactions are no longer at the same rate. As above the reaction will progress until the rates are the same again. Depending on the change, this new equilibrium position may be more to the left or let or more to the right.

It is tempting to think that if you increase temperature, the rate of both reactions will increase the same. This is not the case. Although both reactions will increase in rate, because of the usual collision theory, but the same temperature change can have a much bigger effect on the rate of one reaction than the other.

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